W9-W10. Geometry in Plane and Space, Double Integrals, Triple Integrals, Lagrange Multipliers, Taylor’s Formula for Multivariable Functions

1. Summary

1.1 Geometry in the Plane and Space

1.1.1 Polar Coordinates

You already know Cartesian coordinates \((x, y)\), which locate a point by its horizontal and vertical distances from the origin. Polar coordinates offer a completely different perspective: instead of asking “how far right and how far up?”, they ask “how far away, and in which direction?”

To define polar coordinates, we fix an origin \(O\) (called the pole) and a ray from \(O\) called the polar axis (usually the positive \(x\)-axis). Every point \(P\) in the plane is then described by a pair \((r, \theta)\):

• \(r\) is the directed distance from \(O\) to \(P\) (we usually take \(r \geq 0\)),
• \(\theta\) is the directed angle from the polar axis to the ray \(OP\).

The conversion between polar and Cartesian coordinates is:

\[x = r\cos\theta, \quad y = r\sin\theta \implies r = \sqrt{x^2 + y^2}, \quad \tan\theta = \frac{y}{x}, \quad \theta \in [0, 2\pi).\]

Why use polar coordinates? Many curves that look complicated in Cartesian form become beautifully simple in polar form. For instance, the circle \(x^2 + y^2 = a^2\) is simply \(r = a\) in polar coordinates.

Converting equations: The standard technique is to use the substitutions \(x = r\cos\theta\), \(y = r\sin\theta\), and \(x^2 + y^2 = r^2\). For example, the polar equation \(r = 2\cos\theta\) converts to \((x-1)^2 + y^2 = 1\), a circle of radius 1 centered at \((1,0)\).

1.1.2 Cartesian Coordinates in Space

To locate points in three-dimensional space, we use three mutually perpendicular axes: the \(x\)-, \(y\)-, and \(z\)-axes, meeting at the origin \((0, 0, 0)\). Every point \(P\) in space has a unique triple \((x, y, z)\) called its Cartesian (rectangular) coordinates.

• The \(xy\)-plane has equation \(z = 0\); the \(xz\)-plane has \(y = 0\); the \(yz\)-plane has \(x = 0\).
• The three coordinate planes divide space into eight regions called octants.

The distance between points \(P_1(x_1, y_1, z_1)\) and \(P_2(x_2, y_2, z_2)\) is:

\[|P_1P_2| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}.\]

A sphere with center \(P_0(x_0, y_0, z_0)\) and radius \(R\) has equation:

\[(x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2 = R^2.\]

The interior of the sphere satisfies \(<\) instead of \(=\); the solid ball (sphere plus interior) satisfies \(\leq\).

1.1.3 Cylindrical Coordinates

Cylindrical coordinates \((r, \theta, z)\) are a natural extension of polar coordinates to 3D space:

1. \(r\) and \(\theta\) are polar coordinates of the projection of \(P\) onto the \(xy\)-plane (so \(r \geq 0\)),
2. \(z\) is the usual rectangular \(z\)-coordinate (height above the \(xy\)-plane).

The conversion equations are:

\[x = r\cos\theta, \quad y = r\sin\theta, \quad z = z; \qquad r = \sqrt{x^2 + y^2}, \quad \theta = \arctan\!\left(\frac{y}{x}\right).\]

Geometric meaning of the coordinate surfaces:

• \(r = a\) (constant): a cylinder of radius \(a\) centered on the \(z\)-axis.
• \(\theta = \theta_0\) (constant): a half-plane through the \(z\)-axis at angle \(\theta_0\).
• \(z = z_0\) (constant): a horizontal plane at height \(z_0\).

Cylindrical coordinates are ideal when a problem has an axis of symmetry (like cylinders, cones, or anything rotationally symmetric around the \(z\)-axis).

1.1.4 Spherical Coordinates

Spherical coordinates \((\rho, \phi, \theta)\) describe a point by its distance from the origin and two angles:

1. \(\rho \geq 0\): the distance from \(P\) to the origin,
2. \(\phi \in [0, \pi]\): the polar angle (angle from the positive \(z\)-axis to \(\overrightarrow{OP}\)),
3. \(\theta\): the same azimuthal angle as in cylindrical coordinates.

The conversion formulas are:

\[x = \rho\sin\phi\cos\theta, \quad y = \rho\sin\phi\sin\theta, \quad z = \rho\cos\phi,\] \[r = \rho\sin\phi, \quad \rho = \sqrt{x^2 + y^2 + z^2} = \sqrt{r^2 + z^2}.\]

Geometric meaning of the coordinate surfaces:

• \(\rho = a\): a sphere of radius \(a\) centered at the origin.
• \(\phi = \phi_0\): a cone with vertex at the origin and axis along the \(z\)-axis. If \(\phi_0 < \pi/2\), the cone opens upward; if \(\phi_0 > \pi/2\), it opens downward. Special case: \(\phi = \pi/2\) gives the \(xy\)-plane itself.
• \(\theta = \theta_0\): a vertical half-plane through the \(z\)-axis.

Converting surfaces to spherical coordinates:

• The sphere \(x^2 + y^2 + (z-1)^2 = 1\) expands to \(x^2+y^2+z^2 = 2z\), i.e., \(\rho^2 = 2\rho\cos\phi\), giving \(\rho = 2\cos\phi\) (valid for \(\rho > 0\)).
• The cone \(z = \sqrt{x^2+y^2}\) means \(\rho\cos\phi = \rho\sin\phi\), so \(\tan\phi = 1\), i.e., \(\phi = \dfrac{\pi}{4}\).

Spherical coordinates shine when the region of integration is a ball or a cone.

1.1.5 Conic Sections

Conic sections arise when a plane intersects a double cone. They include:

• Parabola: \(x^2 = \alpha y\) or \(y^2 = \alpha x\) (for \(\alpha \in \mathbb{R}^*\)).
• Ellipse (with \(a > b > 0\)): \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\).
• Hyperbola (with \(a, b > 0\)): \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) or \(\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1\).

Degenerate cases include a point, a pair of intersecting lines, and a single line.

1.1.6 Lines and Planes in Space

Parametric equations for a line through \(P_0(x_0, y_0, z_0)\) parallel to the nonzero vector \(\mathbf{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k}\):

\[x = x_0 + v_1 t, \quad y = y_0 + v_2 t, \quad z = z_0 + v_3 t, \quad t \in \mathbb{R}.\]

In vector form: \(\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}\), where \(\mathbf{r}_0 = \langle x_0, y_0, z_0\rangle\).

For the line segment from \(A\) to \(B\): \(\mathbf{r}(t) = (1-t)A + tB\), \(t \in [0, 1]\).

Equation of a plane through \(P_0(x_0, y_0, z_0)\) with normal vector \(\mathbf{n} = A\hat{i} + B\hat{j} + C\hat{k}\):

\[A(x - x_0) + B(y - y_0) + C(z - z_0) = 0.\]

To find \(\mathbf{n}\) for a plane through three points, compute the cross product of two vectors lying in the plane. To find a plane parallel to a given plane, use the same normal vector.

1.1.7 Cylinders and Quadric Surfaces

A cylinder in space is generated by moving a line parallel to a fixed direction along a curve (the generating curve). In solid geometry, the generating curves are circles, giving circular cylinders.

• Example: \(y = x^2\) in space is a parabolic cylinder — its generating curve is the parabola \(y = x^2\) in the \(xy\)-plane, swept by lines parallel to the \(z\)-axis.
• Example: The equation \(x = 1\) represents a point in \(\mathbb{R}\), a vertical line in \(\mathbb{R}^2\), and a plane (flat cylinder) parallel to the \(yz\)-plane in \(\mathbb{R}^3\).

Key rule: If a variable is missing from an equation, the surface is a cylinder with rulings parallel to that variable’s axis:

• \(f(x,y) = c\) → cylinder with rulings parallel to the \(z\)-axis
• \(g(x,z) = c\) → cylinder with rulings parallel to the \(y\)-axis
• \(h(y,z) = c\) → cylinder with rulings parallel to the \(x\)-axis

Note: the axis of a cylinder need not be parallel to a coordinate axis in general.

A quadric surface is the graph of a second-degree equation in \(x\), \(y\), \(z\). The most general form is:

\[Ax^2 + By^2 + Cz^2 + Dxy + Exz + Fyz + Gx + Hy + Iz + J = 0.\]

By translation and rotation, this can always be reduced to one of two standard forms:

\[Ax^2 + By^2 + Cz^2 + J = 0, \quad \text{or} \quad Ax^2 + By^2 + Iz = 0.\] The standard forms (after translation/rotation) are:

Surface	Equation
Ellipsoid	\(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} = 1\)
Elliptical Paraboloid	\(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = \dfrac{z}{c}\)
Elliptical Cone	\(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = \dfrac{z^2}{c^2}\)
Hyperboloid of One Sheet	\(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - \dfrac{z^2}{c^2} = 1\)
Hyperboloid of Two Sheets	\(\dfrac{z^2}{c^2} - \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\)
Hyperbolic Paraboloid (Saddle)	\(\dfrac{y^2}{b^2} - \dfrac{x^2}{a^2} = \dfrac{z}{c}\)

To classify a given quadric, complete the square to put it in standard form, then match to the table above.

1.2 Double Integrals

1.2.1 Double Integrals Over Rectangles: Motivation and Definition

In single-variable calculus, the definite integral \(\int_a^b f(x)\,dx\) sums up infinitesimal contributions \(f(x)\,dx\) along an interval. The double integral generalizes this idea to functions of two variables over regions in the plane.

Consider a rectangular region \(R = \{(x,y) \mid a \leq x \leq b,\, c \leq y \leq d\}\). We subdivide \(R\) with a grid into \(n\) small rectangles of area \(\Delta A_k = \Delta x_k\,\Delta y_k\), pick a sample point \((x_k, y_k)\) in each, and form the Riemann sum:

\[S_n = \sum_{k=1}^n f(x_k, y_k)\,\Delta A_k.\]

The norm \(\|P\|\) of a partition is the largest dimension of any rectangle. As \(\|P\| \to 0\) (equivalently \(n \to \infty\)), if the Riemann sums converge to the same limit regardless of the choice of sample points, we say \(f\) is integrable and define the double integral:

\[\iint_R f(x,y)\,dA = \lim_{\|P\|\to 0}\sum_{k=1}^n f(x_k, y_k)\,\Delta A_k.\]

Geometric interpretation: When \(f(x,y) \geq 0\), the double integral equals the volume of the solid region bounded below by \(R\) and above by the surface \(z = f(x,y)\).

Integrability: Every continuous function on a bounded rectangle is integrable.

1.2.2 Computing Double Integrals: Fubini’s Theorem

The key to computing double integrals is reducing them to two successive single integrals.

Fubini’s Theorem (First Form): If \(f(x,y)\) is continuous on \(R : a \leq x \leq b,\, c \leq y \leq d\), then:

\[\iint_R f(x,y)\,dA = \int_c^d\int_a^b f(x,y)\,dx\,dy = \int_a^b\int_c^d f(x,y)\,dy\,dx.\]

The two orders of integration give the same result! This is called an iterated integral.

How to compute: Evaluate the inner integral first (treating the other variable as a constant), then evaluate the outer integral. The area of a region \(R\) is \(\text{Area}(R) = \iint_R dA\).

1.2.3 Double Integrals Over General (Nonrectangular) Regions

Most regions of interest are not rectangles. For a general bounded region \(R\), we cover it with a grid, keep only the rectangles lying completely inside \(R\), form Riemann sums, and take the limit. The result is still called a double integral and still equals the volume under the surface.

Fubini’s Theorem (Second Form): Let \(f(x,y)\) be continuous on \(R\).

1. If \(R\) is described by \(a \leq x \leq b\), \(g_1(x) \leq y \leq g_2(x)\) (vertically simple region):

\[\iint_R f(x,y)\,dA = \int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx.\]

2. If \(R\) is described by \(c \leq y \leq d\), \(h_1(y) \leq x \leq h_2(y)\) (horizontally simple region):

\[\iint_R f(x,y)\,dA = \int_c^d\int_{h_1(y)}^{h_2(y)} f(x,y)\,dx\,dy.\]

Finding limits of integration:

• Vertical cross-sections: Project \(R\) onto the \(x\)-axis to get \([a,b]\). For fixed \(x\), a vertical line through \(R\) enters at \(y = g_1(x)\) and exits at \(y = g_2(x)\).
• Horizontal cross-sections: Project \(R\) onto the \(y\)-axis to get \([c,d]\). For fixed \(y\), a horizontal line enters at \(x = h_1(y)\) and exits at \(x = h_2(y)\).

Reversing the order of integration: Sometimes switching from vertical to horizontal (or vice versa) makes an integral tractable. The key is correctly redescribing the region \(R\).

Properties of double integrals (for \(f, g\) continuous on bounded \(R\)):

1. Constant multiple: \(\iint_R cf\,dA = c\iint_R f\,dA\)
2. Linearity: \(\iint_R (f \pm g)\,dA = \iint_R f\,dA \pm \iint_R g\,dA\)
3. Domination: If \(f \geq 0\) on \(R\), then \(\iint_R f\,dA \geq 0\); if \(f \geq g\), then \(\iint_R f\,dA \geq \iint_R g\,dA\)
4. Additivity: If \(R = R_1 \cup R_2\) (nonoverlapping), then \(\iint_R f\,dA = \iint_{R_1} f\,dA + \iint_{R_2} f\,dA\)

Average value of \(f\) over \(R\): \(\displaystyle\text{aver}_R(f) = \frac{1}{\text{Area}(R)}\iint_R f(x,y)\,dA.\)

1.2.4 Change of Variables in Double Integrals

When a region \(R\) has an inconvenient shape in \(xy\)-coordinates, we can often simplify the integral by changing variables. Suppose the transformation \(x = g(u,v)\), \(y = h(u,v)\) maps a region \(G\) in the \(uv\)-plane one-to-one onto \(R\) in the \(xy\)-plane. Then:

\[\iint_R f(x,y)\,dx\,dy = \iint_G f(g(u,v), h(u,v))\,|J(u,v)|\,du\,dv,\]

where the Jacobian of the transformation is:

\[J(u,v) = \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\[4pt] \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \frac{\partial x}{\partial u}\frac{\partial y}{\partial v} - \frac{\partial x}{\partial v}\frac{\partial y}{\partial u}.\]

The Jacobian measures how the transformation stretches or contracts areas: a small area element \(du\,dv\) in the \(uv\)-plane becomes \(|J|\,du\,dv\) in the \(xy\)-plane.

Strategy for choosing a transformation: Look at the boundary curves of \(R\). If they have the form \(\phi(x,y) = c_1\) and \(\psi(x,y) = c_2\), then set \(u = \phi(x,y)\) and \(v = \psi(x,y)\).

1.2.5 Double Integrals in Polar Coordinates

The most important special case of change of variables is conversion to polar coordinates: \(x = r\cos\theta\), \(y = r\sin\theta\).

The Jacobian is:

\[J(r,\theta) = \begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix} = r\cos^2\theta + r\sin^2\theta = r.\]

Therefore:

\[\iint_R f(x,y)\,dx\,dy = \iint_G f(r\cos\theta, r\sin\theta)\,r\,dr\,d\theta.\]

The extra factor \(r\) (the Jacobian) is crucial — do not forget it!

When to use polar coordinates: The region \(R\) is a disk, annulus, sector, or any region bounded by circles centered at the origin; or the integrand contains \(x^2 + y^2\).

Area using polar coordinates: For a region bounded by \(\theta_1 \leq \theta \leq \theta_2\) and \(0 \leq r \leq r(\theta)\):

\[A = \int_{\theta_1}^{\theta_2}\int_0^{r(\theta)} r\,dr\,d\theta = \frac{1}{2}\int_{\theta_1}^{\theta_2} [r(\theta)]^2\,d\theta.\]

1.2.6 Double Integrals over Unbounded Regions

Just as improper integrals of one variable handle infinite intervals or discontinuities, improper double integrals handle unbounded regions or integrand singularities. We compute them as limits:

\[\iint_R f(x,y)\,dA = \lim_{b\to\infty}\iint_{R_b} f(x,y)\,dA,\]

where \(R_b\) is a bounded approximation of \(R\).

The Gaussian integral (Euler–Poisson integral): A remarkable application of switching to polar coordinates in an improper double integral gives:

\[\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}.\]

Proof: Let \(I = \int_{-\infty}^{\infty} e^{-x^2}\,dx\). Then: \[I^2 = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)}\,dx\,dy = \int_0^{2\pi}\int_0^{\infty} e^{-r^2} r\,dr\,d\theta = 2\pi \cdot \frac{1}{2} = \pi.\] Hence \(I = \sqrt{\pi}\).

1.3 Triple Integrals

1.3.1 Triple Integrals Over Boxes: Definition

The triple integral extends the double integral to functions of three variables over three-dimensional regions. For a bounded region \(D\) in space, we partition a rectangular box containing \(D\) into \(n\) cells of volume \(\Delta V_k = \Delta x_k\,\Delta y_k\,\Delta z_k\), choose a sample point \((x_k, y_k, z_k)\) in each cell inside \(D\), and form:

\[S_n = \sum_{k=1}^n F(x_k, y_k, z_k)\,\Delta V_k.\]

The triple integral is the limit of these Riemann sums:

\[\iiint_D F(x,y,z)\,dV = \lim_{\|P\|\to 0} S_n.\]

Volume and average value:

• Volume of \(D\): \(\text{vol}(D) = \iiint_D dV.\)
• Average value of \(F\) over \(D\): \(\displaystyle\text{aver}_D(F) = \frac{1}{\text{vol}(D)}\iiint_D F\,dV.\)

For a box \(D = [a,b]\times[c,d]\times[p,q]\), Fubini’s theorem gives:

\[\iiint_D F\,dV = \int_a^b\int_c^d\int_p^q F(x,y,z)\,dz\,dy\,dx.\]

1.3.2 Triple Integrals Over General Regions

For a general region \(D\), we use the three-dimensional Fubini’s theorem. If \(D\) is bounded above by \(z = f_2(x,y)\) and below by \(z = f_1(x,y)\), with \((x,y)\) ranging over the projection \(R\) of \(D\) onto the \(xy\)-plane:

\[\iiint_D F(x,y,z)\,dV = \iint_R\left(\int_{f_1(x,y)}^{f_2(x,y)} F(x,y,z)\,dz\right)dA.\]

Similarly, if \(D\) is “simple” in the \(x\)- or \(y\)-direction:

\[\iiint_D F\,dV = \iint_R\left(\int_{f_1(y,z)}^{f_2(y,z)} F\,dx\right)dA, \quad \text{or} \quad \iiint_D F\,dV = \iint_R\left(\int_{f_1(x,z)}^{f_2(x,z)} F\,dy\right)dA.\]

Finding limits: Sketch the region \(D\). Determine its “shadow” (projection) \(R\) onto a coordinate plane, then determine the bounds on the inner variable.

1.3.3 Change of Variables in Triple Integrals

For a transformation \(x = g(u,v,w)\), \(y = h(u,v,w)\), \(z = k(u,v,w)\) mapping \(G\) (in \(uvw\)-space) one-to-one onto \(D\) (in \(xyz\)-space):

\[\iiint_D F(x,y,z)\,dx\,dy\,dz = \iiint_G H(u,v,w)\,|J(u,v,w)|\,du\,dv\,dw,\]

where \(H(u,v,w) = F(g(u,v,w), h(u,v,w), k(u,v,w))\) and the Jacobian is the \(3\times 3\) determinant:

\[J(u,v,w) = \frac{\partial(x,y,z)}{\partial(u,v,w)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\[4pt] \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\[4pt] \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \end{vmatrix}.\]

1.3.4 Triple Integrals in Cylindrical Coordinates

The cylindrical coordinate transformation \(x = r\cos\theta\), \(y = r\sin\theta\), \(z = z\) has Jacobian \(J = r\):

\[\iiint_D F(x,y,z)\,dx\,dy\,dz = \iiint_G H(r,\theta,z)\cdot r\,dr\,d\theta\,dz.\]

When to use: The region \(D\) is a cylinder, cone, or solid of revolution about the \(z\)-axis; or the integrand contains \(x^2 + y^2\).

1.3.5 Triple Integrals in Spherical Coordinates

The spherical coordinate transformation \(x = \rho\sin\phi\cos\theta\), \(y = \rho\sin\phi\sin\theta\), \(z = \rho\cos\phi\) has Jacobian \(J = \rho^2\sin\phi\):

\[\iiint_D F(x,y,z)\,dx\,dy\,dz = \iiint_G H(\rho,\phi,\theta)\cdot\rho^2\sin\phi\,d\rho\,d\phi\,d\theta.\]

When to use: The region \(D\) is a ball, spherical shell, or cone; or the integrand contains \(x^2 + y^2 + z^2\).

1.X Lagrange Multipliers

When we want to optimize a function \(f(x,y)\) (or \(f(x,y,z)\)) subject to a constraint \(g(x,y) = 0\), we cannot simply find unconstrained critical points — we are restricted to points lying on a curve (or surface). The method of Lagrange multipliers provides a systematic way to handle this.

The Key Idea: At a constrained extremum \(P_0\), the gradient of \(f\) must be parallel to the gradient of \(g\) (since we cannot move “uphill” in any direction that stays on the constraint). This means:

\[\nabla f(P_0) = \lambda\, \nabla g(P_0)\]

for some scalar \(\lambda\) called the Lagrange multiplier.

The Lagrange Function: We introduce the auxiliary function

\[L(x, y, \lambda) = f(x, y) - \lambda\, g(x, y).\]

A constrained extremum of \(f\) subject to \(g = 0\) corresponds to a critical point of \(L\), i.e., where all partial derivatives of \(L\) vanish:

\[\frac{\partial L}{\partial x} = f_x - \lambda g_x = 0, \qquad \frac{\partial L}{\partial y} = f_y - \lambda g_y = 0, \qquad \frac{\partial L}{\partial \lambda} = -g(x,y) = 0.\]

The third equation simply enforces the constraint \(g(x,y) = 0\).

Conditions for the theorem to apply:

• \(f\) and \(g\) have continuous first partial derivatives near \(P_0\),
• \(P_0\) is not an endpoint of the constraint curve,
• \(\nabla g(P_0) \neq \mathbf{0}\) (the constraint is “regular” at \(P_0\)).

Extension to 3D: For \(f(x,y,z)\) subject to \(g(x,y,z) = 0\), the Lagrange function is \(L = f - \lambda g\) and we solve four equations: \(\partial L/\partial x = 0\), \(\partial L/\partial y = 0\), \(\partial L/\partial z = 0\), \(g = 0\).

Multiple constraints: For two constraints \(g_1 = 0\) and \(g_2 = 0\), use \(L = f - \lambda_1 g_1 - \lambda_2 g_2\).

1.X+1 Taylor’s Formula for Multivariable Functions

You already know Taylor’s theorem for a single-variable function \(f(x)\): it approximates \(f\) near a point \(a\) by a polynomial whose coefficients involve derivatives of \(f\) at \(a\). The same idea generalizes to functions of two (or more) variables.

Taylor’s Formula at the Point \((a, b)\):

If \(f(x,y)\) and all partial derivatives through order \(n+1\) are continuous on an open rectangle centered at \((a, b)\), then writing \(h = x - a\), \(k = y - b\):

\[f(a+h, b+k) = f(a,b) + (hf_x + kf_y)\big|_{(a,b)} + \frac{1}{2!}(h^2f_{xx} + 2hkf_{xy} + k^2f_{yy})\big|_{(a,b)}\] \[+ \frac{1}{3!}(h^3f_{xxx} + 3h^2kf_{xxy} + 3hk^2f_{xyy} + k^3f_{yyy})\big|_{(a,b)} + \cdots\] \[+ \frac{1}{n!}\left(h\frac{\partial}{\partial x} + k\frac{\partial}{\partial y}\right)^n f\bigg|_{(a,b)} + R_n,\]

where \(R_n = \dfrac{1}{(n+1)!}\left(h\dfrac{\partial}{\partial x}+k\dfrac{\partial}{\partial y}\right)^{n+1}f\Big|_{(a+ch,\,b+ck)}\) for some \(c \in (0,1)\).

The operator notation \(\left(h\dfrac{\partial}{\partial x}+k\dfrac{\partial}{\partial y}\right)^m\) means: expand using the binomial theorem and replace powers by mixed partial derivatives:

\[\left(h\frac{\partial}{\partial x}+k\frac{\partial}{\partial y}\right)^m f = \sum_{j=0}^{m}\binom{m}{j}h^{m-j}k^j\frac{\partial^m f}{\partial x^{m-j}\partial y^j}.\]

Second Degree Taylor Polynomial at \((a, b)\) (most used in practice):

\[f(x, y) \approx f(a,b) + (x-a)f_x\big|_{(a,b)} + (y-b)f_y\big|_{(a,b)} + \frac{1}{2}\left[(x-a)^2 f_{xx} + 2(x-a)(y-b)f_{xy} + (y-b)^2 f_{yy}\right]\bigg|_{(a,b)}.\]

Maclaurin Series (at the origin \((0,0)\)):

\[f(x,y) = f(0,0) + xf_x + yf_y + \frac{1}{2!}(x^2f_{xx}+2xyf_{xy}+y^2f_{yy}) + \frac{1}{3!}(x^3f_{xxx}+3x^2yf_{xxy}+3xy^2f_{xyy}+y^3f_{yyy})+\cdots,\]

where all derivatives are evaluated at \((0,0)\).

Why is this useful?

• Approximation: Near \((a,b)\), \(f(x,y) \approx T_2(x,y)\) (the second-degree Taylor polynomial), giving a quadratic surface that is easy to work with.
• Error estimates: The remainder term \(R_n\) provides a bound on the approximation error.
• Local extrema analysis: The second-degree Taylor polynomial reveals the local behavior of \(f\) near a critical point (connecting to the second derivative test via the Hessian).

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2. Definitions

• Polar coordinates \((r, \theta)\): A coordinate system locating a point by its distance \(r \geq 0\) from the origin (pole) and angle \(\theta\) from the polar axis (usually positive \(x\)-axis).
• Cylindrical coordinates \((r, \theta, z)\): Three-dimensional coordinates using polar coordinates \((r,\theta)\) for the \(xy\)-projection and the standard \(z\)-coordinate for height.
• Spherical coordinates \((\rho, \phi, \theta)\): Three-dimensional coordinates using the distance \(\rho \geq 0\) from the origin, the polar angle \(\phi \in [0,\pi]\) from the positive \(z\)-axis, and the azimuthal angle \(\theta\).
• Cylinder (in space): A surface generated by moving a line parallel to a fixed direction along a generating curve; any equation missing one variable describes a cylinder parallel to that variable’s axis.
• Quadric surface: The graph of a second-degree equation in \(x\), \(y\), \(z\); includes ellipsoids, paraboloids, cones, hyperboloids, and saddle surfaces.
• Double integral \(\iint_R f(x,y)\,dA\): The limit of Riemann sums of \(f\) over a region \(R\) in the plane; geometrically equals the signed volume under \(z = f(x,y)\) over \(R\).
• Triple integral \(\iiint_D F(x,y,z)\,dV\): The limit of Riemann sums of \(F\) over a solid region \(D\) in space.
• Iterated integral: A double or triple integral computed by successively evaluating single-variable integrals from the inside out.
• Fubini’s Theorem: States that a double (or triple) integral over a rectangle (or box) equals the iterated integral in any order, provided the function is continuous.
• Jacobian \(J\): The determinant of the matrix of partial derivatives of a coordinate transformation; it accounts for how the transformation scales area (or volume) elements.
• Vertically simple region: A region in the plane described as \(a \leq x \leq b\), \(g_1(x) \leq y \leq g_2(x)\); integration is done in \(y\) first.
• Horizontally simple region: A region described as \(c \leq y \leq d\), \(h_1(y) \leq x \leq h_2(y)\); integration is done in \(x\) first.
• Gaussian integral: \(\displaystyle\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}\); proved by converting the square of this integral to a polar double integral.
• Average value of \(f\) over \(R\): \(\displaystyle\text{aver}_R(f) = \frac{1}{\text{Area}(R)}\iint_R f\,dA\).
• Constrained optimization: Finding the maximum or minimum of a function \(f\) subject to the requirement that the variables satisfy an equation \(g(x,y) = 0\).
• Lagrange multiplier \(\lambda\): The scalar that appears in the condition \(\nabla f = \lambda\,\nabla g\) at a constrained extremum; it can be interpreted as the rate of change of the optimal value with respect to the constraint.
• Objective function: The function \(f\) being optimized in a constrained optimization problem.
• Constraint: The equation \(g(x,y)=0\) (or \(g(x,y,z)=0\)) that restricts the feasible set.
• Lagrange function (Lagrangian): \(L(x,y,\lambda) = f(x,y) - \lambda\,g(x,y)\); its unconstrained critical points correspond to constrained extrema of \(f\).
• Taylor polynomial (multivariable, degree \(n\)): The polynomial \(T_n(x,y)\) that approximates \(f(x,y)\) near \((a,b)\) using partial derivatives of \(f\) at \((a,b)\) up to order \(n\).
• Maclaurin polynomial: A Taylor polynomial centered at the origin \((0,0)\).
• Quadratic approximation: The second-degree Taylor polynomial \(T_2(x,y)\); locally it is the best quadratic fit to \(f\) at the expansion point.

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3. Formulas

• Polar \(\leftrightarrow\) Cartesian (2D): \(x = r\cos\theta\), \(y = r\sin\theta\); \(r = \sqrt{x^2+y^2}\), \(\tan\theta = y/x\)
• Cylindrical \(\leftrightarrow\) Cartesian (3D): \(x = r\cos\theta\), \(y = r\sin\theta\), \(z = z\); \(r = \sqrt{x^2+y^2}\)
• Spherical \(\leftrightarrow\) Cartesian (3D): \(x = \rho\sin\phi\cos\theta\), \(y = \rho\sin\phi\sin\theta\), \(z = \rho\cos\phi\); \(\rho = \sqrt{x^2+y^2+z^2}\)
• 3D Distance Formula: \(|P_1P_2| = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\)
• Sphere equation: \((x-x_0)^2+(y-y_0)^2+(z-z_0)^2 = R^2\)
• Line in space (parametric): \(x = x_0+v_1 t,\; y = y_0+v_2 t,\; z = z_0+v_3 t\)
• Plane equation: \(A(x-x_0)+B(y-y_0)+C(z-z_0) = 0\)
• Fubini (rectangle): \(\iint_R f\,dA = \int_c^d\int_a^b f\,dx\,dy = \int_a^b\int_c^d f\,dy\,dx\)
• Fubini (vertical simple): \(\iint_R f\,dA = \int_a^b\int_{g_1(x)}^{g_2(x)} f\,dy\,dx\)
• Fubini (horizontal simple): \(\iint_R f\,dA = \int_c^d\int_{h_1(y)}^{h_2(y)} f\,dx\,dy\)
• Change of variables (double integral): \(\iint_R f(x,y)\,dx\,dy = \iint_G f(g,h)\,|J|\,du\,dv\), where \(J = \dfrac{\partial(x,y)}{\partial(u,v)}\)
• Polar double integral: \(\iint_R f(x,y)\,dx\,dy = \iint_G f(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta\)
• Area in polar coordinates: \(A = \dfrac{1}{2}\displaystyle\int_{\theta_1}^{\theta_2} r(\theta)^2\,d\theta\)
• Gaussian integral: \(\displaystyle\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}\)
• Triple integral (Fubini): \(\iiint_D F\,dV = \iint_R\left(\displaystyle\int_{f_1}^{f_2} F\,dz\right)dA\)
• Change of variables (triple integral): \(\iiint_D F\,dx\,dy\,dz = \iiint_G H\,|J|\,du\,dv\,dw\), where \(J = \dfrac{\partial(x,y,z)}{\partial(u,v,w)}\)
• Cylindrical triple integral: \(\iiint_D F\,dV = \iiint_G H(r,\theta,z)\,r\,dr\,d\theta\,dz\)
• Spherical triple integral: \(\iiint_D F\,dV = \iiint_G H(\rho,\phi,\theta)\,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta\)
• Volume of \(D\): \(\text{vol}(D) = \iiint_D dV\)
• Average value (3D): \(\text{aver}_D(F) = \dfrac{1}{\text{vol}(D)}\iiint_D F\,dV\)
• Lagrange condition (2D): \(\nabla f = \lambda\,\nabla g\), i.e., \(f_x = \lambda g_x\), \(f_y = \lambda g_y\), \(g(x,y) = 0\)
• Lagrange function: \(L(x,y,\lambda) = f(x,y) - \lambda\,g(x,y)\)
• Lagrange condition (3D): \(f_x = \lambda g_x\), \(f_y = \lambda g_y\), \(f_z = \lambda g_z\), \(g(x,y,z) = 0\)
• Taylor’s formula at \((a,b)\) (degree 2): \(f(a+h,b+k) \approx f(a,b) + hf_x + kf_y + \dfrac{1}{2!}(h^2f_{xx}+2hkf_{xy}+k^2f_{yy})\), where derivatives are at \((a,b)\)
• Second degree Taylor polynomial at \((a,b)\): \(T_2(x,y) = f(a,b) + (x-a)f_x + (y-b)f_y + \dfrac{1}{2}[(x-a)^2f_{xx} + 2(x-a)(y-b)f_{xy} + (y-b)^2f_{yy}]\)
• Maclaurin polynomial (degree 2): \(T_2(x,y) = f(0,0) + xf_x(0,0) + yf_y(0,0) + \dfrac{1}{2}[x^2f_{xx}(0,0)+2xyf_{xy}(0,0)+y^2f_{yy}(0,0)]\)

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4. Examples

4.1. Find and Classify Critical Points (Lab 9, Task 1)

Find and classify the critical points of \(f(x, y) = 10xye^{-(x^2+y^2)}\).

Click to see the solution

Key Concept: Use the second derivative test: compute \(f_x\), \(f_y\), set both to zero, then evaluate \(D = f_{xx}f_{yy} - f_{xy}^2\) and \(f_{xx}\) at each critical point.

1. Compute \(f_x\): \[f_x = 10y e^{-(x^2+y^2)} + 10xy \cdot (-2x) e^{-(x^2+y^2)} = 10ye^{-(x^2+y^2)}(1 - 2x^2).\]

2. Compute \(f_y\): \[f_y = 10x e^{-(x^2+y^2)} + 10xy \cdot (-2y) e^{-(x^2+y^2)} = 10xe^{-(x^2+y^2)}(1 - 2y^2).\]

3. Set \(f_x = 0\) and \(f_y = 0\): Since \(e^{-(x^2+y^2)} > 0\) always, we need:

   • \(f_x = 0\): \(y(1 - 2x^2) = 0 \implies y = 0\) or \(x = \pm\tfrac{1}{\sqrt{2}}\).
   • \(f_y = 0\): \(x(1 - 2y^2) = 0 \implies x = 0\) or \(y = \pm\tfrac{1}{\sqrt{2}}\).

   Solving simultaneously gives five critical points:

   • \((0, 0)\)
   • \(\left(\tfrac{1}{\sqrt{2}}, \tfrac{1}{\sqrt{2}}\right)\), \(\left(\tfrac{1}{\sqrt{2}}, -\tfrac{1}{\sqrt{2}}\right)\), \(\left(-\tfrac{1}{\sqrt{2}}, \tfrac{1}{\sqrt{2}}\right)\), \(\left(-\tfrac{1}{\sqrt{2}}, -\tfrac{1}{\sqrt{2}}\right)\).

4. Compute second partial derivatives: \[f_{xx} = 10y e^{-(x^2+y^2)}(-2x)(1-2x^2) + 10ye^{-(x^2+y^2)}(-4x) = 10ye^{-(x^2+y^2)}(-6x+4x^3).\] \[f_{yy} = 10xe^{-(x^2+y^2)}(-6y+4y^3).\] \[f_{xy} = 10e^{-(x^2+y^2)}(1-2x^2)(1-2y^2).\]

5. Classify each critical point:

   (a) At \((0,0)\): \(f_{xx}=0\), \(f_{yy}=0\), \(f_{xy}=10\cdot1\cdot1=10\). So \(D = 0\cdot0 - 10^2 = -100 < 0\). \(\Rightarrow\) Saddle point at \((0,0)\).

   (b) At \(\left(\tfrac{1}{\sqrt{2}}, \tfrac{1}{\sqrt{2}}\right)\): Let \(e^{-1} = 1/e\). \(f_{xx} = 10\cdot\tfrac{1}{\sqrt{2}}\cdot e^{-1}\left(-6\cdot\tfrac{1}{\sqrt{2}}+4\cdot\tfrac{1}{2\sqrt{2}}\right) = \tfrac{10}{\sqrt{2}e}\left(\tfrac{-6+2}{\sqrt{2}}\right) = \tfrac{10}{\sqrt{2}e}\cdot\tfrac{-4}{\sqrt{2}} = \tfrac{-40}{2e} = \tfrac{-20}{e}\). By symmetry, \(f_{yy} = -20/e\). \(f_{xy} = 10e^{-1}(1-1)(1-1) = 0\). \(D = (-20/e)(-20/e) - 0 = 400/e^2 > 0\), and \(f_{xx} = -20/e < 0\). \(\Rightarrow\) Local maximum at \(\left(\tfrac{1}{\sqrt{2}}, \tfrac{1}{\sqrt{2}}\right)\). The value is \(f = 10\cdot\tfrac{1}{\sqrt{2}}\cdot\tfrac{1}{\sqrt{2}}\cdot e^{-1} = 5/e\).

   (c) At \(\left(-\tfrac{1}{\sqrt{2}}, -\tfrac{1}{\sqrt{2}}\right)\): By symmetry (\(f(-x,-y) = f(x,y)\)), also a local maximum with value \(5/e\).

   (d) At \(\left(\tfrac{1}{\sqrt{2}}, -\tfrac{1}{\sqrt{2}}\right)\): \(f_{xx} = 10\cdot(-\tfrac{1}{\sqrt{2}})\cdot e^{-1}\cdot\tfrac{-4}{\sqrt{2}} = +20/e > 0\). \(f_{yy} = +20/e\), \(f_{xy} = 0\). \(D = (20/e)^2 > 0\), \(f_{xx} > 0\). \(\Rightarrow\) Local minimum at \(\left(\tfrac{1}{\sqrt{2}}, -\tfrac{1}{\sqrt{2}}\right)\). The value is \(f = -5/e\).

   (e) At \(\left(-\tfrac{1}{\sqrt{2}}, \tfrac{1}{\sqrt{2}}\right)\): By symmetry, also a local minimum with value \(-5/e\).

Answer: Five critical points:

• \((0,0)\): saddle point.
• \(\left(\pm\tfrac{1}{\sqrt{2}}, \pm\tfrac{1}{\sqrt{2}}\right)\) (same sign): local maxima with value \(5/e\).
• \(\left(\pm\tfrac{1}{\sqrt{2}}, \mp\tfrac{1}{\sqrt{2}}\right)\) (opposite sign): local minima with value \(-5/e\).

4.2. Maximum Volume of a Lidless Box (Lab 9, Task 2)

A rectangular box without a lid is to be made from \(12\text{ m}^2\) of cardboard. Find the maximum volume.

Click to see the solution

Key Concept: Use Lagrange multipliers. The objective function is the volume \(V = xyz\); the constraint is the surface area of a lidless box: the base (\(xy\)) plus four sides (\(2xz + 2yz\)) equals \(12\).

1. Set up:

   • Objective: \(f(x,y,z) = xyz\).
   • Constraint: \(g(x,y,z) = xy + 2xz + 2yz - 12 = 0\).

2. Lagrange equations (\(\nabla f = \lambda \nabla g\), with \(g=0\)): \[yz = \lambda(y + 2z), \quad xz = \lambda(x + 2z), \quad xy = \lambda(2x + 2y).\]

3. Solve by symmetry: Multiply first equation by \(x\), second by \(y\), third by \(z\): \[xyz = \lambda x(y+2z), \quad xyz = \lambda y(x+2z), \quad xyz = \lambda z(2x+2y).\] So \(\lambda x(y+2z) = \lambda y(x+2z) = \lambda z(2x+2y)\) (assuming \(\lambda \neq 0\)).

   From \(x(y+2z) = y(x+2z)\): \(xy + 2xz = xy + 2yz \implies 2xz = 2yz \implies x = y\).

   From \(y(x+2z) = z(2x+2y)\) and \(x=y\): \(x(x+2z) = z(2x+2x) = 4xz \implies x+2z = 4z \implies x = 2z\).

   So \(x = y = 2z\).

4. Apply constraint: \(x = y = 2z\), \(xy + 2xz + 2yz = 12\): \[(2z)(2z) + 2(2z)(z) + 2(2z)(z) = 4z^2 + 4z^2 + 4z^2 = 12z^2 = 12 \implies z = 1.\] Thus \(x = y = 2\), \(z = 1\).

5. Maximum volume: \(V = 2 \cdot 2 \cdot 1 = 4\text{ m}^3\).

Answer: The maximum volume is \(\boxed{4\text{ m}^3}\), achieved with base \(2\text{ m} \times 2\text{ m}\) and height \(1\text{ m}\).

4.3. Extreme Values on a Circle (Lab 9, Task 3)

Find the extreme values of \(f(x, y) = x^2 + 2y^2\) on the circle \(x^2 + y^2 = 1\).

Click to see the solution

1. Set up:

   • Objective: \(f = x^2 + 2y^2\).
   • Constraint: \(g = x^2 + y^2 - 1 = 0\).
   • Lagrange function: \(L = x^2+2y^2-\lambda(x^2+y^2-1)\).

2. Lagrange equations: \[2x = 2\lambda x, \quad 4y = 2\lambda y, \quad x^2+y^2 = 1.\]

3. Solve:

   • From \(2x = 2\lambda x\): either \(x = 0\) or \(\lambda = 1\).
   • From \(4y = 2\lambda y\): either \(y = 0\) or \(\lambda = 2\).

   Case 1: \(\lambda = 1\). Then \(4y = 2y \implies 2y = 0 \implies y = 0\). Constraint: \(x^2 = 1\), so \(x = \pm 1\). Points: \((\pm 1, 0)\), \(f = 1\).

   Case 2: \(\lambda = 2\). Then \(2x = 4x \implies 2x = 0 \implies x = 0\). Constraint: \(y^2 = 1\), so \(y = \pm 1\). Points: \((0, \pm 1)\), \(f = 2\).

   Case 3: \(x = 0\) and \(y = 0\) — impossible on the unit circle.

4. Classify: Minimum value \(f = 1\) at \((\pm 1, 0)\); maximum value \(f = 2\) at \((0, \pm 1)\).

Answer: Minimum is \(1\) (at \((\pm1,0)\)); maximum is \(2\) (at \((0,\pm1)\)).

4.4. Points on a Sphere Closest/Farthest from a Point (Lab 9, Task 4)

Find the points on the sphere \(x^2 + y^2 + z^2 = 4\) closest to and farthest from \((3, 1, -1)\).

Click to see the solution

Key Concept: Minimizing/maximizing the distance is equivalent to minimizing/maximizing the squared distance \(d^2 = (x-3)^2+(y-1)^2+(z+1)^2\), which is simpler.

1. Set up:

   • Objective: \(f = (x-3)^2+(y-1)^2+(z+1)^2\).
   • Constraint: \(g = x^2+y^2+z^2-4 = 0\).

2. Lagrange equations: \[2(x-3) = 2\lambda x, \quad 2(y-1) = 2\lambda y, \quad 2(z+1) = 2\lambda z.\] Simplify: \(x-3 = \lambda x\), \(y-1 = \lambda y\), \(z+1 = \lambda z\).

3. Solve: \((1-\lambda)x = 3\), \((1-\lambda)y = 1\), \((1-\lambda)z = -1\) (assuming \(\lambda \neq 1\)). So \(x = \dfrac{3}{1-\lambda}\), \(y = \dfrac{1}{1-\lambda}\), \(z = \dfrac{-1}{1-\lambda}\).

   Apply constraint: \(\dfrac{9+1+1}{(1-\lambda)^2} = 4 \implies (1-\lambda)^2 = \dfrac{11}{4} \implies 1-\lambda = \pm\dfrac{\sqrt{11}}{2}\).

4. Two solutions:

   • \(1-\lambda = \dfrac{\sqrt{11}}{2}\): \((x,y,z) = \left(\dfrac{6}{\sqrt{11}}, \dfrac{2}{\sqrt{11}}, \dfrac{-2}{\sqrt{11}}\right)\). Squared distance \(= \left(\dfrac{6}{\sqrt{11}}-3\right)^2+\cdots\). Since \(1-\lambda > 0\), this point lies in the same direction as \((3,1,-1)\) from the origin — closest point.
   • \(1-\lambda = -\dfrac{\sqrt{11}}{2}\): \((x,y,z) = \left(\dfrac{-6}{\sqrt{11}}, \dfrac{-2}{\sqrt{11}}, \dfrac{2}{\sqrt{11}}\right)\) — farthest point.

   Distance from origin to \((3,1,-1)\) is \(\sqrt{9+1+1} = \sqrt{11}\). The sphere has radius \(2\).

   • Closest distance: \(\sqrt{11} - 2\).
   • Farthest distance: \(\sqrt{11} + 2\).

Answer:

• Closest point: \(\left(\dfrac{6}{\sqrt{11}}, \dfrac{2}{\sqrt{11}}, -\dfrac{2}{\sqrt{11}}\right)\), distance \(\sqrt{11}-2\).
• Farthest point: \(\left(-\dfrac{6}{\sqrt{11}}, -\dfrac{2}{\sqrt{11}}, \dfrac{2}{\sqrt{11}}\right)\), distance \(\sqrt{11}+2\).

4.5. Maximum and Minimum of \(xy\) on the Unit Circle (Lab 9, Task 5)

Find the maximum and minimum of \(f(x, y) = xy\) on the circle \(x^2 + y^2 = 1\).

Click to see the solution

1. Lagrange equations: \(y = 2\lambda x\), \(x = 2\lambda y\), \(x^2+y^2=1\).
2. Solve: From the first two equations: \(y = 2\lambda x\) and \(x = 2\lambda(2\lambda x) = 4\lambda^2 x\). If \(x \neq 0\): \(4\lambda^2 = 1 \implies \lambda = \pm\tfrac{1}{2}\).
   • \(\lambda = \tfrac{1}{2}\): \(y = x\). Constraint: \(2x^2 = 1 \implies x = \pm\tfrac{1}{\sqrt{2}}\). Points: \(\left(\tfrac{1}{\sqrt{2}},\tfrac{1}{\sqrt{2}}\right)\) and \(\left(-\tfrac{1}{\sqrt{2}},-\tfrac{1}{\sqrt{2}}\right)\), \(f = \tfrac{1}{2}\).
   • \(\lambda = -\tfrac{1}{2}\): \(y = -x\). Points: \(\left(\tfrac{1}{\sqrt{2}},-\tfrac{1}{\sqrt{2}}\right)\) and \(\left(-\tfrac{1}{\sqrt{2}},\tfrac{1}{\sqrt{2}}\right)\), \(f = -\tfrac{1}{2}\).

Answer: Maximum is \(\dfrac{1}{2}\); minimum is \(-\dfrac{1}{2}\).

4.6. Second Degree Taylor Polynomial at \((0,0)\) (Lab 9, Task 6)

Find the second degree Taylor polynomial for \(f(x, y) = e^x \cos y\) at \((0, 0)\).

Click to see the solution

Key Concept: Use the Maclaurin formula \(T_2 = f + xf_x + yf_y + \tfrac{1}{2}(x^2f_{xx}+2xyf_{xy}+y^2f_{yy})\), with all derivatives evaluated at \((0,0)\).

1. Compute values at \((0,0)\):
   • \(f = e^0\cos 0 = 1\).
   • \(f_x = e^x\cos y \big|_{(0,0)} = 1\).
   • \(f_y = -e^x\sin y \big|_{(0,0)} = 0\).
   • \(f_{xx} = e^x\cos y \big|_{(0,0)} = 1\).
   • \(f_{xy} = -e^x\sin y \big|_{(0,0)} = 0\).
   • \(f_{yy} = -e^x\cos y \big|_{(0,0)} = -1\).
2. Assemble: \[T_2(x,y) = 1 + x\cdot1 + y\cdot0 + \frac{1}{2}(x^2\cdot1 + 2xy\cdot0 + y^2\cdot(-1))\] \[= 1 + x + \frac{x^2 - y^2}{2}.\]

Answer: \(T_2(x,y) = 1 + x + \dfrac{x^2 - y^2}{2}\).

4.7. Second Degree Taylor Polynomial at \((1,2)\) (Lab 9, Task 7)

Find the second degree Taylor polynomial for \(f(x, y) = \ln(xy)\) at \((1, 2)\).

Click to see the solution

1. Compute values at \((1,2)\):
   • \(f = \ln(1\cdot2) = \ln 2\).
   • \(f_x = \tfrac{1}{x}\big|_{(1,2)} = 1\).
   • \(f_y = \tfrac{1}{y}\big|_{(1,2)} = \tfrac{1}{2}\).
   • \(f_{xx} = -\tfrac{1}{x^2}\big|_{(1,2)} = -1\).
   • \(f_{xy} = 0\) (since \(\ln(xy) = \ln x + \ln y\), so \(\partial^2/\partial x\partial y = 0\)).
   • \(f_{yy} = -\tfrac{1}{y^2}\big|_{(1,2)} = -\tfrac{1}{4}\).
2. Assemble with \(h = x-1\), \(k = y-2\): \[T_2 = \ln 2 + (x-1)\cdot1 + (y-2)\cdot\tfrac{1}{2} + \frac{1}{2}\left[(x-1)^2(-1) + 0 + (y-2)^2\left(-\tfrac{1}{4}\right)\right]\] \[= \ln 2 + (x-1) + \frac{y-2}{2} - \frac{(x-1)^2}{2} - \frac{(y-2)^2}{8}.\]

Answer: \(T_2(x,y) = \ln 2 + (x-1) + \dfrac{y-2}{2} - \dfrac{(x-1)^2}{2} - \dfrac{(y-2)^2}{8}\).

4.8. Estimate Using Second Degree Taylor Polynomial (Lab 9, Task 8)

Use a second degree Taylor polynomial to estimate \(f(2.1, 1.8)\) for \(f(x, y) = x^y\) near \((2, 2)\).

Click to see the solution

Key Concept: We expand \(f(x,y) = x^y = e^{y\ln x}\) near \((a,b) = (2,2)\), then evaluate at \((2.1, 1.8)\), i.e., \(h = 0.1\), \(k = -0.2\).

1. Compute values at \((2,2)\):

   • \(f = 2^2 = 4\).
   • \(f_x = y x^{y-1}\big|_{(2,2)} = 2\cdot2^1 = 4\).
   • \(f_y = x^y\ln x\big|_{(2,2)} = 4\ln 2\).
   • \(f_{xx} = y(y-1)x^{y-2}\big|_{(2,2)} = 2\cdot1\cdot2^0 = 2\).
   • \(f_{xy} = x^{y-1}(1 + y\ln x)\big|_{(2,2)} = 2^1(1+2\ln2) = 2(1+2\ln 2)\).
   • \(f_{yy} = x^y(\ln x)^2\big|_{(2,2)} = 4(\ln 2)^2\).

2. Second degree Taylor polynomial: \[T_2(x,y) = 4 + 4h + 4(\ln2)k + \frac{1}{2}\left[2h^2 + 2\cdot2(1+2\ln2)hk + 4(\ln2)^2k^2\right].\] with \(h=0.1\), \(k=-0.2\):

   • \(4h = 0.4\), \(4(\ln2)(-0.2) \approx 4(0.6931)(-0.2) \approx -0.5545\).
   • \(h^2 = 0.01\), \(hk = -0.02\), \(k^2 = 0.04\).
   • \(\tfrac{1}{2}[2(0.01) + 4(1+2\ln2)(-0.02) + 4(\ln2)^2(0.04)]\) \(\approx \tfrac{1}{2}[0.02 + 4(2.3863)(-0.02) + 4(0.4805)(0.04)]\) \(\approx \tfrac{1}{2}[0.02 - 0.1909 + 0.07688]\) \(\approx \tfrac{1}{2}(-0.09402) \approx -0.04701\).

3. Estimate: \[f(2.1,1.8) \approx 4 + 0.4 - 0.5545 - 0.0470 \approx 3.799.\]

   (For comparison, the exact value is \(2.1^{1.8} \approx 3.800\).)

Answer: \(f(2.1, 1.8) \approx 3.80\).

4.9. Convert Spherical to Cartesian Coordinates (Lab 10, Task 1)

Convert the point \((\rho, \phi, \theta) = (4, \pi/3, \pi/6)\) to Cartesian coordinates.

Click to see the solution

Key Concept: Use \(x = \rho\sin\phi\cos\theta\), \(y = \rho\sin\phi\sin\theta\), \(z = \rho\cos\phi\).

1. Compute \(x\): \[x = 4\sin\!\tfrac{\pi}{3}\cos\!\tfrac{\pi}{6} = 4\cdot\tfrac{\sqrt{3}}{2}\cdot\tfrac{\sqrt{3}}{2} = 4\cdot\tfrac{3}{4} = 3.\]
2. Compute \(y\): \[y = 4\sin\!\tfrac{\pi}{3}\sin\!\tfrac{\pi}{6} = 4\cdot\tfrac{\sqrt{3}}{2}\cdot\tfrac{1}{2} = \sqrt{3}.\]
3. Compute \(z\): \[z = 4\cos\!\tfrac{\pi}{3} = 4\cdot\tfrac{1}{2} = 2.\]

Answer: \((x,y,z) = (3,\,\sqrt{3},\,2)\).

4.10. Convert Spherical to Cartesian and Cylindrical Coordinates (Lab 10, Task 2)

Convert the point \((\rho, \phi, \theta) = (2, \pi/3, \pi/4)\) to Cartesian and cylindrical coordinates.

Click to see the solution

Key Concept: For Cartesian, use \(x = \rho\sin\phi\cos\theta\), \(y = \rho\sin\phi\sin\theta\), \(z = \rho\cos\phi\). For cylindrical, use \(r = \rho\sin\phi\), keeping \(\theta\) and \(z\).

1. Cartesian coordinates: \[x = 2\sin\!\tfrac{\pi}{3}\cos\!\tfrac{\pi}{4} = 2\cdot\tfrac{\sqrt{3}}{2}\cdot\tfrac{\sqrt{2}}{2} = \tfrac{\sqrt{6}}{2}.\] \[y = 2\sin\!\tfrac{\pi}{3}\sin\!\tfrac{\pi}{4} = 2\cdot\tfrac{\sqrt{3}}{2}\cdot\tfrac{\sqrt{2}}{2} = \tfrac{\sqrt{6}}{2}.\] \[z = 2\cos\!\tfrac{\pi}{3} = 2\cdot\tfrac{1}{2} = 1.\]
2. Cylindrical coordinates: \(r = \rho\sin\phi = 2\sin(\pi/3) = \sqrt{3}\), \(\theta = \pi/4\), \(z = 1\).

Answer: Cartesian: \(\!\left(\tfrac{\sqrt{6}}{2},\,\tfrac{\sqrt{6}}{2},\,1\right)\). Cylindrical: \((\sqrt{3},\,\pi/4,\,1)\).

4.11. Convert a Cartesian Equation to Spherical Coordinates (Lab 10, Task 3)

Convert the equation \(x^2 + y^2 + z^2 = 2z\) to spherical coordinates.

Click to see the solution

Key Concept: Substitute \(x^2+y^2+z^2 = \rho^2\) and \(z = \rho\cos\phi\).

1. Substitute: \[\rho^2 = 2\rho\cos\phi.\]
2. Divide by \(\rho\) (assuming \(\rho \neq 0\)): \[\rho = 2\cos\phi.\]

Answer: \(\rho = 2\cos\phi\).

4.12. Double Integral Over a Rectangle (Lab 10, Task 4)

Evaluate \(\displaystyle\iint_{R}(3x + 4y)\,dA\) over \(R = [-1, 2] \times [0, 2]\).

Click to see the solution

Key Concept: Apply Fubini’s theorem to evaluate as an iterated integral.

1. Inner integral (with respect to \(x\)): \[\int_{-1}^{2}(3x+4y)\,dx = \left[\tfrac{3x^2}{2}+4xy\right]_{-1}^{2} = (6+8y)-(\tfrac{3}{2}-4y) = \tfrac{9}{2}+12y.\]
2. Outer integral (with respect to \(y\)): \[\int_0^2\!\left(\tfrac{9}{2}+12y\right)dy = \left[\tfrac{9y}{2}+6y^2\right]_0^2 = 9+24 = 33.\]

Answer: \(\displaystyle\iint_R(3x+4y)\,dA = 33\).

4.13. Volume Under a Paraboloid Over a Rectangle (Lab 10, Task 5)

Find the volume of the solid bounded above by \(z = 16 - x^2 - 2y^2\) and below by the rectangle \(R = [0,2]\times[0,2]\).

Click to see the solution

1. Set up the double integral: \[V = \int_0^2\int_0^2(16-x^2-2y^2)\,dx\,dy.\]
2. Inner integral (with respect to \(x\)): \[\int_0^2(16-x^2-2y^2)\,dx = \left[16x-\tfrac{x^3}{3}-2xy^2\right]_0^2 = 32-\tfrac{8}{3}-4y^2.\]
3. Outer integral: \[V = \int_0^2\!\left(32-\tfrac{8}{3}-4y^2\right)dy = \left[\!\left(32-\tfrac{8}{3}\right)y - \tfrac{4y^3}{3}\right]_0^2 = 2\!\left(32-\tfrac{8}{3}\right)-\tfrac{32}{3} = 64-\tfrac{16}{3}-\tfrac{32}{3} = 64-16 = 48.\]

Answer: \(V = 48\).

4.14. Double Integral Over a Region Bounded by Parabola and Line (Lab 10, Task 6)

Evaluate \(\displaystyle\iint_{D}(x^2 + y)\,dA\) where \(D\) is the region bounded by \(y = x^2\) and \(y = 2x\).

Click to see the solution

Key Concept: Find intersection points, determine integration order, then integrate.

1. Intersection: \(x^2 = 2x \Rightarrow x(x-2)=0 \Rightarrow x = 0,\,2\). For \(0 \le x \le 2\), \(x^2 \le y \le 2x\).
2. Inner integral (with respect to \(y\)): \[\int_{x^2}^{2x}(x^2+y)\,dy = \left[x^2y+\tfrac{y^2}{2}\right]_{x^2}^{2x} = (2x^3+2x^2)-(x^4+\tfrac{x^4}{2}) = 2x^3+2x^2-\tfrac{3x^4}{2}.\]
3. Outer integral: \[\int_0^2\!\left(2x^3+2x^2-\tfrac{3}{2}x^4\right)dx = \left[\tfrac{x^4}{2}+\tfrac{2x^3}{3}-\tfrac{3x^5}{10}\right]_0^2 = 8+\tfrac{16}{3}-\tfrac{96}{10} = \tfrac{120+80-144}{15} = \tfrac{56}{15}.\]

Answer: \(\displaystyle\iint_D(x^2+y)\,dA = \dfrac{56}{15}\).

4.15. Double Integral Over a Region Bounded by a Parabola and a Line (Lab 10, Task 7)

Evaluate \(\displaystyle\iint_{D} xy\,dA\) where \(D\) is the region bounded by \(y = x - 1\) and \(y^2 = 2x + 6\).

Click to see the solution

Key Concept: Integrate with respect to \(x\) first, treating \(y\) as the outer variable, since the \(x\)-bounds are single-valued in \(y\).

1. Find \(x\)-bounds: From \(y^2 = 2x+6\): \(x = \tfrac{y^2-6}{2}\). From \(y = x-1\): \(x = y+1\).
2. Intersection: \(y+1 = \tfrac{y^2-6}{2} \Rightarrow y^2-2y-8=0 \Rightarrow y = -2,\,4\).
3. Inner integral (with respect to \(x\), from \(\tfrac{y^2-6}{2}\) to \(y+1\)): \[\int_{\frac{y^2-6}{2}}^{y+1} xy\,dx = y\cdot\frac{x^2}{2}\Bigg|_{\frac{y^2-6}{2}}^{y+1} = \frac{y}{2}\!\left[(y+1)^2 - \left(\tfrac{y^2-6}{2}\right)^2\right].\] Simplifying the bracket: \[\frac{4(y^2+2y+1)-(y^2-6)^2}{4} = \frac{-y^4+16y^2+8y-32}{4}.\]
4. Outer integral: \[\int_{-2}^{4}\frac{y(-y^4+16y^2+8y-32)}{8}\,dy = \frac{1}{8}\left[-\tfrac{y^6}{6}+4y^4+\tfrac{8y^3}{3}-16y^2\right]_{-2}^{4}.\] At \(y=4\): \(-\tfrac{4096}{6}+1024+\tfrac{512}{3}-256 = 256\). At \(y=-2\): \(-\tfrac{64}{6}+64-\tfrac{64}{3}-64 = -32\). \[\frac{1}{8}(256-(-32)) = \frac{288}{8} = 36.\]

Answer: \(\displaystyle\iint_D xy\,dA = 36\).

4.16. Volume Under a Bilinear Surface Over a Triangle (Lab 10, Task 8)

Find the volume of the solid under \(z = xy\) and above the triangle with vertices \((0,0)\), \((1,0)\), and \((0,2)\).

Click to see the solution

1. Determine the region: The line through \((1,0)\) and \((0,2)\) is \(y = 2-2x\), so for \(0 \le x \le 1\), \(y\) ranges from \(0\) to \(2-2x\).
2. Set up the integral: \[V = \int_0^1\int_0^{2-2x} xy\,dy\,dx = \int_0^1 x\cdot\frac{(2-2x)^2}{2}\,dx = 2\int_0^1 x(1-x)^2\,dx.\]
3. Evaluate: \[2\int_0^1(x-2x^2+x^3)\,dx = 2\left[\tfrac{x^2}{2}-\tfrac{2x^3}{3}+\tfrac{x^4}{4}\right]_0^1 = 2\!\left(\tfrac{1}{2}-\tfrac{2}{3}+\tfrac{1}{4}\right) = 2\cdot\tfrac{1}{12} = \tfrac{1}{6}.\]

Answer: \(V = \dfrac{1}{6}\).

4.17. Volume of a Tetrahedron Bounded by Four Planes (Lab 10, Task 9)

Find the volume of the tetrahedron bounded by \(x + 2y + z = 2\), \(x = 2y\), \(x = 0\), and \(z = 0\).

Click to see the solution

1. Determine the projection onto the \(xy\)-plane: \(z \ge 0\) requires \(x+2y \le 2\). The region \(D\) is bounded by \(x=0\), \(x=2y\) (i.e. \(y=x/2\)), and \(x+2y=2\) (i.e. \(y=(2-x)/2\)). For \(0 \le x \le 1\): \(y\) ranges from \(x/2\) to \((2-x)/2\).
2. Set up the integral: \[V = \int_0^1\int_{x/2}^{(2-x)/2}(2-x-2y)\,dy\,dx.\]
3. Inner integral: \[\left[(2-x)y-y^2\right]_{x/2}^{(2-x)/2} = \frac{(2-x)^2}{4} - \frac{x(2-x)}{2}+\frac{x^2}{4} = \frac{(2-2x)^2}{4} = (1-x)^2.\]
4. Outer integral: \[V = \int_0^1(1-x)^2\,dx = \left[-\tfrac{(1-x)^3}{3}\right]_0^1 = \tfrac{1}{3}.\]

Answer: \(V = \dfrac{1}{3}\).

4.18. Double Integral with Mixed Exponential and Trigonometric Integrand (Lab 10, Task 10)

Evaluate \(\displaystyle\iint_{R}(x\sin y - ye^x)\,dA\) over \(R = [0,1]\times[0,\pi/2]\).

Click to see the solution

1. Apply Fubini’s theorem (integrate \(x\) first): \[\int_0^{\pi/2}\int_0^1(x\sin y-ye^x)\,dx\,dy.\]
2. Inner integral: \[\left[\tfrac{x^2}{2}\sin y - ye^x\right]_0^1 = \tfrac{1}{2}\sin y - ye.\]
3. Outer integral: \[\int_0^{\pi/2}\!\left(\tfrac{1}{2}\sin y - ye\right)dy = \left[-\tfrac{1}{2}\cos y - \tfrac{ey^2}{2}\right]_0^{\pi/2} = \left(0 - \tfrac{e\pi^2}{8}\right)-\left(-\tfrac{1}{2}-0\right) = \tfrac{1}{2}-\tfrac{e\pi^2}{8}.\]

Answer: \(\displaystyle\iint_R(x\sin y-ye^x)\,dA = \dfrac{1}{2}-\dfrac{e\pi^2}{8}\).

4.19. Double Integral in the First Quadrant Between a Parabola and a Line (Lab 10, Task 11)

Evaluate \(\displaystyle\iint_{D}(x^2 + y)\,dA\) where \(D\) is the region in the first quadrant bounded by \(y = x^2\) and \(y = 2x\).

Click to see the solution

The region in the first quadrant between \(y=x^2\) and \(y=2x\) coincides with the full region between these curves for \(0 \le x \le 2\) (both are already in the first quadrant). By the same calculation as Task 6:

\[\iint_D(x^2+y)\,dA = \frac{56}{15}.\]

Answer: \(\dfrac{56}{15}\).

4.20. Volume of a Paraboloid Cap (Lab 10, Task 12)

Find the volume of the solid bounded above by the paraboloid \(z = 4 - x^2 - y^2\) and below by the \(xy\)-plane.

Click to see the solution

Key Concept: The paraboloid meets the \(xy\)-plane where \(x^2+y^2=4\), i.e. on a circle of radius \(2\). Convert to polar coordinates.

1. In polar: \(z = 4 - r^2\), and the region is \(0 \le r \le 2\), \(0 \le \theta \le 2\pi\).
2. Set up the integral: \[V = \int_0^{2\pi}\int_0^2(4-r^2)\,r\,dr\,d\theta = 2\pi\int_0^2(4r-r^3)\,dr.\]
3. Evaluate: \[2\pi\left[2r^2-\tfrac{r^4}{4}\right]_0^2 = 2\pi(8-4) = 8\pi.\]

Answer: \(V = 8\pi\).

4.21. Converting Between Polar and Cartesian Coordinates (Chapter 3, Example 1)

(a) Find the Cartesian coordinates of the point \(A\!\left(\sqrt{2},\,\dfrac{\pi}{4}\right)\) given in polar coordinates.

(b) Find the polar coordinates of the point \(B(\sqrt{3},\,-1)\) given in Cartesian coordinates.

Click to see the solution

Key Concept: Use \(x = r\cos\theta\), \(y = r\sin\theta\) to go from polar to Cartesian, and \(r = \sqrt{x^2+y^2}\), \(\tan\theta = y/x\) to go the other way.

(a) Polar → Cartesian for \(A\!\left(\sqrt{2},\,\pi/4\right)\):

1. Compute \(x\): \(x = r\cos\theta = \sqrt{2}\cdot\cos\!\left(\dfrac{\pi}{4}\right) = \sqrt{2}\cdot\dfrac{\sqrt{2}}{2} = 1\)
2. Compute \(y\): \(y = r\sin\theta = \sqrt{2}\cdot\sin\!\left(\dfrac{\pi}{4}\right) = \sqrt{2}\cdot\dfrac{\sqrt{2}}{2} = 1\)

Answer (a): \(A(1,\,1)\) in Cartesian coordinates.

(b) Cartesian → Polar for \(B(\sqrt{3},\,-1)\):

1. Compute \(r\): \(r = \sqrt{(\sqrt{3})^2+(-1)^2} = \sqrt{3+1} = 2\)
2. Compute \(\theta\): \(\tan\theta = \dfrac{-1}{\sqrt{3}}\), so the reference angle is \(\pi/6\). Since \(x > 0\) and \(y < 0\), the point is in the fourth quadrant, so \(\theta = 2\pi - \pi/6 = \dfrac{5\pi}{3}\).

Answer (b): \(B\!\left(2,\,\dfrac{5\pi}{3}\right)\) in polar coordinates.

4.22. Converting a Polar Equation to Cartesian Form (Chapter 3, Example 2)

Show that \(r = 2\cos\theta\) represents a circle, and identify its center and radius.

Click to see the solution

Key Concept: Multiply both sides by \(r\) to use the identity \(r^2 = x^2 + y^2\) and \(r\cos\theta = x\).

1. Multiply both sides by \(r\): \(r^2 = 2r\cos\theta\)
2. Substitute \(r^2 = x^2+y^2\) and \(r\cos\theta = x\): \(x^2+y^2 = 2x\)
3. Complete the square: \(x^2 - 2x + 1 + y^2 = 1 \implies (x-1)^2 + y^2 = 1\)

Answer: The equation \(r = 2\cos\theta\) represents a circle of radius \(1\) centered at \((1,\,0)\).

4.23. Converting an Ellipse to Polar Form (Chapter 3, Example 3)

Find the polar equation of the ellipse \(\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1\).

Click to see the solution

1. Substitute \(x = r\cos\theta\) and \(y = r\sin\theta\): \[\frac{r^2\cos^2\theta}{9} + \frac{r^2\sin^2\theta}{4} = 1\]
2. Factor out \(r^2\): \[r^2\!\left(\frac{\cos^2\theta}{9} + \frac{\sin^2\theta}{4}\right) = 1\]
3. Find a common denominator inside the parentheses: \[r^2 \cdot \frac{4\cos^2\theta + 9\sin^2\theta}{36} = 1 \implies r^2(4\cos^2\theta + 9\sin^2\theta) = 36\]

Answer: \(r^2(4\cos^2\theta + 9\sin^2\theta) = 36\).

4.24. Parametric Equations for a Line in Space (Chapter 3, Example 4)

Find the parametric equations for the line through \(A(4,2,-1)\) and \(B(0,1,-4)\). Also write the parametric equations for the line segment from \(A\) to \(B\).

Click to see the solution

Key Concept: A direction vector for the line is \(\vec{AB}\). For the segment, \(t\) is restricted to \([0,1]\).

1. Find the direction vector: \(\vec{AB} = B - A = (0-4)\hat{i}+(1-2)\hat{j}+(-4+1)\hat{k} = -4\hat{i}-\hat{j}-3\hat{k}\)

2. Write the parametric equations using point \(A\) and direction \(\vec{AB}\): \[x = 4 - 4t, \quad y = 2 - t, \quad z = -1 - 3t, \quad t \in \mathbb{R}.\]

3. For the line segment (same equations, restricted domain): \[x = 4 - 4t, \quad y = 2 - t, \quad z = -1 - 3t, \quad t \in [0,1].\]

   (At \(t=0\) we get point \(A\); at \(t=1\) we get point \(B\).)

Answer: Line: \(x = 4-4t,\; y = 2-t,\; z = -1-3t\) for \(t \in \mathbb{R}\). Segment: same with \(t \in [0,1]\).

4.25. Equation of a Plane Through Three Points (Chapter 3, Example 5)

Find an equation for the plane through \(A(1,2,3)\), \(B(-1,5,2)\), and \(C(0,3,2)\).

Click to see the solution

Key Concept: The normal vector \(\mathbf{n}\) to the plane equals \(\vec{AB} \times \vec{AC}\).

1. Find two vectors in the plane: \[\vec{AB} = (-2,3,-1), \quad \vec{AC} = (-1,1,-1).\]
2. Compute the cross product: \[\mathbf{n} = \vec{AB}\times\vec{AC} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-2&3&-1\\-1&1&-1\end{vmatrix} = \hat{i}(3\cdot(-1)-(-1)\cdot 1) - \hat{j}((-2)(-1)-(-1)(-1)) + \hat{k}((-2)(1)-3(-1))\] \[= \hat{i}(-3+1) - \hat{j}(2-1) + \hat{k}(-2+3) = -2\hat{i} - \hat{j} + \hat{k}.\]
3. Write the plane equation using point \(A(1,2,3)\) and \(\mathbf{n} = (-2,-1,1)\): \[-2(x-1) - (y-2) + (z-3) = 0 \implies -2x - y + z + 1 = 0.\]

Answer: \(-2x - y + z + 1 = 0\).

4.26. Classifying a Quadric Surface (Chapter 3, Example 6)

Classify the surface \(4x^2 + 4y^2 + z^2 + 8y - 4z = -4\) and identify its center and semi-axes.

Click to see the solution

Key Concept: Complete the square in each variable, then match to the standard form.

1. Group by variable and complete the square in \(y\): \[4x^2 + 4(y^2 + 2y) + (z^2 - 4z) = -4\] \[4x^2 + 4(y+1)^2 - 4 + (z-2)^2 - 4 = -4\] \[4x^2 + 4(y+1)^2 + (z-2)^2 = 4.\]
2. Divide through by 4: \[x^2 + (y+1)^2 + \frac{(z-2)^2}{4} = 1.\]
3. Identify: This matches \(\dfrac{x^2}{1^2} + \dfrac{(y+1)^2}{1^2} + \dfrac{(z-2)^2}{2^2} = 1\) — an ellipsoid centered at \((0,-1,2)\) with semi-axes \(a = 1\), \(b = 1\), \(c = 2\).

Answer: Ellipsoid centered at \((0,-1,2)\) with semi-axes \(1\), \(1\), and \(2\).

4.27. Double Integral Over a Rectangle (Chapter 3, Example 7)

Calculate \(I = \displaystyle\iint_R xye^{xy^2}\,dA\), where \(R: 0 \leq x \leq 2,\; 0 \leq y \leq 1\).

Click to see the solution

Key Concept: Integrate first with respect to \(y\) (inner), using substitution \(u = xy^2\).

1. Set up the iterated integral (integrate \(y\) first): \[I = \int_0^2 x\left(\int_0^1 ye^{xy^2}\,dy\right)dx.\]
2. Inner integral: Let \(u = xy^2\), \(du = 2xy\,dy\), so \(y\,dy = \dfrac{du}{2x}\): \[\int_0^1 ye^{xy^2}\,dy = \frac{1}{2x}\int_0^x e^u\,du = \frac{1}{2x}\left[e^{xy^2}\right]_0^1 = \frac{e^x - 1}{2x}.\]
3. Outer integral: \[I = \int_0^2 x \cdot \frac{e^x - 1}{2x}\,dx = \frac{1}{2}\int_0^2 (e^x - 1)\,dx = \frac{1}{2}\left[e^x - x\right]_0^2 = \frac{1}{2}\left[(e^2 - 2) - (1 - 0)\right].\]

Answer: \(I = \dfrac{e^2 - 3}{2}\).

4.28. Volume Under an Elliptical Paraboloid (Chapter 3, Example 8)

Find the volume of the region bounded above by \(z = 10 + x^2 + 3y^2\) and below by \(R: 0 \leq x \leq 1,\; 0 \leq y \leq 2\).

Click to see the solution

1. Set up as a double integral: \[V = \iint_R (10 + x^2 + 3y^2)\,dA = \int_0^1\int_0^2 (10+x^2+3y^2)\,dy\,dx.\]
2. Inner integral (with respect to \(y\)): \[\int_0^2 (10+x^2+3y^2)\,dy = \left[10y + x^2 y + y^3\right]_0^2 = 20 + 2x^2 + 8 = 28 + 2x^2.\]
3. Outer integral: \[V = \int_0^1 (28 + 2x^2)\,dx = \left[28x + \frac{2x^3}{3}\right]_0^1 = 28 + \frac{2}{3} = \frac{86}{3}.\]

Answer: \(V = \dfrac{86}{3}\).

4.29. Reversing the Order of Integration (Chapter 3, Example 9)

Calculate \(I = \displaystyle\iint_R \frac{\sin x}{x}\,dA\), where \(R\) is the triangle bounded by the \(x\)-axis, \(y = x\), and \(x = 1\).

Click to see the solution

Key Concept: The integrand \((\sin x)/x\) has no elementary antiderivative with respect to \(x\). Switch to integrating \(y\) first (vertically simple).

1. Describe \(R\) as a vertically simple region: For fixed \(x \in [0,1]\), \(y\) ranges from \(0\) to \(x\). So: \[I = \int_0^1\int_0^x \frac{\sin x}{x}\,dy\,dx.\]
2. Inner integral (the integrand does not depend on \(y\)): \[\int_0^x \frac{\sin x}{x}\,dy = \frac{\sin x}{x}\cdot x = \sin x.\]
3. Outer integral: \[I = \int_0^1 \sin x\,dx = [-\cos x]_0^1 = -\cos 1 + 1 = 1 - \cos 1.\]

Answer: \(I = 1 - \cos 1\).

4.30. Volume of a Tetrahedron (Chapter 3, Example 10)

Find the volume of the tetrahedron bounded by the coordinate planes and the plane \(z = 4 - 4x - 2y\).

Click to see the solution

Key Concept: The plane \(z = 4-4x-2y\) is non-negative on \(R\), so volume \(= \iint_R (4-4x-2y)\,dA\).

1. Identify the region \(R\): The plane intersects the \(xy\)-plane when \(z=0\): \(4-4x-2y=0 \implies y = 2-2x\). The region \(R\) is the triangle with vertices \((0,0)\), \((1,0)\), and \((0,2)\), bounded by \(x=0\), \(y=0\), and \(y=2-2x\).
2. Set up the iterated integral: \[V = \int_0^1\int_0^{2-2x} (4-4x-2y)\,dy\,dx.\]
3. Inner integral: \[\int_0^{2-2x}(4-4x-2y)\,dy = \left[(4-4x)y - y^2\right]_0^{2-2x} = (4-4x)(2-2x) - (2-2x)^2\] \[= (2-2x)\left[(4-4x) - (2-2x)\right] = (2-2x)(2-2x) = (2-2x)^2 = 4(1-x)^2.\]
4. Outer integral: \[V = \int_0^1 4(1-x)^2\,dx = 4\left[-\frac{(1-x)^3}{3}\right]_0^1 = 4\cdot\frac{1}{3} = \frac{4}{3}.\]

Answer: \(V = \dfrac{4}{3}\).

4.31. Volume Under a Paraboloid Over a Circular Region (Chapter 3, Example 11)

Find the volume of the solid bounded by the cylinder \(x^2+y^2 = 4\), the plane \(y+z=4\), and the plane \(z=0\).

Click to see the solution

Key Concept: The solid is bounded above by \(z = 4-y\) and below by \(z=0\) over the disk \(R: x^2+y^2\leq 4\).

1. Set up the integral: \[V = \iint_{x^2+y^2\leq 4}(4-y)\,dA = \int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}(4-y)\,dy\,dx.\]
2. Inner integral: \[\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}(4-y)\,dy = \left[4y - \frac{y^2}{2}\right]_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} = 8\sqrt{4-x^2}.\] (The \(y^2/2\) terms cancel since \((\sqrt{4-x^2})^2 = (-\sqrt{4-x^2})^2\).)
3. Outer integral: \[V = \int_{-2}^2 8\sqrt{4-x^2}\,dx = 8\cdot\frac{\pi\cdot 4}{2} = 16\pi.\] (We used \(\int_{-2}^2\sqrt{4-x^2}\,dx = \pi\cdot 2^2/2 = 2\pi\), the area of a semicircle of radius 2.)

Answer: \(V = 16\pi\).

4.32. Change of Variables: Hyperbolic Region (Chapter 3, Example 12)

Calculate \(I_1 = \displaystyle\iint_R e^{xy}\,dA\), where \(R\) is the region enclosed by \(y = \tfrac{1}{2}x\), \(y = x\), \(xy = 1\), and \(xy = 2\).

Click to see the solution

Key Concept: The boundary curves suggest the substitution \(u = y/x\), \(v = xy\).

1. Choose the transformation: Set \(u = y/x\) and \(v = xy\). Then: \[x = \sqrt{v/u}, \quad y = \sqrt{uv}.\] The boundary becomes: \(u = 1/2\), \(u = 1\), \(v = 1\), \(v = 2\) — a rectangle in \(uv\)-space!
2. Compute the Jacobian: \[J = \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix}\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}\\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}\end{vmatrix} = \begin{vmatrix}-\frac{1}{2}\sqrt{v/u^3}&\frac{1}{2\sqrt{uv}}\\\frac{1}{2}\sqrt{v/u}&\frac{1}{2}\sqrt{u/v}\end{vmatrix} = -\frac{1}{2u}.\] So \(|J| = \dfrac{1}{2u}\).
3. Transform the integral: Note \(xy = v\), so \(e^{xy} = e^v\): \[I_1 = \int_1^2\int_{1/2}^1 e^v\cdot\frac{1}{2u}\,du\,dv = \frac{1}{2}\int_1^2 e^v\,dv \cdot \int_{1/2}^1\frac{du}{u}.\]
4. Evaluate: \[= \frac{1}{2}(e^2 - e)\cdot[\ln u]_{1/2}^1 = \frac{1}{2}(e^2-e)\cdot(0-\ln\tfrac{1}{2}) = \frac{1}{2}(e^2-e)\ln 2.\]

Answer: \(I_1 = \dfrac{1}{2}(e^2 - e)\ln 2\).

4.33. Change of Variables: Linear Transformation (Chapter 3, Example 13)

Calculate \(I_2 = \displaystyle\iint_R \frac{x-y}{x+y}\,dA\), where \(R\) is the region enclosed by \(x-y=0\), \(x-y=1\), \(x+y=1\), and \(x+y=3\).

Click to see the solution

1. Choose the transformation: Let \(u = x+y\) and \(v = x-y\). Then \(x = \tfrac{1}{2}(u+v)\), \(y = \tfrac{1}{2}(u-v)\). The new region \(G\) is the rectangle \(1\leq u\leq 3\), \(0\leq v\leq 1\).
2. Compute the Jacobian: \(J = \begin{vmatrix}1/2&1/2\\1/2&-1/2\end{vmatrix} = -\tfrac{1}{2}\), so \(|J| = \tfrac{1}{2}\).
3. Transform: The integrand \(\dfrac{x-y}{x+y} = \dfrac{v}{u}\): \[I_2 = \int_0^1\int_1^3 \frac{v}{u}\cdot\frac{1}{2}\,du\,dv = \frac{1}{2}\int_0^1 v\,dv\cdot\int_1^3\frac{du}{u}.\]
4. Evaluate: \[= \frac{1}{2}\cdot\frac{1}{2}\cdot\ln 3 = \frac{\ln 3}{4}.\]

Answer: \(I_2 = \dfrac{\ln 3}{4}\).

4.34. Polar Coordinates: Semicircle (Chapter 3, Example 14)

Calculate \(I_1 = \displaystyle\iint_R e^{x^2+y^2}\,dA\), where \(R\) is the semicircular region \(y \geq 0\), \(x^2+y^2 \leq 1\).

Click to see the solution

Key Concept: The integrand contains \(x^2+y^2 = r^2\), and the region is a semicircle — perfect for polar coordinates.

1. Describe \(R\) in polar coordinates: \(0 \leq r \leq 1\), \(0 \leq \theta \leq \pi\).
2. Convert the integral: \[I_1 = \int_0^\pi\int_0^1 e^{r^2}\cdot r\,dr\,d\theta.\]
3. Inner integral: Let \(u = r^2\), \(du = 2r\,dr\): \[\int_0^1 re^{r^2}\,dr = \frac{1}{2}\left[e^{r^2}\right]_0^1 = \frac{e-1}{2}.\]
4. Outer integral: \[I_1 = \int_0^\pi\frac{e-1}{2}\,d\theta = \frac{(e-1)\pi}{2}.\]

Answer: \(I_1 = \dfrac{\pi(e-1)}{2}\).

4.35. Area of a Lemniscate (Chapter 3, Example 15)

Find the area enclosed by the lemniscate \(r^2 = 4\cos 2\theta\).

Click to see the solution

Key Concept: The lemniscate has two symmetric loops. We compute the area of one loop in the first quadrant and multiply by 4. The first-quadrant loop exists for \(0 \leq \theta \leq \pi/4\) (where \(\cos 2\theta \geq 0\)).

1. Set up: \[A = 4\int_0^{\pi/4}\int_0^{\sqrt{4\cos 2\theta}} r\,dr\,d\theta = 4\int_0^{\pi/4}\frac{r^2}{2}\bigg|_0^{\sqrt{4\cos 2\theta}}d\theta = 4\int_0^{\pi/4} 2\cos 2\theta\,d\theta.\]
2. Evaluate: \[A = 8\int_0^{\pi/4}\cos 2\theta\,d\theta = 8\cdot\frac{\sin 2\theta}{2}\bigg|_0^{\pi/4} = 4[\sin(\pi/2) - \sin 0] = 4.\]

Answer: \(A = 4\).

4.36. Area Between a Cardioid and Circle (Chapter 3, Example 16)

Find the area of the region in the right-half plane between the cardioid \(r = 1+\cos\theta\) and the unit circle \(r = 1\).

Click to see the solution

Key Concept: In the right-half plane \(\theta \in [-\pi/2, \pi/2]\), the cardioid lies outside the unit circle (since \(1+\cos\theta \geq 1\) for \(|\theta| \leq \pi/2\)). The area is the region between \(r=1\) (inner) and \(r=1+\cos\theta\) (outer).

1. Set up: \[A = \int_{-\pi/2}^{\pi/2}\int_1^{1+\cos\theta} r\,dr\,d\theta = \int_{-\pi/2}^{\pi/2}\frac{(1+\cos\theta)^2 - 1}{2}\,d\theta.\]
2. Expand: \((1+\cos\theta)^2 - 1 = 2\cos\theta + \cos^2\theta\).
3. Use symmetry (the integrand is even in \(\theta\)): \[A = 2\int_0^{\pi/2}\frac{2\cos\theta+\cos^2\theta}{2}\,d\theta = \int_0^{\pi/2}\!\left(2\cos\theta + \frac{1+\cos 2\theta}{2}\right)d\theta.\]
4. Evaluate: \[= \left[2\sin\theta + \frac{\theta}{2} + \frac{\sin 2\theta}{4}\right]_0^{\pi/2} = 2 + \frac{\pi}{4} + 0.\]

Answer: \(A = 2 + \dfrac{\pi}{4}\).

4.37. Improper Double Integral (Chapter 3, Example 17)

Calculate \(I_1 = \displaystyle\int_1^\infty\int_{e^{-x}}^1 \frac{1}{x^3 y}\,dy\,dx\).

Click to see the solution

1. Inner integral (with respect to \(y\)): \[\int_{e^{-x}}^1\frac{dy}{x^3 y} = \frac{1}{x^3}[\ln y]_{e^{-x}}^1 = \frac{1}{x^3}(0-(-x)) = \frac{1}{x^2}.\]
2. Outer integral (improper): \[I_1 = \int_1^\infty\frac{dx}{x^2} = \lim_{b\to\infty}\left[-\frac{1}{x}\right]_1^b = \lim_{b\to\infty}\!\left(1 - \frac{1}{b}\right) = 1.\]

Answer: \(I_1 = 1\).

4.38. The Gaussian Integral (Chapter 3, Example 18)

Using a double integral in polar coordinates, prove that \(\displaystyle\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}\).

Click to see the solution

Key Concept: Square the integral \(I\) to create a double integral over \(\mathbb{R}^2\), then convert to polar.

1. Square the integral: \[I^2 = \left(\int_{-\infty}^\infty e^{-x^2}\,dx\right)\!\left(\int_{-\infty}^\infty e^{-y^2}\,dy\right) = \iint_{\mathbb{R}^2} e^{-(x^2+y^2)}\,dx\,dy.\]
2. Convert to polar (\(x^2+y^2 = r^2\), \(r\in[0,\infty)\), \(\theta\in[0,2\pi)\)): \[I^2 = \int_0^{2\pi}\int_0^\infty e^{-r^2}\,r\,dr\,d\theta.\]
3. Inner integral (let \(u = r^2\)): \[\int_0^\infty re^{-r^2}\,dr = \frac{1}{2}\left[-e^{-r^2}\right]_0^\infty = \frac{1}{2}.\]
4. Outer integral: \[I^2 = \int_0^{2\pi}\frac{1}{2}\,d\theta = \pi.\]
5. Conclude: Since \(I > 0\), we get \(I = \sqrt{\pi}\).

Answer: \(\displaystyle\int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\pi}\).

4.39. Average Value of a Triple Integral (Chapter 3, Example 19)

Find the average value of \(F(x,y,z) = xyz\) over the box \(D = [0,2]\times[0,2]\times[0,2]\).

Click to see the solution

1. Volume of \(D\): \(\text{vol}(D) = 2\cdot 2\cdot 2 = 8\).
2. Compute the triple integral (the integrand separates into a product): \[\iiint_D xyz\,dV = \int_0^2 x\,dx\cdot\int_0^2 y\,dy\cdot\int_0^2 z\,dz = \left[\frac{x^2}{2}\right]_0^2\!\cdot\!\left[\frac{y^2}{2}\right]_0^2\!\cdot\!\left[\frac{z^2}{2}\right]_0^2 = 2\cdot 2\cdot 2 = 8.\]
3. Average value: \[\text{aver}_D(F) = \frac{1}{8}\cdot 8 = 1.\]

Answer: The average value is \(1\).

4.40. Volume Between Two Paraboloids (Chapter 3, Example 20)

Find the volume of the solid \(D\) enclosed by \(z = x^2 + 3y^2\) (bottom) and \(z = 8 - x^2 - y^2\) (top).

Click to see the solution

Key Concept: Find where the surfaces intersect to determine the projection \(R\), then integrate \(dz\).

1. Find intersection: Set \(x^2+3y^2 = 8-x^2-y^2\): \(2x^2+4y^2 = 8\), i.e., \(x^2+2y^2 = 4\) (an ellipse). This is the boundary of the projection \(R\) onto the \(xy\)-plane.
2. Set up the integral: \[\text{Vol} = \iint_R (8-x^2-y^2 - x^2-3y^2)\,dA = \iint_R (8-2x^2-4y^2)\,dA.\] The region \(R\) is the ellipse \(x^2+2y^2\leq 4\), so \(-2\leq x\leq 2\) and \(-\sqrt{(4-x^2)/2}\leq y\leq \sqrt{(4-x^2)/2}\).
3. Evaluate (after integration — details involve completing the square and using the ellipse area formula): \[\text{Vol} = 8\sqrt{2}\,\pi.\]

Answer: \(\text{Vol} = 8\sqrt{2}\,\pi\).

4.41. Triple Integral by General Transformation (Chapter 3, Example 21)

Calculate \(I = \displaystyle\int_0^3\int_0^4\int_{y/2}^{y/2+1}\!\left(\frac{2x-y}{2}+\frac{z}{3}\right)dx\,dy\,dz\) using the substitution \(u = \dfrac{2x-y}{2}\), \(v = \dfrac{y}{2}\), \(w = \dfrac{z}{3}\).

Click to see the solution

1. Find the inverse transformation: \(x = u+v\), \(y = 2v\), \(z = 3w\).
2. Find the new limits: The original limits \(x\in[y/2,\,y/2+1]\) become \(u\in[0,1]\); \(y\in[0,4]\) becomes \(v\in[0,2]\); \(z\in[0,3]\) becomes \(w\in[0,1]\).
3. Compute the Jacobian: \[J = \frac{\partial(x,y,z)}{\partial(u,v,w)} = \begin{vmatrix}1&1&0\\0&2&0\\0&0&3\end{vmatrix} = 1\cdot(2\cdot 3) = 6.\]
4. The integrand becomes \(u + w\) (since \(\dfrac{2x-y}{2} = u\) and \(\dfrac{z}{3} = w\)): \[I = \int_0^1\int_0^2\int_0^1 (u+w)\cdot 6\,du\,dv\,dw.\]
5. Evaluate: \[= 6\int_0^1\int_0^2\left[\frac{u^2}{2}+uw\right]_0^1 dv\,dw = 6\int_0^1\int_0^2\!\left(\frac{1}{2}+w\right)dv\,dw\] \[= 6\int_0^1\left(\frac{v}{2}+vw\right)_0^2 dw = 6\int_0^1(1+2w)\,dw = 6\left[w+w^2\right]_0^1 = 6\cdot 2 = 12.\]

Answer: \(I = 12\).

4.42. Triple Integral in Cylindrical Coordinates (Chapter 3, Example 22)

Calculate \(I = \displaystyle\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{\sqrt{x^2+y^2}}^2 (x^2+y^2)\,dz\,dy\,dx\).

Click to see the solution

Key Concept: The region is bounded below by the cone \(z = \sqrt{x^2+y^2}\) and above by \(z=2\) over the disk \(x^2+y^2\leq 4\). The integrand \(x^2+y^2 = r^2\) — ideal for cylindrical coordinates.

1. Describe \(D\) in cylindrical coordinates: \(0\leq\theta\leq 2\pi\), \(0\leq r\leq 2\), \(r\leq z\leq 2\).
2. Convert: \(x^2+y^2 = r^2\), \(dV = r\,dr\,d\theta\,dz\): \[I = \int_0^{2\pi}\int_0^2\int_r^2 r^2\cdot r\,dz\,dr\,d\theta = \int_0^{2\pi}d\theta\int_0^2 r^3(2-r)\,dr.\]
3. Radial integral: \[\int_0^2 r^3(2-r)\,dr = \int_0^2(2r^3 - r^4)\,dr = \left[\frac{r^4}{2}-\frac{r^5}{5}\right]_0^2 = 8 - \frac{32}{5} = \frac{8}{5}.\]
4. Full integral: \[I = 2\pi\cdot\frac{8}{5} = \frac{16\pi}{5}.\]

Answer: \(I = \dfrac{16\pi}{5}\).

4.43. Volume of an Ice Cream Cone in Spherical Coordinates (Chapter 3, Example 23)

Find the volume of the solid \(D\) bounded above by the sphere \(\rho = 1\) and below by the cone \(\phi = \dfrac{\pi}{3}\).

Click to see the solution

Key Concept: In spherical coordinates, the region is \(0\leq\rho\leq 1\), \(0\leq\phi\leq\pi/3\), \(0\leq\theta\leq 2\pi\) — a simple box in spherical space.

1. Set up the integral: \[V = \int_0^{2\pi}\int_0^{\pi/3}\int_0^1 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta.\]
2. Innermost integral (\(\rho\)): \[\int_0^1\rho^2\,d\rho = \frac{1}{3}.\]
3. Middle integral (\(\phi\)): \[\int_0^{\pi/3}\frac{\sin\phi}{3}\,d\phi = \frac{1}{3}\left[-\cos\phi\right]_0^{\pi/3} = \frac{1}{3}\!\left(-\frac{1}{2}+1\right) = \frac{1}{6}.\]
4. Outer integral (\(\theta\)): \[V = \int_0^{2\pi}\frac{1}{6}\,d\theta = \frac{2\pi}{6} = \frac{\pi}{3}.\]

Answer: \(V = \dfrac{\pi}{3}\).

4.44. Plane Parallel to a Given Plane (Chapter 3, Example 24)

Find an equation for the plane through \((1, -1, 3)\) parallel to the plane \(3x + y + z = 7\).

Click to see the solution

Key Concept: Parallel planes share the same normal vector. The plane \(3x + y + z = 7\) has normal \(\mathbf{n} = (3, 1, 1)\).

1. Use the same normal \(\mathbf{n} = (3, 1, 1)\) and the given point \((1, -1, 3)\): \[3(x - 1) + 1(y + 1) + 1(z - 3) = 0\] \[3x + y + z - 3 + 1 - 3 = 0 \implies 3x + y + z - 5 = 0.\]

Answer: \(3x + y + z - 5 = 0\).

4.45. Plane Perpendicular to Another Plane (Chapter 3, Example 25)

Find an equation for the plane through \(P_1(1, 2, 3)\) and \(P_2(3, 2, 1)\) and perpendicular to the plane \(4x - y + 2z = 7\).

Click to see the solution

Key Concept: The desired plane contains vector \(\vec{P_1P_2}\) and is perpendicular to the given plane, so its normal \(\mathbf{n}\) must be perpendicular to both \(\vec{P_1P_2}\) and the normal of the given plane.

1. Direction vector of the line: \(\vec{P_1P_2} = (2, 0, -2)\).
2. Normal of the given plane: \(\mathbf{n}_1 = (4, -1, 2)\).
3. Normal of desired plane \(= \mathbf{n}_1 \times \vec{P_1P_2}\): \[\mathbf{n} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\4&-1&2\\2&0&-2\end{vmatrix} = \hat{i}((-1)(-2)-2\cdot0) - \hat{j}(4(-2)-2\cdot2) + \hat{k}(4\cdot0-(-1)\cdot2)\] \[= 2\hat{i} + 12\hat{j} + 2\hat{k}.\]
4. Plane equation using point \(P_1(1,2,3)\) and \(\mathbf{n} = (2, 12, 2)\) (divide by 2: \((1,6,1)\)): \[(x-1) + 6(y-2) + (z-3) = 0 \implies x + 6y + z - 16 = 0.\]

Answer: \(x + 6y + z - 16 = 0\).

4.46. Improper Double Integral with Exponential (Chapter 3, Example 26)

Calculate \(I_2 = \displaystyle\int_0^\infty\int_0^\infty xe^{-(x+2y)}\,dx\,dy\).

Click to see the solution

1. Integrate \(y\) first (inner integral): \[\int_0^\infty e^{-(x+2y)}\,dy = e^{-x}\int_0^\infty e^{-2y}\,dy = e^{-x}\cdot\frac{1}{2}.\]
2. Outer integral: \[I_2 = \int_0^\infty x\cdot\frac{e^{-x}}{2}\,dx = \frac{1}{2}\int_0^\infty xe^{-x}\,dx.\]
3. Integration by parts (\(u=x\), \(dv=e^{-x}dx\)): \[\int_0^\infty xe^{-x}\,dx = [-xe^{-x}]_0^\infty + \int_0^\infty e^{-x}\,dx = 0 + 1 = 1.\]
4. Result: \(I_2 = \dfrac{1}{2}\).

Answer: \(I_2 = \dfrac{1}{2}\).

4.47. Improper Double Integral with Singularity (Chapter 3, Example 27)

Calculate \(I_3 = \displaystyle\int_0^1\int_0^3\frac{x^2}{(y-1)^{2/3}}\,dy\,dx\).

Click to see the solution

Key Concept: The integrand has a singularity at \(y=1\in[0,3]\), making this an improper integral. Switch the order to integrate \(x\) first.

1. Integrate \(x\) first: \[I_3 = \int_0^3\frac{1}{(y-1)^{2/3}}\left(\int_0^1 x^2\,dx\right)dy = \frac{1}{3}\int_0^3(y-1)^{-2/3}\,dy.\]
2. Split at the singularity \(y=1\): \[= \frac{1}{3}\left(\lim_{\varepsilon\to0^+}\int_0^{1-\varepsilon}(y-1)^{-2/3}\,dy + \lim_{\delta\to0^+}\int_{1+\delta}^3(y-1)^{-2/3}\,dy\right).\]
3. Antiderivative of \((y-1)^{-2/3}\) is \(3(y-1)^{1/3}\): \[\lim_{\varepsilon\to0^+}\left[3(y-1)^{1/3}\right]_0^{1-\varepsilon} = \lim_{\varepsilon\to0^+}(3(-\varepsilon)^{1/3}+3) = 3,\] \[\lim_{\delta\to0^+}\left[3(y-1)^{1/3}\right]_{1+\delta}^3 = 3\sqrt[3]{2}.\]
4. Result: \[I_3 = \frac{1}{3}\left(3 + 3\sqrt[3]{2}\right) = 1+\sqrt[3]{2}.\]

Answer: \(I_3 = 1 + \sqrt[3]{2}\).

4.48. Convert to Cartesian Polar Coordinates (Chapter 3, Task 1)

Find the Cartesian coordinates of the following points given in polar coordinates: \((4,\,4\pi/3)\), \((2,\,2\pi/3)\), \((1,\,\pi)\), \((-\sqrt{2},\,\pi/4)\).

Click to see the solution

Key Concept: Use \(x = r\cos\theta\), \(y = r\sin\theta\). Note: negative \(r\) means the point is in the opposite direction.

(a) \((4,\,4\pi/3)\): \(x = 4\cos(4\pi/3) = 4(-1/2) = -2\), \(y = 4\sin(4\pi/3) = 4(-\sqrt{3}/2) = -2\sqrt{3}\). Point: \((-2,\,-2\sqrt{3})\).

(b) \((2,\,2\pi/3)\): \(x = 2\cos(2\pi/3) = 2(-1/2) = -1\), \(y = 2\sin(2\pi/3) = 2(\sqrt{3}/2) = \sqrt{3}\). Point: \((-1,\,\sqrt{3})\).

(c) \((1,\,\pi)\): \(x = 1\cdot\cos\pi = -1\), \(y = 1\cdot\sin\pi = 0\). Point: \((-1,\,0)\).

(d) \((-\sqrt{2},\,\pi/4)\): A negative \(r\) means we reflect through the origin: actual point is at \(r = \sqrt{2}\), \(\theta = \pi/4 + \pi = 5\pi/4\). \(x = \sqrt{2}\cos(5\pi/4) = \sqrt{2}(-\sqrt{2}/2) = -1\), \(y = \sqrt{2}\sin(5\pi/4) = -1\). Point: \((-1,\,-1)\).

Answer: \((-2,-2\sqrt{3})\); \((-1,\sqrt{3})\); \((-1,0)\); \((-1,-1)\).

4.49. Reduce to Standard Form and Classify Quadric Surfaces (Chapter 3, Task 2)

Classify each surface by reducing to standard form:

(a) \(4x^2 - y + 2z^2 = 0\)

(b) \(y^2 = x^2 + 4z^2 + 4\)

(c) \(x^2 + y^2 - 2x - 6y - z + 10 = 0\)

Click to see the solution

(a) \(4x^2 - y + 2z^2 = 0\): Rearrange: \(y = 4x^2 + 2z^2\), i.e., \(\dfrac{x^2}{1/4} + \dfrac{z^2}{1/2} = \dfrac{y}{1}\). This matches \(\dfrac{x^2}{a^2}+\dfrac{z^2}{b^2} = \dfrac{y}{c}\). Elliptical paraboloid opening in the positive \(y\)-direction.

(b) \(y^2 = x^2 + 4z^2 + 4\): Rearrange: \(y^2 - x^2 - 4z^2 = 4\), i.e., \(\dfrac{y^2}{4} - \dfrac{x^2}{4} - z^2 = 1\). This matches \(\dfrac{y^2}{c^2} - \dfrac{x^2}{a^2} - \dfrac{z^2}{b^2} = 1\). Hyperboloid of two sheets (along the \(y\)-axis).

(c) \(x^2 + y^2 - 2x - 6y - z + 10 = 0\): Complete the square: \((x-1)^2 - 1 + (y-3)^2 - 9 - z + 10 = 0\), so \(z = (x-1)^2 + (y-3)^2\). Elliptical (circular) paraboloid with vertex at \((1,3,0)\), opening upward.

Answer: (a) Elliptical paraboloid; (b) Hyperboloid of two sheets; (c) Circular paraboloid with vertex \((1,3,0)\).

4.50. Double Integrals Over a Rectangle (Chapter 3, Task 3)

Calculate the following integrals:

(a) \(\displaystyle\iint_R y\sin(x+y)\,dA\), \(R: -\pi\leq x\leq 0,\; 0\leq y\leq\pi\)

(b) \(\displaystyle\iint_R \frac{xy^3}{1+x^2}\,dA\), \(R: 0\leq x\leq 1,\; 0\leq y\leq 2\)

Click to see the solution

(a) Integrate \(x\) first using integration by parts:

\[I = \int_0^\pi y\left(\int_{-\pi}^0 \sin(x+y)\,dx\right)dy = \int_0^\pi y\left[-\cos(x+y)\right]_{-\pi}^0 dy\] \[= \int_0^\pi y(-\cos y + \cos(y-\pi))\,dy = \int_0^\pi y(-\cos y - \cos y)\,dy = -2\int_0^\pi y\cos y\,dy.\]

Integration by parts (\(u=y\), \(dv=\cos y\,dy\)): \[-2\left([y\sin y]_0^\pi - \int_0^\pi\sin y\,dy\right) = -2\left(0 + [-\cos y]_0^\pi\right) = -2(1-(-1)) = \boxed{-4}.\]

(b) The integrand factors: \(\dfrac{xy^3}{1+x^2}= \dfrac{x}{1+x^2}\cdot y^3\).

\[I = \int_0^1\frac{x}{1+x^2}\,dx\cdot\int_0^2 y^3\,dy = \left[\frac{\ln(1+x^2)}{2}\right]_0^1\cdot\left[\frac{y^4}{4}\right]_0^2 = \frac{\ln 2}{2}\cdot 4 = \boxed{2\ln 2}.\]

Answer: (a) \(-4\); (b) \(2\ln 2\).

4.51. Volume Under a Plane (Chapter 3, Task 4)

Find the volume of the region bounded above by \(z = 2 - x - y\) and below by the square \([0,1]\times[0,1]\).

Click to see the solution

1. Set up: \[V = \int_0^1\int_0^1(2-x-y)\,dy\,dx.\]
2. Inner integral: \[\int_0^1(2-x-y)\,dy = \left[2y - xy - \frac{y^2}{2}\right]_0^1 = 2-x-\frac{1}{2} = \frac{3}{2}-x.\]
3. Outer integral: \[V = \int_0^1\!\left(\frac{3}{2}-x\right)dx = \left[\frac{3}{2}x - \frac{x^2}{2}\right]_0^1 = \frac{3}{2} - \frac{1}{2} = 1.\]

Answer: \(V = 1\).

4.52. Reversing Order of Integration (Chapter 3, Task 5)

Compute the following integrals by switching the order of integration:

(a) \(I_5 = \displaystyle\int_0^\pi\int_x^\pi \frac{\sin y}{y}\,dy\,dx\)

(b) \(I_7 = \displaystyle\int_0^1\int_y^1 x^2 e^{xy}\,dx\,dy\)

Click to see the solution

Key Concept: The inner integrand has no closed-form antiderivative with respect to the inner variable. Switch the order and re-describe the region.

(a) The region is \(\{(x,y): 0\leq x\leq\pi,\; x\leq y\leq\pi\} = \{(x,y): 0\leq y\leq\pi,\; 0\leq x\leq y\}\):

\[I_5 = \int_0^\pi\int_0^y\frac{\sin y}{y}\,dx\,dy = \int_0^\pi\sin y\,dy = [-\cos y]_0^\pi = 2.\]

(b) The region is \(\{(x,y): 0\leq y\leq 1,\; y\leq x\leq 1\} = \{(x,y): 0\leq x\leq 1,\; 0\leq y\leq x\}\):

\[I_7 = \int_0^1\int_0^x x^2 e^{xy}\,dy\,dx = \int_0^1 x^2\left[\frac{e^{xy}}{x}\right]_0^x dx = \int_0^1 x(e^{x^2}-1)\,dx.\] \[= \int_0^1 xe^{x^2}\,dx - \int_0^1 x\,dx = \frac{e-1}{2} - \frac{1}{2} = \frac{e-2}{2}.\]

Answer: (a) \(I_5 = 2\); (b) \(I_7 = \dfrac{e-2}{2}\).

4.53. Change of Variables: Polar, Linear (Chapter 3, Task 6)

Calculate the following integrals using an appropriate change of variables:

(a) \(\displaystyle\int_{-1}^0\int_{-\sqrt{1-x^2}}^0 \frac{2}{1+\sqrt{x^2+y^2}}\,dy\,dx\)

(b) \(\displaystyle\iint_R \frac{\sin(x-y)}{\cos(x+y)}\,dA\) where \(R\) is the triangle with \(y=0\), \(y=x\), \(x+y=\pi/4\).

Click to see the solution

(a) The region is the quarter-disk in the third quadrant (\(x\leq 0\), \(y\leq 0\)), \(x^2+y^2\leq 1\). Use polar: \(x = r\cos\theta\), \(y = r\sin\theta\), \(0\leq r\leq 1\), \(\theta\in[\pi,\,3\pi/2]\).

\[\int_\pi^{3\pi/2}\int_0^1\frac{2}{1+r}\cdot r\,dr\,d\theta = \frac{\pi}{2}\int_0^1\frac{2r}{1+r}\,dr.\] \[= \pi\int_0^1\!\left(1 - \frac{1}{1+r}\right)dr = \pi\left[r - \ln(1+r)\right]_0^1 = \pi(1-\ln 2).\]

(b) The boundary curves suggest \(u = x+y\), \(v = x-y\), so \(x = (u+v)/2\), \(y = (u-v)/2\), \(|J| = 1/2\).

The region \(R\) (triangle with \(y=0\), \(y=x\), \(x+y=\pi/4\)) maps to: \(v\) ranges from \(0\) to \(u\) (from \(y=0\) giving \(v=x=u\) to \(y=x\) giving \(v=0\)), and \(u\) from \(0\) to \(\pi/4\).

\[I = \int_0^{\pi/4}\int_0^u\frac{\sin v}{\cos u}\cdot\frac{1}{2}\,dv\,du = \frac{1}{2}\int_0^{\pi/4}\frac{[-\cos v]_0^u}{\cos u}\,du = \frac{1}{2}\int_0^{\pi/4}\frac{1-\cos u}{\cos u}\,du\] \[= \frac{1}{2}\int_0^{\pi/4}(\sec u - 1)\,du = \frac{1}{2}\left[\ln|\sec u+\tan u| - u\right]_0^{\pi/4} = \frac{1}{2}\left(\ln(1+\sqrt{2}) - \frac{\pi}{4}\right).\]

Answer: (a) \(\pi(1-\ln 2)\); (b) \(\dfrac{1}{2}\!\left(\ln(1+\sqrt{2}) - \dfrac{\pi}{4}\right)\).

4.54. Area of an Ellipse (Chapter 3, Task 7)

Show that the area of the ellipse \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} = 1\) is \(\pi ab\).

Click to see the solution

Key Concept: Use the substitution \(x = au\), \(y = bv\) to transform the ellipse into the unit disk.

1. Transformation: \(x = au\), \(y = bv\). The Jacobian is \(J = ab\).
2. The ellipse maps to the unit disk: \(u^2+v^2\leq 1\).
3. Area: \[A = \iint_{u^2+v^2\leq 1} ab\,du\,dv = ab\cdot(\text{area of unit disk}) = ab\cdot\pi = \pi ab.\]

Answer: Area \(= \pi ab\). \(\square\)

4.55. Triple Integrals by Switching Order (Chapter 3, Task 8)

Calculate \(\displaystyle\int_0^4\int_0^1\int_{2y}^2\frac{4\cos(x^2)}{2\sqrt{z}}\,dx\,dy\,dz\).

Click to see the solution

Key Concept: \(\cos(x^2)\) has no elementary antiderivative with respect to \(x\). Switch the order of \(x\) and \(y\) integration.

The inner limits \(x: 2y\to 2\) (for fixed \(y\)) describe the region \(0\leq y\leq 1\), \(2y\leq x\leq 2\). In reversed order: \(0\leq x\leq 2\), \(0\leq y\leq x/2\).

\[I = \int_0^4\frac{1}{2\sqrt{z}}\left(\int_0^2\int_0^{x/2}4\cos(x^2)\,dy\,dx\right)dz\] \[= \int_0^4\frac{1}{2\sqrt{z}}\left(\int_0^2 4\cos(x^2)\cdot\frac{x}{2}\,dx\right)dz\] \[= \int_0^4\frac{1}{2\sqrt{z}}\cdot\left[2\cdot\frac{\sin(x^2)}{2}\right]_0^2 dz = \int_0^4\frac{\sin 4}{2\sqrt{z}}\,dz\] \[= \sin 4\cdot\left[\sqrt{z}\right]_0^4 = 2\sin 4.\]

Answer: \(2\sin 4\).

4.56. Volume in Cylindrical Coordinates (Chapter 3, Task 9)

Find the volume of the solid region that lies inside the sphere \(x^2+y^2+z^2 = 2\) and outside the cylinder \(x^2+y^2 = 1\).

Click to see the solution

Key Concept: Use cylindrical coordinates. The sphere becomes \(r^2+z^2 = 2\), so \(z = \pm\sqrt{2-r^2}\). The region outside the cylinder means \(r\geq 1\); inside the sphere means \(r\leq\sqrt{2}\).

1. Set up: \[V = \int_0^{2\pi}\int_1^{\sqrt{2}}\int_{-\sqrt{2-r^2}}^{\sqrt{2-r^2}} r\,dz\,dr\,d\theta.\]
2. Inner integral (\(z\)): \[\int_{-\sqrt{2-r^2}}^{\sqrt{2-r^2}} dz = 2\sqrt{2-r^2}.\]
3. Remaining: \[V = 2\pi\int_1^{\sqrt{2}} 2r\sqrt{2-r^2}\,dr.\] Let \(u = 2-r^2\), \(du = -2r\,dr\). When \(r=1\), \(u=1\); when \(r=\sqrt{2}\), \(u=0\): \[= 4\pi\int_1^0(-\sqrt{u})\,\frac{du}{2} = 2\pi\int_0^1\sqrt{u}\,du = 2\pi\left[\frac{2}{3}u^{3/2}\right]_0^1 = \frac{4\pi}{3}.\]

Answer: \(V = \dfrac{4\pi}{3}\).

4.57. Volume in Spherical Coordinates (Chapter 3, Task 10)

Use spherical coordinates to find the volume of the solid that lies above the cone \(z = \sqrt{x^2+y^2}\) and below the sphere \(x^2+y^2+z^2 = z\).

Click to see the solution

Key Concept: First convert the surfaces to spherical coordinates.

1. Convert the sphere: \(x^2+y^2+z^2 = z\) becomes \(\rho^2 = \rho\cos\phi\), so \(\rho = \cos\phi\).
2. Convert the cone: \(z = \sqrt{x^2+y^2}\) becomes \(\rho\cos\phi = \rho\sin\phi\), so \(\tan\phi = 1\), \(\phi = \pi/4\).
3. Region: \(0\leq\rho\leq\cos\phi\), \(0\leq\phi\leq\pi/4\), \(0\leq\theta\leq 2\pi\).
4. Set up: \[V = \int_0^{2\pi}\int_0^{\pi/4}\int_0^{\cos\phi}\rho^2\sin\phi\,d\rho\,d\phi\,d\theta.\]
5. Innermost: \[\int_0^{\cos\phi}\rho^2\,d\rho = \frac{\cos^3\phi}{3}.\]
6. Middle: \[\int_0^{\pi/4}\frac{\cos^3\phi}{3}\sin\phi\,d\phi = \frac{1}{3}\int_0^{\pi/4}\cos^3\phi\sin\phi\,d\phi.\] Let \(u=\cos\phi\): \(= \dfrac{1}{3}\int_{1/\sqrt{2}}^1 u^3(-du)/(-1) = \dfrac{1}{3}\left[\dfrac{u^4}{4}\right]_{1/\sqrt{2}}^1 = \dfrac{1}{3}\cdot\dfrac{1}{4}\!\left(1-\dfrac{1}{4}\right) = \dfrac{1}{16}\).
7. Outer: \[V = 2\pi\cdot\frac{1}{16} = \frac{\pi}{8}.\]

Answer: \(V = \dfrac{\pi}{8}\).

4.58. Classify Quadric Surfaces (continued) (Chapter 3, Task 11)

Reduce to standard form and classify:

(d) \(4x^2 + y^2 + z^2 - 24x - 8y + 4z + 55 = 0\)

(e) \(x^2 - y^2 - z^2 - 4x - 2z + 3 = 0\)

Click to see the solution

(d) Complete the square: \[4(x^2 - 6x) + (y^2 - 8y) + (z^2 + 4z) = -55\] \[4(x-3)^2 - 36 + (y-4)^2 - 16 + (z+2)^2 - 4 = -55\] \[4(x-3)^2 + (y-4)^2 + (z+2)^2 = 1\] \[\frac{(x-3)^2}{1/4} + \frac{(y-4)^2}{1} + \frac{(z+2)^2}{1} = 1.\]

Ellipsoid centered at \((3, 4, -2)\) with semi-axes \(a = 1/2\), \(b = 1\), \(c = 1\).

(e) Complete the square: \[(x^2 - 4x) - y^2 - (z^2 + 2z) = -3\] \[(x-2)^2 - 4 - y^2 - (z+1)^2 + 1 = -3\] \[(x-2)^2 - y^2 - (z+1)^2 = 0\] \[\frac{(x-2)^2}{1} = \frac{y^2}{1} + \frac{(z+1)^2}{1}.\]

This matches \(\frac{x^2}{a^2} = \frac{y^2}{b^2} + \frac{z^2}{c^2}\). Elliptical cone with vertex at \((2, 0, -1)\), axis along the \(x\)-direction.

Answer: (d) Ellipsoid centered at \((3,4,-2)\); (e) Elliptical cone with vertex at \((2,0,-1)\).

4.59. Remaining Rectangle Integrals (Chapter 3, Task 12)

Calculate:

(a) \(\displaystyle\iint_R \frac{y}{1 + x^2y^2}\,dA\), \(R: 0\leq x\leq 1,\; 0\leq y\leq 1\)

(b) \(\displaystyle\iint_R \frac{\ln x}{xy}\,dA\), \(R: 1\leq x\leq e,\; 1\leq y\leq 4\)

(c) \(\displaystyle\iint_R xy\cos y\,dA\), \(R: -1\leq x\leq 1,\; 0\leq y\leq\pi\)

Click to see the solution

(a) Integrate \(y\) first. Let \(u = xy\): \[I = \int_0^1\left(\int_0^1\frac{y}{1+x^2y^2}\,dy\right)dx.\] Inner: let \(u = xy\), \(du = x\,dy\), so \(y\,dy = u\,du/x^2\)… alternatively, let \(t = x^2y^2\), \(dt = 2x^2y\,dy\): \[\int_0^1\frac{y\,dy}{1+x^2y^2} = \frac{1}{2x^2}\left[\ln(1+x^2y^2)\right]_0^1 = \frac{\ln(1+x^2)}{2x^2}.\] Outer: \[I = \int_0^1\frac{\ln(1+x^2)}{2x^2}\,dx.\] Integration by parts: \(u = \ln(1+x^2)\), \(dv = x^{-2}dx\), \(v = -1/x\), \(du = 2x/(1+x^2)\,dx\): \[= \frac{1}{2}\left[-\frac{\ln(1+x^2)}{x}\right]_0^1 + \int_0^1\frac{1}{1+x^2}\,dx = \frac{1}{2}\left(-\ln 2 + 0\right) + \frac{\pi}{4} = \frac{\pi}{4} - \frac{\ln 2}{2}.\]

(b) The integrand factors: \[I = \int_1^e\frac{\ln x}{x}\,dx\cdot\int_1^4\frac{dy}{y} = \left[\frac{(\ln x)^2}{2}\right]_1^e\cdot[\ln y]_1^4 = \frac{1}{2}\cdot\ln 4 = \ln 2.\]

(c) The integrand factors: \(xy\cos y = x\cdot y\cos y\), and \(\int_{-1}^1 x\,dx = 0\) (odd function on symmetric interval): \[I = \left(\int_{-1}^1 x\,dx\right)\cdot\left(\int_0^\pi y\cos y\,dy\right) = 0\cdot(\ldots) = 0.\]

Answer: (a) \(\dfrac{\pi}{4} - \dfrac{\ln 2}{2}\); (b) \(\ln 2\); (c) \(0\).

4.60. Volume Under a Trigonometric Surface (Chapter 3, Task 13)

Find the volume of the region bounded above by \(z = 2\sin x\cos y\) and below by \(R: 0\leq x\leq \tfrac{\pi}{2},\; 0\leq y\leq\tfrac{\pi}{4}\).

Click to see the solution

1. Set up (the integrand factors): \[V = \int_0^{\pi/2}\int_0^{\pi/4} 2\sin x\cos y\,dy\,dx = 2\int_0^{\pi/2}\sin x\,dx\cdot\int_0^{\pi/4}\cos y\,dy.\]
2. Evaluate each factor: \[\int_0^{\pi/2}\sin x\,dx = [-\cos x]_0^{\pi/2} = 1, \quad \int_0^{\pi/4}\cos y\,dy = [\sin y]_0^{\pi/4} = \frac{\sqrt{2}}{2}.\]
3. Combine: \[V = 2\cdot 1\cdot\frac{\sqrt{2}}{2} = \sqrt{2}.\]

Answer: \(V = \sqrt{2}\).

4.61. Find a Constant from a Definite Integral (Chapter 3, Task 14)

Find the value of the constant \(k\) such that \(\displaystyle\int_1^2\int_0^3 kx^2y\,dx\,dy = 1\).

Click to see the solution

1. Compute the double integral in terms of \(k\): \[\int_1^2\int_0^3 kx^2y\,dx\,dy = k\int_1^2 y\,dy\cdot\int_0^3 x^2\,dx = k\left[\frac{y^2}{2}\right]_1^2\cdot\left[\frac{x^3}{3}\right]_0^3.\]
2. Evaluate: \[= k\cdot\frac{3}{2}\cdot 9 = \frac{27k}{2}.\]
3. Set equal to 1: \(\dfrac{27k}{2} = 1 \implies k = \dfrac{2}{27}\).

Answer: \(k = \dfrac{2}{27}\).

4.62. More Integrals by Switching Order (Chapter 3, Task 15)

Compute by switching the order of integration:

(a) \(I_6 = \displaystyle\int_0^2\int_x^2 2y^2\sin(xy)\,dy\,dx\)

(b) \(I_8 = \displaystyle\int_0^2\int_0^{4-x^2}\frac{xe^{2y}}{4-y}\,dy\,dx\)

Click to see the solution

(a) Region: \(\{0\leq x\leq 2,\; x\leq y\leq 2\}\) = \(\{0\leq y\leq 2,\; 0\leq x\leq y\}\): \[I_6 = \int_0^2\int_0^y 2y^2\sin(xy)\,dx\,dy = \int_0^2 2y^2\left[-\frac{\cos(xy)}{y}\right]_0^y dy = \int_0^2 2y(1-\cos y^2)\,dy.\] \[= \left[y^2 - \sin(y^2)\right]_0^2 = 4 - \sin 4.\]

(b) Region: \(\{0\leq x\leq 2,\; 0\leq y\leq 4-x^2\}\). Rewrite: for fixed \(y\), \(x\) ranges from \(0\) to \(\sqrt{4-y}\) (since \(y\leq 4-x^2 \iff x^2\leq 4-y\)), and \(y\) from \(0\) to \(4\): \[I_8 = \int_0^4\int_0^{\sqrt{4-y}}\frac{xe^{2y}}{4-y}\,dx\,dy = \int_0^4\frac{e^{2y}}{4-y}\cdot\frac{4-y}{2}\,dy = \frac{1}{2}\int_0^4 e^{2y}\,dy = \frac{1}{4}(e^8-1).\]

Answer: (a) \(4 - \sin 4\); (b) \(\dfrac{e^8-1}{4}\).

4.63. Double Integral Over a Diamond Region (Chapter 3, Task 16)

Calculate \(\displaystyle\iint_R (y - 2x^2)\,dA\) where \(R = \{(x,y)\mid |x|+|y|\leq 1\}\).

Click to see the solution

Key Concept: The diamond \(|x|+|y|\leq 1\) is symmetric in both \(x\) and \(y\). The term \(-2x^2\) is even in \(x\) and even in \(y\); the term \(y\) is odd in \(y\) on a region symmetric in \(y\), so \(\iint_R y\,dA = 0\).

1. Split: \(\iint_R (y-2x^2)\,dA = \underbrace{\iint_R y\,dA}_{=0} - 2\iint_R x^2\,dA.\)
2. Compute \(\iint_R x^2\,dA\) using vertical cross-sections (for fixed \(x\in[-1,1]\), \(y\) ranges from \(-(1-|x|)\) to \(1-|x|\)): \[\iint_R x^2\,dA = \int_{-1}^1 x^2\cdot 2(1-|x|)\,dx = 4\int_0^1 x^2(1-x)\,dx = 4\left(\frac{1}{3}-\frac{1}{4}\right) = \frac{1}{3}.\]
3. Result: \(0 - 2\cdot\dfrac{1}{3} = -\dfrac{2}{3}\).

Answer: \(-\dfrac{2}{3}\).

4.64. Double Integral Over a Disk (Chapter 3, Task 17)

Calculate \(\displaystyle\iint_R (3x-2y)\,dA\) where \(R\) is the disk \(x^2+y^2\leq 1\).

Click to see the solution

Key Concept: Both \(3x\) and \(-2y\) are odd functions, and the disk is symmetric about both axes.

• \(\iint_R 3x\,dA = 0\) (odd in \(x\), symmetric region).
• \(\iint_R (-2y)\,dA = 0\) (odd in \(y\), symmetric region).

Answer: \(0\).

4.65. Double Integral in the First Quadrant (Chapter 3, Task 18)

Calculate \(\displaystyle\iint_R y\,dA\) where \(R\) is the region in the first quadrant enclosed between the circle \(x^2+y^2=25\) and the line \(x+y=5\).

Click to see the solution

Key Concept: Use horizontal cross-sections. In the first quadrant, the circle \(x^2+y^2=25\) bounds \(x\) up to \(\sqrt{25-y^2}\) and the line \(x+y=5\) gives \(x=5-y\). The two curves meet at \((0,5)\) and \((5,0)\).

For fixed \(y\in[0,5]\): \(x\) runs from the line \(x=5-y\) (left boundary) to the circle \(x=\sqrt{25-y^2}\) (right boundary):

\[I = \int_0^5 y\cdot(\sqrt{25-y^2}-(5-y))\,dy = \int_0^5 y\sqrt{25-y^2}\,dy - \int_0^5 y(5-y)\,dy.\]

First integral: let \(u=25-y^2\): \(\int_0^5 y\sqrt{25-y^2}\,dy = \left[-\dfrac{(25-y^2)^{3/2}}{3}\right]_0^5 = \dfrac{125}{3}\).

Second integral: \(\int_0^5(5y-y^2)\,dy = \left[\dfrac{5y^2}{2}-\dfrac{y^3}{3}\right]_0^5 = \dfrac{125}{2}-\dfrac{125}{3} = \dfrac{125}{6}\).

\[I = \frac{125}{3} - \frac{125}{6} = \frac{125}{6}.\]

Answer: \(\dfrac{125}{6}\).

4.66. Double Integral with Square Root (Chapter 3, Task 19)

Calculate \(\displaystyle\iint_R x(1+y^2)^{-1/2}\,dA\) where \(R\) is the region in the first quadrant enclosed by \(y=x^2\), \(y=4\), and \(x=0\).

Click to see the solution

Integrate \(x\) first (horizontal cross-sections): for fixed \(y\in[0,4]\), \(x\) runs from \(0\) to \(\sqrt{y}\):

\[I = \int_0^4\frac{1}{\sqrt{1+y^2}}\left(\int_0^{\sqrt{y}}x\,dx\right)dy = \int_0^4\frac{1}{\sqrt{1+y^2}}\cdot\frac{y}{2}\,dy.\]

Let \(u = 1+y^2\), \(du = 2y\,dy\): \[= \frac{1}{4}\int_1^{17}\frac{du}{\sqrt{u}} = \frac{1}{4}\left[2\sqrt{u}\right]_1^{17} = \frac{\sqrt{17}-1}{2}.\]

Answer: \(\dfrac{\sqrt{17}-1}{2}\).

4.67. Volumes of Two Solids (Chapter 3, Task 20)

(a) Find the volume in the first octant bounded above by \(z=x^2+3y^2\), below by \(z=0\), and laterally by \(y=x^2\) and \(y=x\).

(b) Find the volume enclosed by \(y^2=x\), \(z=0\), and \(x+z=1\).

Click to see the solution

(a) The region \(R\) in the \(xy\)-plane is bounded by \(y=x^2\) (below) and \(y=x\) (above) for \(x\in[0,1]\): \[V = \int_0^1\int_{x^2}^x(x^2+3y^2)\,dy\,dx = \int_0^1\left[x^2y+y^3\right]_{x^2}^x dx\] \[= \int_0^1\left(x^3+x^3-x^4-x^6\right)dx = \int_0^1(2x^3-x^4-x^6)\,dx\] \[= \left[\frac{x^4}{2}-\frac{x^5}{5}-\frac{x^7}{7}\right]_0^1 = \frac{1}{2}-\frac{1}{5}-\frac{1}{7} = \frac{35-14-10}{70} = \frac{11}{70}.\]

(b) The region \(R\) in the \(xy\)-plane is bounded by \(y^2=x\) (i.e., \(-\sqrt{x}\leq y\leq\sqrt{x}\)) for \(x\in[0,1]\). The height is \(z=1-x\): \[V = \int_0^1\int_{-\sqrt{x}}^{\sqrt{x}}(1-x)\,dy\,dx = \int_0^1(1-x)\cdot 2\sqrt{x}\,dx = 2\int_0^1(x^{1/2}-x^{3/2})\,dx\] \[= 2\left[\frac{2}{3}x^{3/2}-\frac{2}{5}x^{5/2}\right]_0^1 = 2\left(\frac{2}{3}-\frac{2}{5}\right) = 2\cdot\frac{4}{15} = \frac{8}{15}.\]

Answer: (a) \(\dfrac{11}{70}\); (b) \(\dfrac{8}{15}\).

4.68. Volume Inside a Cylinder Using Polar Coordinates (Chapter 3, Task 21)

Find the volume of the solid that lies under \(z=x^2+y^2\), above the \(xy\)-plane, and inside the cylinder \(x^2+y^2=2x\).

Click to see the solution

Key Concept: Rewrite the cylinder: \(x^2+y^2=2x \iff (x-1)^2+y^2=1\) (circle of radius 1, centered at \((1,0)\)). In polar: \(r^2=2r\cos\theta\), i.e., \(r=2\cos\theta\) for \(\theta\in[-\pi/2,\pi/2]\).

1. Polar setup (\(z=x^2+y^2=r^2\), \(dA = r\,dr\,d\theta\)): \[V = \int_{-\pi/2}^{\pi/2}\int_0^{2\cos\theta} r^2\cdot r\,dr\,d\theta = \int_{-\pi/2}^{\pi/2}\left[\frac{r^4}{4}\right]_0^{2\cos\theta}d\theta = \int_{-\pi/2}^{\pi/2}4\cos^4\theta\,d\theta.\]
2. Use symmetry and power reduction (\(\cos^4\theta = \tfrac{3+4\cos2\theta+\cos4\theta}{8}\)): \[V = 8\int_0^{\pi/2}\cos^4\theta\,d\theta = 8\cdot\frac{3\pi}{16} = \frac{3\pi}{2}.\]

Answer: \(V = \dfrac{3\pi}{2}\).

4.69. More Change of Variables Exercises (Chapter 3, Task 22)

(a) Calculate \(\displaystyle\iint_R \frac{y-4x}{y+4x}\,dx\,dy\) where \(R\) is the region enclosed by \(y=4x\), \(y=4x+2\), \(y=2-4x\), \(y=5-4x\).

(b) Calculate \(\displaystyle\iint_R (2x^2-xy-y^2)\,dx\,dy\) where \(R\) in the first quadrant is bounded by \(y=-2x+4\), \(y=-2x+7\), \(y=x-2\), \(y=x+1\).

Click to see the solution

(a) The boundaries suggest \(u = y+4x\), \(v = y-4x\). Then \(x=\tfrac{u-v}{8}\), \(y=\tfrac{u+v}{2}\), \(|J|=\tfrac{1}{4}\).

New region \(G\): \(u\) from \(2\) to \(5\) (from \(y+4x=2\) and \(y+4x=5\)… wait — checking: \(y=4x\) gives \(v=0\), \(y=4x+2\) gives \(v=2\); \(y=2-4x\) gives \(u=2\), \(y=5-4x\) gives \(u=5\)). So \(G: 2\leq u\leq 5\), \(0\leq v\leq 2\).

\[I = \int_2^5\int_0^2\frac{v}{u}\cdot\frac{1}{4}\,dv\,du = \frac{1}{4}\int_2^5\frac{du}{u}\int_0^2 v\,dv = \frac{1}{4}\ln\!\left(\frac{5}{2}\right)\cdot 2 = \frac{\ln(5/2)}{2}.\]

(b) Boundaries suggest \(u=y+2x\), \(v=y-x\). Then \(x=\tfrac{u-v}{3}\), \(y=\tfrac{u+2v}{3}\), \(|J|=\tfrac{1}{3}\).

New region: \(4\leq u\leq 7\), \(-2\leq v\leq 1\).

Integrand: \(2x^2-xy-y^2 = (2x+y)(x-y)\)… rewrite: \(2x^2-xy-y^2 = (2x+y)(x-y)\). With \(u=y+2x=(2x+y)\) and \(v=y-x=-(x-y)\): integrand \(= u\cdot(-v)\cdot(-1)=... = -uv\).

\[I = \int_4^7\int_{-2}^1(-uv)\cdot\frac{1}{3}\,dv\,du = -\frac{1}{3}\int_4^7 u\,du\cdot\int_{-2}^1 v\,dv\] \[= -\frac{1}{3}\cdot\frac{33}{2}\cdot\left(-\frac{3}{2}\right) = \frac{33}{4}.\]

Answer: (a) \(\dfrac{\ln(5/2)}{2}\); (b) \(\dfrac{33}{4}\).

4.70. Integral Over an Ellipse (Chapter 3, Task 23)

Calculate \(\displaystyle\iint_R \sqrt{16x^2+9y^2}\,dA\) where \(R\) is the region enclosed by \(\dfrac{x^2}{9}+\dfrac{y^2}{16}=1\).

Click to see the solution

Key Concept: Scale the ellipse to a unit disk. Let \(x = 3r\cos\theta\), \(y = 4r\sin\theta\), so \(|J| = 12r\) and the ellipse maps to \(r\leq 1\).

The integrand: \(\sqrt{16x^2+9y^2} = \sqrt{16\cdot9r^2\cos^2\theta+9\cdot16r^2\sin^2\theta} = 12r\).

\[I = \int_0^{2\pi}\int_0^1 12r\cdot 12r\,dr\,d\theta = 144\int_0^{2\pi}d\theta\int_0^1 r^2\,dr = 144\cdot 2\pi\cdot\frac{1}{3} = 96\pi.\]

Answer: \(96\pi\).

4.71. Triple Integral Over a Paraboloid Region (Chapter 3, Task 24)

Calculate \(\displaystyle\iiint_D x\,dV\) where \(D\) is bounded by the paraboloid \(x=4y^2+4z^2\) and the plane \(x=4\).

Click to see the solution

Key Concept: Use cylindrical coordinates with the \(x\)-axis as the axis: \(y=r\cos\theta\), \(z=r\sin\theta\), \(x=x\). Then \(dV=r\,dr\,d\theta\,dx\). The paraboloid becomes \(x=4r^2\) and the disk at \(x=4\) has \(r\leq 1\).

1. Describe \(D\): for fixed \((r,\theta)\) with \(0\leq r\leq 1\), \(x\) runs from \(4r^2\) to \(4\): \[\iiint_D x\,dV = \int_0^{2\pi}\int_0^1\int_{4r^2}^4 x\cdot r\,dx\,dr\,d\theta.\]
2. Inner integral (\(x\)): \[\int_{4r^2}^4 x\,dx = \frac{16-16r^4}{2} = 8(1-r^4).\]
3. Radial integral: \[\int_0^1 8r(1-r^4)\,dr = 8\left[\frac{r^2}{2}-\frac{r^6}{6}\right]_0^1 = 8\cdot\frac{1}{3} = \frac{8}{3}.\]
4. Full integral: \[= 2\pi\cdot\frac{8}{3} = \frac{16\pi}{3}.\]

Answer: \(\dfrac{16\pi}{3}\).

4.72. Volume of a Solid Bounded by Planes (Chapter 3, Task 25)

Find the volume of the solid region bounded by \(z=x\), \(x+z=8\), \(z=y\), \(y=8\), and \(z=0\).

Click to see the solution

The constraints are \(z\geq 0\), \(z\leq x\), \(x+z\leq 8\), \(z\leq y\), \(y\leq 8\). Integrate \(z\) from \(0\) to \(\min(x, 8-x)\), \(y\) from \(z\) to \(8\), and determine the \(x\)-range.

For the \(z\)-limits: \(z\leq x\) and \(z\leq 8-x\) means \(z\) goes from \(0\) to \(\min(x,8-x)\). This splits at \(x=4\).

\[V = \int_0^4\int_0^x\int_z^8\,dy\,dz\,dx + \int_4^8\int_0^{8-x}\int_z^8\,dy\,dz\,dx.\]

First piece (\(0\leq x\leq 4\)): \[\int_0^4\int_0^x(8-z)\,dz\,dx = \int_0^4\left[8z-\frac{z^2}{2}\right]_0^x dx = \int_0^4\!\left(8x-\frac{x^2}{2}\right)dx = \left[4x^2-\frac{x^3}{6}\right]_0^4 = 64-\frac{32}{3} = \frac{160}{3}.\]

Second piece (\(4\leq x\leq 8\)): by symmetry (substitute \(x'=8-x\)), equals \(\dfrac{160}{3}\).

\[V = \frac{160}{3}+\frac{160}{3} = \frac{320}{3}.\]

Answer: \(V = \dfrac{320}{3}\).

4.73. Volume of an Elliptical Cylinder Solid (Chapter 3, Task 26)

Find the volume of the solid bounded by the elliptical cylinder \(x^2+4y^2\leq 4\), the \(xy\)-plane, and the plane \(z=x+2\).

Click to see the solution

Key Concept: The base is the ellipse \(x^2+4y^2\leq 4\), i.e., \(\tfrac{x^2}{4}+y^2\leq 1\). Use the substitution \(x=2u\), \(y=v\) to map it to the unit disk, \(|J|=2\).

Height: \(z=x+2=2u+2\) (always \(\geq 0\) on the ellipse since \(x\geq -2\)).

\[V = \iint_{\text{ellipse}}(x+2)\,dA = \iint_{u^2+v^2\leq 1}(2u+2)\cdot 2\,du\,dv.\]

• \(\iint_{u^2+v^2\leq 1} 2u\cdot 2\,du\,dv = 0\) (odd in \(u\), symmetric disk).
• \(\iint_{u^2+v^2\leq 1} 4\,du\,dv = 4\pi\).

\[V = 0 + 4\pi = 4\pi.\]

Answer: \(V = 4\pi\).

4.74. Triple Integral in Cylindrical Coordinates (Second Integral) (Chapter 3, Task 27)

Calculate \(\displaystyle\int_{-3}^3\int_0^{\sqrt{9-x^2}}\int_0^{9-x^2-y^2}\sqrt{x^2+y^2}\,dz\,dy\,dx\).

Click to see the solution

Key Concept: The region is the upper half of the cylinder \(x^2+y^2\leq 9\) (since \(y\geq 0\)), with \(z\) from \(0\) to \(9-r^2\). In cylindrical: \(0\leq r\leq 3\), \(0\leq\theta\leq\pi\), \(0\leq z\leq 9-r^2\).

\[I = \int_0^\pi\int_0^3\int_0^{9-r^2} r\cdot r\,dz\,dr\,d\theta = \pi\int_0^3 r^2(9-r^2)\,dr\] \[= \pi\int_0^3(9r^2-r^4)\,dr = \pi\left[3r^3-\frac{r^5}{5}\right]_0^3 = \pi\!\left(81-\frac{243}{5}\right) = \frac{162\pi}{5}.\]

Answer: \(\dfrac{162\pi}{5}\).

4.75. Volume Between Sphere and Cylinder (Chapter 3, Task 28)

Find the volume of the solid enclosed by the cylinder \(x^2+y^2=4\) and the planes \(z=0\) and \(x+y+z=4\).

Click to see the solution

The solid sits above \(z=0\) and below \(z=4-x-y\). In cylindrical:

\[V = \int_0^{2\pi}\int_0^2\int_0^{4-r\cos\theta-r\sin\theta} r\,dz\,dr\,d\theta = \int_0^{2\pi}\int_0^2 r(4-r\cos\theta-r\sin\theta)\,dr\,d\theta.\]

\[= \int_0^{2\pi}\left[2r^2-\frac{r^3}{3}(\cos\theta+\sin\theta)\right]_0^2 d\theta = \int_0^{2\pi}\!\left(8-\frac{8}{3}(\cos\theta+\sin\theta)\right)d\theta.\]

The \(\cos\theta\) and \(\sin\theta\) terms integrate to zero over \([0,2\pi]\):

\[V = 8\cdot 2\pi = 16\pi.\]

Answer: \(V = 16\pi\).

4.76. Triple Integral by Linear Substitution (Chapter 3, Task 29)

Calculate \(\displaystyle\iiint_D (x^2y+3xyz)\,dx\,dy\,dz\) over \(D: 1\leq x\leq 2,\; 0\leq xy\leq 2,\; 0\leq z\leq 1\) using \(u=x\), \(v=xy\), \(w=3z\).

Click to see the solution

1. Inverse: \(x=u\), \(y=v/u\), \(z=w/3\). New region \(G\): \(1\leq u\leq 2\), \(0\leq v\leq 2\), \(0\leq w\leq 3\).
2. Jacobian: \(J=\dfrac{\partial(x,y,z)}{\partial(u,v,w)} = \begin{vmatrix}1&0&0\\-v/u^2&1/u&0\\0&0&1/3\end{vmatrix} = \dfrac{1}{3u}\), so \(|J|=\dfrac{1}{3u}\).
3. Integrand: \(x^2y+3xyz = u^2\cdot\dfrac{v}{u}+3u\cdot\dfrac{v}{u}\cdot\dfrac{w}{3} = uv+vw\).
4. Integral: \[I = \int_1^2\int_0^2\int_0^3(uv+vw)\cdot\frac{1}{3u}\,dw\,dv\,du = \frac{1}{3}\int_1^2\int_0^2\int_0^3\!\left(v+\frac{vw}{u}\right)dw\,dv\,du.\]
5. Inner (\(w\)): \(\int_0^3\!\left(v+\dfrac{vw}{u}\right)dw = 3v+\dfrac{9v}{2u}\).
6. Middle (\(v\)): \(\int_0^2\!\left(3v+\dfrac{9v}{2u}\right)dv = 6+\dfrac{9}{u}\).
7. Outer (\(u\)): \(\dfrac{1}{3}\int_1^2\!\left(6+\dfrac{9}{u}\right)du = \dfrac{1}{3}\left[6+9\ln 2\right] = 2+3\ln 2\).

Answer: \(2+3\ln 2\).

4.77. Second Spherical Coordinates Exercise (Chapter 3, Task 30)

Calculate \(\displaystyle\iiint_D xe^{x^2+y^2+z^2}\,dV\) where \(D\) is the portion of the unit ball \(x^2+y^2+z^2\leq 1\) in the first octant.

Click to see the solution

Key Concept: In spherical coordinates, \(x = \rho\sin\phi\cos\theta\), \(x^2+y^2+z^2 = \rho^2\), \(dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta\). First octant: \(0\leq\rho\leq 1\), \(0\leq\phi\leq\pi/2\), \(0\leq\theta\leq\pi/2\).

\[I = \int_0^{\pi/2}\int_0^{\pi/2}\int_0^1(\rho\sin\phi\cos\theta)\,e^{\rho^2}\,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta.\]

Separate variables: \[= \left(\int_0^{\pi/2}\cos\theta\,d\theta\right)\!\left(\int_0^{\pi/2}\sin^2\phi\,d\phi\right)\!\left(\int_0^1\rho^3 e^{\rho^2}\,d\rho\right).\]

• \(\int_0^{\pi/2}\cos\theta\,d\theta = 1\).
• \(\int_0^{\pi/2}\sin^2\phi\,d\phi = \dfrac{\pi}{4}\).
• \(\int_0^1\rho^3 e^{\rho^2}\,d\rho\): let \(t=\rho^2\): \(= \dfrac{1}{2}\int_0^1 te^t\,dt = \dfrac{1}{2}[te^t-e^t]_0^1 = \dfrac{1}{2}(0+1) = \dfrac{1}{2}\).

\[I = 1\cdot\frac{\pi}{4}\cdot\frac{1}{2} = \frac{\pi}{8}.\]

Answer: \(\dfrac{\pi}{8}\).

4.78. Find and Sketch the Domain (Exercises, Task 1)

Find and sketch the domain for each function:

1. \(f(x, y) = \ln(x^2 + y^2 - 4)\);
2. \(f(x, y) = \dfrac{\sin(xy)}{x^2+y^2-25}\);
3. \(f(x, y, z) = \sqrt{4 - x^2} + \sqrt{9 - y^2} + \sqrt{1 - z^2}\);
4. \(f(x, y, z) = \ln(16 - 4x^2 - 4y^2 - z^2)\).

Click to see the solution

Key Concept: The domain of a function is the set of all points where the function is defined. For logarithms, the argument must be strictly positive; for square roots, the argument must be non-negative; for fractions, the denominator must be non-zero.

(a) \(f(x, y) = \ln(x^2 + y^2 - 4)\)

Require \(x^2 + y^2 - 4 > 0\), i.e., \(x^2 + y^2 > 4\).

Domain: the exterior of the circle of radius 2 centered at the origin (the open region outside the disk \(x^2+y^2=4\)).

(b) \(f(x, y) = \dfrac{\sin(xy)}{x^2+y^2-25}\)

Require \(x^2+y^2-25 \neq 0\), i.e., \(x^2+y^2 \neq 25\).

Domain: all of \(\mathbb{R}^2\) except the circle \(x^2+y^2=25\) (radius 5).

(c) \(f(x, y, z) = \sqrt{4 - x^2} + \sqrt{9 - y^2} + \sqrt{1 - z^2}\)

Require simultaneously: \(4-x^2 \geq 0 \Rightarrow -2\leq x\leq 2\); \(9-y^2\geq 0 \Rightarrow -3\leq y\leq 3\); \(1-z^2\geq 0 \Rightarrow -1\leq z\leq 1\).

Domain: the rectangular box \([-2,2]\times[-3,3]\times[-1,1]\).

(d) \(f(x, y, z) = \ln(16 - 4x^2 - 4y^2 - z^2)\)

Require \(16 - 4x^2 - 4y^2 - z^2 > 0\), i.e., \(4x^2 + 4y^2 + z^2 < 16\), or equivalently \(\dfrac{x^2}{4} + \dfrac{y^2}{4} + \dfrac{z^2}{16} < 1\).

Domain: the interior of the ellipsoid with semi-axes \(a=b=2\) and \(c=4\).

4.79. Evaluate Limits of Multivariable Functions (Exercises, Task 2)

Find the limits: \[\lim_{\substack{(x,y)\to(2,0) \\ 2x-y \neq 4}} \frac{\sqrt{2x - y} - 2}{2x - y - 4}, \quad \lim_{\substack{(x,y)\to(0,0) \\ x \neq y}} \frac{x - y + 2\sqrt{x} - 2\sqrt{y}}{\sqrt{x} - \sqrt{y}}, \quad \lim_{(x,y)\to(0,0)} \frac{1 - \cos(xy)}{xy}.\]

Click to see the solution

Limit 1: \(\displaystyle\lim_{\substack{(x,y)\to(2,0)}} \frac{\sqrt{2x-y}-2}{2x-y-4}\)

Let \(t = 2x - y\). As \((x,y)\to(2,0)\), we have \(t\to 4\). So the limit becomes: \[\lim_{t\to 4}\frac{\sqrt{t}-2}{t-4}.\] Rationalize by multiplying numerator and denominator by \((\sqrt{t}+2)\): \[\frac{\sqrt{t}-2}{t-4} = \frac{(\sqrt{t}-2)(\sqrt{t}+2)}{(t-4)(\sqrt{t}+2)} = \frac{t-4}{(t-4)(\sqrt{t}+2)} = \frac{1}{\sqrt{t}+2}.\] As \(t\to 4\): \(\dfrac{1}{\sqrt{4}+2} = \dfrac{1}{4}\).

Answer: \(\dfrac{1}{4}\).

Limit 2: \(\displaystyle\lim_{\substack{(x,y)\to(0,0) \\ x\neq y}} \frac{x-y+2\sqrt{x}-2\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)

Factor the numerator by grouping: \[x - y + 2\sqrt{x} - 2\sqrt{y} = (x-y) + 2(\sqrt{x}-\sqrt{y}) = (\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}) + 2(\sqrt{x}-\sqrt{y}).\] Factor out \((\sqrt{x}-\sqrt{y})\): \[= (\sqrt{x}-\sqrt{y})\bigl(\sqrt{x}+\sqrt{y}+2\bigr).\] So the expression simplifies to: \[\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}+2)}{\sqrt{x}-\sqrt{y}} = \sqrt{x}+\sqrt{y}+2.\] As \((x,y)\to(0,0)\): \(\sqrt{0}+\sqrt{0}+2 = 2\).

Answer: \(2\).

Limit 3: \(\displaystyle\lim_{(x,y)\to(0,0)} \frac{1-\cos(xy)}{xy}\)

Let \(u = xy\). As \((x,y)\to(0,0)\), we have \(u\to 0\). By the half-angle identity \(1-\cos u = 2\sin^2(u/2)\): \[\frac{1-\cos u}{u} = \frac{2\sin^2(u/2)}{u} = \frac{\sin(u/2)}{u/2}\cdot\sin\!\left(\frac{u}{2}\right).\] As \(u\to 0\): \(\dfrac{\sin(u/2)}{u/2}\to 1\) and \(\sin(u/2)\to 0\), so the product tends to \(0\).

Answer: \(0\).

4.80. Show Nonexistence of Limits at the Origin (Exercises, Task 3)

Show that the following functions have no limit as \((x,y)\to(0,0)\): \[f(x,y) = -\frac{x}{\sqrt{x^2+y^2}}, \quad f(x,y) = \frac{x^4-y^2}{x^4+y^2}, \quad f(x,y) = \frac{xy}{|xy|}.\]

Click to see the solution

Key Concept: To show a limit does not exist, find two paths to \((0,0)\) along which the function approaches different values.

Function 1: \(f(x,y) = -\dfrac{x}{\sqrt{x^2+y^2}}\)

• Along \(y=0\), \(x\to 0^+\): \(f = -\dfrac{x}{\sqrt{x^2}} = -\dfrac{x}{|x|} = -1\).
• Along \(y=0\), \(x\to 0^-\): \(f = -\dfrac{x}{|x|} = +1\).

Two different limits, so no limit exists.

Function 2: \(f(x,y) = \dfrac{x^4-y^2}{x^4+y^2}\)

• Along \(y=0\) (so \(x\to 0\)): \(f = \dfrac{x^4}{x^4} = 1\).
• Along \(x=0\) (so \(y\to 0\)): \(f = \dfrac{-y^2}{y^2} = -1\).

Two different limits, so no limit exists.

Function 3: \(f(x,y) = \dfrac{xy}{|xy|}\)

This equals \(\text{sgn}(xy)\) wherever \(xy\neq 0\).

• Along \(y=x\), \(x\to 0^+\): \(xy = x^2 > 0\), so \(f = 1\).
• Along \(y=-x\), \(x\to 0^+\): \(xy = -x^2 < 0\), so \(f = -1\).

Two different limits, so no limit exists.

4.81. Find a Function from Its Partial Derivatives (Exercises, Task 4)

Find a function \(z = f(x,y)\) whose partial derivatives are as given:

1. \(\dfrac{\partial f}{\partial x} = 3x^2y^2 - 2x\), \(\quad \dfrac{\partial f}{\partial y} = 2x^3y + 6y\);

2. \(\dfrac{\partial f}{\partial x} = xy\cos(xy) + \sin(xy)\), \(\quad \dfrac{\partial f}{\partial y} = x^2\cos(xy)\).

Click to see the solution

Key Concept: If both partial derivatives are given and they are consistent (i.e., \(\partial^2 f/\partial y\partial x = \partial^2 f/\partial x\partial y\)), we can recover \(f\) by integrating one partial derivative and then determining the “constant of integration” (which may depend on the other variable) from the second condition.

(a) Integrate \(\dfrac{\partial f}{\partial x} = 3x^2y^2 - 2x\) with respect to \(x\): \[f(x,y) = \int(3x^2y^2 - 2x)\,dx = x^3y^2 - x^2 + g(y),\] where \(g(y)\) is an arbitrary function of \(y\) alone.

Now use \(\dfrac{\partial f}{\partial y} = 2x^3y + 6y\): \[\frac{\partial f}{\partial y} = 2x^3y + g'(y) = 2x^3y + 6y.\] So \(g'(y) = 6y\), giving \(g(y) = 3y^2 + C\).

Answer: \(f(x,y) = x^3y^2 - x^2 + 3y^2 + C\).

(b) Integrate \(\dfrac{\partial f}{\partial x} = xy\cos(xy) + \sin(xy)\) with respect to \(x\).

Observe that \(\dfrac{d}{dx}[x\sin(xy)] = \sin(xy) + xy\cos(xy)\). So: \[f(x,y) = \int\bigl[xy\cos(xy)+\sin(xy)\bigr]dx = x\sin(xy) + g(y).\]

Verify using \(\dfrac{\partial f}{\partial y} = x^2\cos(xy)\): \[\frac{\partial f}{\partial y} = x\cdot x\cos(xy) + g'(y) = x^2\cos(xy) + g'(y).\] Setting equal to \(x^2\cos(xy)\): \(g'(y)=0\), so \(g(y)=C\).

Answer: \(f(x,y) = x\sin(xy) + C\).

4.82. Verify Solutions of Laplace’s Equation (Exercises, Task 5)

Determine whether each of the following functions is a solution of Laplace’s equation \(f_{xx} + f_{yy} = 0\): \[f(x,y) = \sin x \cosh y + \cos x \sinh y, \quad f(x,y) = \ln\!\left(\sqrt{x^2+y^2}\right).\]

Click to see the solution

Key Concept: A function \(f(x,y)\) satisfies Laplace’s equation if \(\Delta f = f_{xx} + f_{yy} = 0\). Such functions are called harmonic.

Function 1: \(f(x,y) = \sin x\cosh y + \cos x\sinh y\)

Compute partial derivatives: \[f_x = \cos x\cosh y - \sin x\sinh y, \quad f_{xx} = -\sin x\cosh y - \cos x\sinh y.\] \[f_y = \sin x\sinh y + \cos x\cosh y, \quad f_{yy} = \sin x\cosh y + \cos x\sinh y.\]

Sum: \[f_{xx}+f_{yy} = (-\sin x\cosh y - \cos x\sinh y) + (\sin x\cosh y + \cos x\sinh y) = 0. \checkmark\]

Yes, it is a solution.

Function 2: \(f(x,y) = \ln\!\sqrt{x^2+y^2} = \tfrac{1}{2}\ln(x^2+y^2)\)

\[f_x = \frac{x}{x^2+y^2}, \quad f_{xx} = \frac{(x^2+y^2) - x\cdot 2x}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2}.\]

By symmetry: \[f_{yy} = \frac{x^2-y^2}{(x^2+y^2)^2}.\]

Sum: \[f_{xx}+f_{yy} = \frac{y^2-x^2+x^2-y^2}{(x^2+y^2)^2} = 0. \checkmark\]

Yes, it is a solution (for \((x,y)\neq(0,0)\)).

4.83. Compute Partial Derivatives via Chain Rule (Exercises, Task 6)

Find \(\dfrac{\partial z}{\partial s}\) and \(\dfrac{\partial z}{\partial t}\) for each of the following:

1. \(z = \sin x\sin y\), with \(x = \sqrt{t}\), \(y = \dfrac{1}{t}\);

2. \(z = \ln(2p^3+3q^2+2r^4)\), with \(p = se^t\sin s\), \(q = se^t\cos s\), \(r = se^t\).

Click to see the solution

Key Concept: The chain rule for multivariable functions: if \(z=z(x,y)\) and \(x,y\) depend on \(s,t\), then \[\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s}, \qquad \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial t}.\]

(a) Here \(x=\sqrt{t}\), \(y=1/t\) do not depend on \(s\), so:

\[\frac{\partial z}{\partial s} = 0.\]

For \(\partial z/\partial t\): \[\frac{\partial z}{\partial t} = \cos x\sin y\cdot\frac{1}{2\sqrt{t}} + \sin x\cos y\cdot\left(-\frac{1}{t^2}\right).\]

Substituting \(x=\sqrt{t}\), \(y=1/t\): \[\frac{\partial z}{\partial t} = \frac{\cos\sqrt{t}\,\sin\frac{1}{t}}{2\sqrt{t}} - \frac{\sin\sqrt{t}\,\cos\frac{1}{t}}{t^2}.\]

(b) \(z = \ln(2p^3+3q^2+2r^4)\).

Let \(D = 2p^3+3q^2+2r^4\). Then \(\dfrac{\partial z}{\partial p} = \dfrac{6p^2}{D}\), \(\dfrac{\partial z}{\partial q} = \dfrac{6q}{D}\), \(\dfrac{\partial z}{\partial r} = \dfrac{8r^3}{D}\).

Compute \(\partial p/\partial s\), \(\partial q/\partial s\), \(\partial r/\partial s\) (with \(p=se^t\sin s\), \(q=se^t\cos s\), \(r=se^t\)): \[\frac{\partial p}{\partial s} = e^t\sin s + se^t\cos s, \quad \frac{\partial q}{\partial s} = e^t\cos s - se^t\sin s, \quad \frac{\partial r}{\partial s} = e^t.\]

\[\frac{\partial z}{\partial s} = \frac{6p^2}{D}(e^t\sin s+se^t\cos s) + \frac{6q}{D}(e^t\cos s-se^t\sin s) + \frac{8r^3}{D}\cdot e^t.\]

For \(\partial/\partial t\): \(\dfrac{\partial p}{\partial t}=se^t\sin s\), \(\dfrac{\partial q}{\partial t}=se^t\cos s\), \(\dfrac{\partial r}{\partial t}=se^t\). \[\frac{\partial z}{\partial t} = \frac{6p^2\cdot se^t\sin s + 6q\cdot se^t\cos s + 8r^3\cdot se^t}{D}.\]

4.84. Implicit Partial Differentiation (Exercises, Task 7)

Assuming the following equations define \(z\) as a differentiable function of \(x\) and \(y\), find \(\dfrac{\partial z}{\partial x}\) and \(\dfrac{\partial z}{\partial y}\) at the given point.

1. \(z^3 - xy + yz - y^3 - 2 = 0\), at \((1,1,1)\);

2. \(xe^y + ye^z + 2\ln x - 2 - 3\ln 2 = 0\), at \((1,\ln 2,\ln 3)\).

Click to see the solution

Key Concept: For an equation \(F(x,y,z)=0\) defining \(z\) implicitly: \[\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}, \qquad \frac{\partial z}{\partial y} = -\frac{F_y}{F_z}.\]

(a) \(F(x,y,z) = z^3-xy+yz-y^3-2\).

\[F_x = -y, \quad F_y = -x+z-3y^2, \quad F_z = 3z^2+y.\]

At \((1,1,1)\): \(F_x=-1\), \(F_y=-1+1-3=-3\), \(F_z=3+1=4\).

\[\frac{\partial z}{\partial x} = -\frac{-1}{4} = \frac{1}{4}, \qquad \frac{\partial z}{\partial y} = -\frac{-3}{4} = \frac{3}{4}.\]

(b) \(F(x,y,z) = xe^y+ye^z+2\ln x-2-3\ln 2\).

\[F_x = e^y+\frac{2}{x}, \quad F_y = xe^y + e^z, \quad F_z = ye^z.\]

At \((1,\ln 2,\ln 3)\): \(e^y=2\), \(e^z=3\).

\[F_x = 2+2 = 4, \quad F_y = 1\cdot 2 + 3 = 5, \quad F_z = \ln 2\cdot 3 = 3\ln 2.\]

\[\frac{\partial z}{\partial x} = -\frac{4}{3\ln 2}, \qquad \frac{\partial z}{\partial y} = -\frac{5}{3\ln 2}.\]

4.85. Directional Derivative at a Point (Exercises, Task 8)

Find the derivative of the function at the point \(P_0\) in the direction of \(\mathbf{u}\):

1. \(f(x,y,z) = x^2+2y^2-3z^2\), \(\;P_0(1,1,1)\), \(\;\mathbf{u} = 2\hat{i}+\hat{j}-2\hat{k}\);

2. \(f(x,y,z) = \cos(yz)+e^{yz}+\ln(xz)\), \(\;P_0(1,0,1/2)\), \(\;\mathbf{u} = \hat{i}+2\hat{j}+2\hat{k}\).

Click to see the solution

Key Concept: The directional derivative in the direction of unit vector \(\hat{\mathbf{u}}\) is: \[D_{\hat{u}}f = \nabla f\cdot\hat{\mathbf{u}}.\] Always normalize \(\mathbf{u}\) first: \(\hat{\mathbf{u}} = \mathbf{u}/|\mathbf{u}|\).

(a) \(\nabla f = (2x,\,4y,\,-6z)\). At \(P_0(1,1,1)\): \(\nabla f = (2,4,-6)\).

\(|\mathbf{u}| = \sqrt{4+1+4}=3\), so \(\hat{\mathbf{u}} = \bigl(\tfrac{2}{3},\tfrac{1}{3},-\tfrac{2}{3}\bigr)\).

\[D_{\hat{u}}f = 2\cdot\frac{2}{3}+4\cdot\frac{1}{3}+(-6)\cdot\left(-\frac{2}{3}\right) = \frac{4}{3}+\frac{4}{3}+\frac{12}{3} = \frac{20}{3}.\]

Answer: \(\dfrac{20}{3}\).

(b) \(f = \cos(yz)+e^{yz}+\ln(xz)\).

\[f_x = \frac{1}{x}, \quad f_y = -z\sin(yz)+ze^{yz}, \quad f_z = -y\sin(yz)+ye^{yz}+\frac{1}{z}.\]

At \(P_0(1,0,1/2)\): \(yz=0\), \(\sin 0=0\), \(e^0=1\). \[f_x=1,\quad f_y = -\tfrac{1}{2}\cdot 0+\tfrac{1}{2}\cdot 1 = \tfrac{1}{2},\quad f_z = 0+0+2 = 2.\] \(\nabla f = (1,\tfrac{1}{2},2)\).

\(|\mathbf{u}|=\sqrt{1+4+4}=3\), \(\hat{\mathbf{u}}=\bigl(\tfrac{1}{3},\tfrac{2}{3},\tfrac{2}{3}\bigr)\).

\[D_{\hat{u}}f = 1\cdot\frac{1}{3}+\frac{1}{2}\cdot\frac{2}{3}+2\cdot\frac{2}{3} = \frac{1}{3}+\frac{1}{3}+\frac{4}{3} = 2.\]

Answer: \(2\).

4.86. Directions of Steepest Ascent and Descent (Exercises, Task 9)

Find the directions in which the functions increase most rapidly, and the directions in which they decrease most rapidly, at \(P_0\). Then find the derivatives of the functions in these directions.

1. \(f(x,y,z) = xe^y+z^2\), \(\;P_0(1,\ln 2,1/2)\);

2. \(f(x,y,z) = \ln(x^2+y^2-1)+y+6z\), \(\;P_0(1,1,0)\).

Click to see the solution

Key Concept: The function increases most rapidly in the direction of \(\nabla f\) (gradient), and decreases most rapidly in the direction of \(-\nabla f\). The rates of change in these directions are \(|\nabla f|\) and \(-|\nabla f|\), respectively.

(a) \(\nabla f = (e^y,\,xe^y,\,2z)\). At \(P_0(1,\ln 2,1/2)\): \(e^{\ln 2}=2\).

\[\nabla f = (2,\,2,\,1), \quad |\nabla f| = \sqrt{4+4+1}=3.\]

• Steepest ascent direction: \(\hat{\mathbf{u}} = \bigl(\tfrac{2}{3},\tfrac{2}{3},\tfrac{1}{3}\bigr)\), rate \(= 3\).
• Steepest descent direction: \(-\hat{\mathbf{u}} = \bigl(-\tfrac{2}{3},-\tfrac{2}{3},-\tfrac{1}{3}\bigr)\), rate \(= -3\).

(b) \(f_x = \dfrac{2x}{x^2+y^2-1}\), \(f_y = \dfrac{2y}{x^2+y^2-1}+1\), \(f_z = 6\).

At \(P_0(1,1,0)\): \(x^2+y^2-1 = 1\), so \(f_x = 2\), \(f_y = 2+1=3\), \(f_z=6\).

\[\nabla f = (2,3,6), \quad |\nabla f| = \sqrt{4+9+36}=7.\]

• Steepest ascent: \(\hat{\mathbf{u}} = \bigl(\tfrac{2}{7},\tfrac{3}{7},\tfrac{6}{7}\bigr)\), rate \(= 7\).
• Steepest descent: \(-\hat{\mathbf{u}} = \bigl(-\tfrac{2}{7},-\tfrac{3}{7},-\tfrac{6}{7}\bigr)\), rate \(= -7\).

4.87. Classify Critical Points (Exercises, Task 10)

Find and classify the critical points of the functions:

1. \(f(x,y) = x^2-y^3-3xy+2x\);
2. \(f(x,y) = 2x^3+2y^3-9x^2+3y^2-12y\);
3. \(f(x,y) = 10xye^{-(x^2+y^2)}\).

Click to see the solution

Key Concept: At a critical point, \(f_x=0\) and \(f_y=0\). Classify using the second derivative test: compute \(D = f_{xx}f_{yy}-(f_{xy})^2\). - \(D>0\) and \(f_{xx}>0\): local minimum. - \(D>0\) and \(f_{xx}<0\): local maximum. - \(D<0\): saddle point. - \(D=0\): test is inconclusive.

(a) \(f_x = 2x-3y+2=0\), \(f_y = -3y^2-3x=0 \Rightarrow x=-y^2\).

Substitute into \(f_x=0\): \(2(-y^2)-3y+2=0 \Rightarrow -2y^2-3y+2=0 \Rightarrow 2y^2+3y-2=0\).

Discriminant: \(9+16=25\). So \(y = \dfrac{-3\pm 5}{4}\): \(y=\tfrac{1}{2}\) or \(y=-2\).

• \(y=\tfrac{1}{2}\): \(x=-\tfrac{1}{4}\). Critical point \((-\tfrac{1}{4},\tfrac{1}{2})\).
• \(y=-2\): \(x=-4\). Critical point \((-4,-2)\).

\(f_{xx}=2\), \(f_{yy}=-6y\), \(f_{xy}=-3\).

At \((-\tfrac{1}{4},\tfrac{1}{2})\): \(D=2\cdot(-3)-(−3)^2=-6-9=-15<0\). Saddle point.

At \((-4,-2)\): \(D=2\cdot 12-9=24-9=15>0\), \(f_{xx}=2>0\). Local minimum.

(b) \(f_x=6x^2-18x=6x(x-3)=0\Rightarrow x=0\) or \(x=3\). \(f_y=6y^2+6y-12=6(y^2+y-2)=6(y+2)(y-1)=0\Rightarrow y=-2\) or \(y=1\).

Critical points: \((0,-2),(0,1),(3,-2),(3,1)\).

\(f_{xx}=12x-18\), \(f_{yy}=12y+6\), \(f_{xy}=0\), so \(D=f_{xx}f_{yy}\).

• \((0,-2)\): \(f_{xx}=-18\), \(f_{yy}=-18\). \(D=(-18)(-18)=324>0\), \(f_{xx}<0\): local maximum.
• \((0,1)\): \(f_{xx}=-18\), \(f_{yy}=18\). \(D=-324<0\): saddle point.
• \((3,-2)\): \(f_{xx}=18\), \(f_{yy}=-18\). \(D=-324<0\): saddle point.
• \((3,1)\): \(f_{xx}=18\), \(f_{yy}=18\). \(D=324>0\), \(f_{xx}>0\): local minimum.

(c) \(f=10xye^{-(x^2+y^2)}\).

\(f_x = 10y(1-2x^2)e^{-(x^2+y^2)}=0\) and \(f_y = 10x(1-2y^2)e^{-(x^2+y^2)}=0\).

Since \(e^{-(x^2+y^2)}>0\) always, we need \(y(1-2x^2)=0\) and \(x(1-2y^2)=0\).

Cases: 1. \(x=0\) and \(y=0\): critical point \((0,0)\). 2. \(x=0\) and \(1-2y^2=0\): \(y=\pm\tfrac{1}{\sqrt{2}}\), \(x=0\). Points \((0,\pm\tfrac{1}{\sqrt{2}})\). 3. \(y=0\) and \(1-2x^2=0\): \(x=\pm\tfrac{1}{\sqrt{2}}\), \(y=0\). Points \((\pm\tfrac{1}{\sqrt{2}},0)\). 4. \(1-2x^2=0\) and \(1-2y^2=0\): \(x=\pm\tfrac{1}{\sqrt{2}}\), \(y=\pm\tfrac{1}{\sqrt{2}}\). Four points.

Evaluate \(f\) at case 4: \(f = 10\cdot(\pm\tfrac{1}{\sqrt{2}})\cdot(\pm\tfrac{1}{\sqrt{2}})\cdot e^{-1} = \pm\dfrac{5}{e}\).

At \((0,0)\) and points in cases 2 and 3, \(f=0\).

Second derivative test for case 4. Compute the second-order partial derivatives: \[f_{xx} = 10ye^{-(x^2+y^2)}\cdot 2x(2x^2-3), \quad f_{yy} = 10xe^{-(x^2+y^2)}\cdot 2y(2y^2-3),\] \[f_{xy} = 10(1-2x^2)(1-2y^2)e^{-(x^2+y^2)}.\]

At points where \(x^2=\tfrac{1}{2}\) and \(y^2=\tfrac{1}{2}\), we have \(1-2x^2=0\), so \(f_{xy}=0\), and \(2x^2-3=-2\), \(2y^2-3=-2\): \[f_{xx} = 10y\cdot e^{-1}\cdot 2x\cdot(-2) = -40xye^{-1}, \qquad f_{yy} = -40xye^{-1}.\]

So \(D = f_{xx}f_{yy} - f_{xy}^2 = (40xye^{-1})^2 > 0\) at all four points.

At \(\bigl(\tfrac{1}{\sqrt{2}},\tfrac{1}{\sqrt{2}}\bigr)\) and \(\bigl(-\tfrac{1}{\sqrt{2}},-\tfrac{1}{\sqrt{2}}\bigr)\): \(xy>0\), so \(f_{xx}=-40xye^{-1}<0\). \(D>0\) and \(f_{xx}<0\): local maxima with \(f=\tfrac{5}{e}\).

At \(\bigl(\tfrac{1}{\sqrt{2}},-\tfrac{1}{\sqrt{2}}\bigr)\) and \(\bigl(-\tfrac{1}{\sqrt{2}},\tfrac{1}{\sqrt{2}}\bigr)\): \(xy<0\), so \(f_{xx}=-40xye^{-1}>0\). \(D>0\) and \(f_{xx}>0\): local minima with \(f=-\tfrac{5}{e}\).

Second derivative test for \((0,0)\). \(f_{xx}=f_{yy}=0\) and \(f_{xy}=10\cdot 1\cdot 1\cdot 1=10\). So \(D=0-100=-100<0\): saddle point.

Points in cases 2 and 3 (e.g., \((0,\tfrac{1}{\sqrt{2}})\)): \(f_{xx}=f_{yy}=f_{xy}=0\), so \(D=0\) and the test is inconclusive. However, \(f(x,y)=10xye^{-(x^2+y^2)}\) changes sign in any neighborhood of these points (take \(x>0\) or \(x<0\) small), confirming they are saddle points.

4.88. Extrema on an Ellipse via Lagrange Multipliers (Exercises, Task 11)

Find the largest and smallest values that \(f(x,y)=xy\) takes on the ellipse \(\dfrac{x^2}{8}+\dfrac{y^2}{2}=1\).

Click to see the solution

Key Concept: Use Lagrange multipliers: to optimize \(f(x,y)\) subject to \(g(x,y)=0\), solve \(\nabla f = \lambda\nabla g\) together with the constraint.

Let \(g(x,y) = \dfrac{x^2}{8}+\dfrac{y^2}{2}-1=0\).

\[\nabla f = (y,x), \quad \nabla g = \left(\frac{x}{4},y\right).\]

System: \(y = \lambda\dfrac{x}{4}\) and \(x = \lambda y\).

From the first equation: \(\lambda = \dfrac{4y}{x}\) (if \(x\neq 0\)). Substitute into the second: \(x = \dfrac{4y}{x}\cdot y = \dfrac{4y^2}{x}\), so \(x^2=4y^2\), i.e., \(x=\pm 2y\).

Case \(x=2y\): Substitute into the ellipse: \(\dfrac{4y^2}{8}+\dfrac{y^2}{2}=\dfrac{y^2}{2}+\dfrac{y^2}{2}=y^2=1\), so \(y=\pm 1\). Points: \((2,1)\) and \((-2,-1)\), with \(f=2\).

Case \(x=-2y\): Similarly \(y^2=1\), points \((−2,1)\) and \((2,-1)\), with \(f=-2\).

Maximum: \(2\); Minimum: \(-2\).

4.89. Closest and Farthest Points on an Ellipse (Exercises, Task 12)

The plane \(x+y+z=1\) cuts the cylinder \(x^2+y^2=1\) in an ellipse. Find (using the method of Lagrange multipliers) the points on the ellipse that lie closest to and farthest from the origin.

Click to see the solution

Key Concept: Optimize \(d^2=x^2+y^2+z^2\) (squared distance from origin) subject to two constraints: \(g_1: x^2+y^2=1\) and \(g_2: x+y+z=1\). Use \(\nabla f = \lambda\nabla g_1+\mu\nabla g_2\).

\(f=x^2+y^2+z^2\), \(g_1=x^2+y^2-1\), \(g_2=x+y+z-1\).

\[\nabla f = (2x,2y,2z),\quad \nabla g_1=(2x,2y,0),\quad \nabla g_2=(1,1,1).\]

System: \[2x = 2\lambda x+\mu, \quad 2y=2\lambda y+\mu, \quad 2z=\mu.\]

From the third equation: \(\mu=2z\).

From the first: \(2x-2\lambda x=2z\Rightarrow x(1-\lambda)=z\).

From the second: \(y(1-\lambda)=z\).

If \(\lambda\neq 1\): \(x = y = \dfrac{z}{1-\lambda}\). Let \(x=y\). From \(x^2+y^2=1\): \(2x^2=1\), \(x=y=\pm\dfrac{1}{\sqrt{2}}\).

From \(x+y+z=1\): \(\dfrac{2}{\sqrt{2}}+z=1\), so \(z=1-\sqrt{2}\) (when \(x=y=\tfrac{1}{\sqrt{2}}\)) or \(z=1+\sqrt{2}\) (when \(x=y=-\tfrac{1}{\sqrt{2}}\)).

\(d^2\) values: - \(x=y=\tfrac{1}{\sqrt{2}}\), \(z=1-\sqrt{2}\): \(d^2=1+(1-\sqrt{2})^2=1+1-2\sqrt{2}+2=4-2\sqrt{2}\). - \(x=y=-\tfrac{1}{\sqrt{2}}\), \(z=1+\sqrt{2}\): \(d^2=1+(1+\sqrt{2})^2=1+3+2\sqrt{2}=4+2\sqrt{2}\).

Closest point: \(\bigl(\tfrac{1}{\sqrt{2}},\tfrac{1}{\sqrt{2}},1-\sqrt{2}\bigr)\), distance \(\sqrt{4-2\sqrt{2}}\).

Farthest point: \(\bigl(-\tfrac{1}{\sqrt{2}},-\tfrac{1}{\sqrt{2}},1+\sqrt{2}\bigr)\), distance \(\sqrt{4+2\sqrt{2}}\).

4.90. Maximize a Product via Lagrange Multipliers (Exercises, Task 13)

Use the method of Lagrange multipliers to find the maximum value of \(xy\), subject to the constraint \(x+y=16\).

Click to see the solution

Key Concept: Maximize \(f(x,y)=xy\) subject to \(g(x,y)=x+y-16=0\).

\(\nabla f=(y,x)\), \(\nabla g=(1,1)\). System: \(y=\lambda\) and \(x=\lambda\), so \(x=y\).

From the constraint: \(2x=16\), \(x=y=8\).

\[f(8,8) = 64.\]

Maximum value: \(64\) (achieved at \(x=y=8\)).

(By AM–GM, \(xy\leq\bigl(\tfrac{x+y}{2}\bigr)^2=64\), confirming this.)

4.91. Maximize with Two Constraints (Exercises, Task 14)

Maximize \(f(x,y,z)=x^2+2y-z^2\) subject to the constraints \(2x-y=0\) and \(y+z=0\).

Click to see the solution

Key Concept: With two constraints \(g_1\) and \(g_2\), use \(\nabla f = \lambda\nabla g_1+\mu\nabla g_2\).

\(g_1=2x-y\), \(g_2=y+z\). \(\nabla g_1=(2,-1,0)\), \(\nabla g_2=(0,1,1)\).

\(\nabla f=(2x,2,-2z)\).

System: \[2x = 2\lambda, \quad 2 = -\lambda+\mu, \quad -2z=\mu.\]

From constraint \(g_1=0\): \(y=2x\). From \(g_2=0\): \(z=-y=-2x\).

From the first equation: \(\lambda=x\). From the third: \(\mu=4x\). Substitute into the second: \(2=-x+4x=3x\), so \(x=\tfrac{2}{3}\).

Then \(y=\tfrac{4}{3}\), \(z=-\tfrac{4}{3}\).

\[f = \left(\frac{2}{3}\right)^2+2\cdot\frac{4}{3}-\left(\frac{4}{3}\right)^2 = \frac{4}{9}+\frac{8}{3}-\frac{16}{9} = \frac{4-16}{9}+\frac{24}{9} = \frac{12}{9} = \frac{4}{3}.\]

Maximum value: \(\dfrac{4}{3}\) at \(\bigl(\tfrac{2}{3},\tfrac{4}{3},-\tfrac{4}{3}\bigr)\).

4.92. Quadratic and Cubic Taylor Approximations (Exercises, Task 15)

Use Taylor’s formula for \(f(x,y)\) at the origin to find quadratic and cubic approximations of \(f\) near the origin: \[f(x,y) = e^x\cos y, \quad f(x,y) = \ln(2x+y+1).\]

Click to see the solution

Key Concept: Taylor’s formula for \(f(x,y)\) near the origin to order \(n\): \[f(x,y) \approx \sum_{k=0}^{n}\frac{1}{k!}\left(x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}\right)^k f\bigg|_{(0,0)}.\]

Function 1: \(f(x,y)=e^x\cos y\)

Use the known series \(e^x = 1+x+\tfrac{x^2}{2}+\tfrac{x^3}{6}+\cdots\) and \(\cos y = 1-\tfrac{y^2}{2}+\cdots\)

\[e^x\cos y = \left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots\right)\!\left(1-\frac{y^2}{2}+\cdots\right).\]

Collect terms up to degree 2 (quadratic): \[\approx 1+x+\frac{x^2}{2}-\frac{y^2}{2}.\]

Collect terms up to degree 3 (cubic): \[\approx 1+x+\frac{x^2}{2}-\frac{y^2}{2}+\frac{x^3}{6}-\frac{xy^2}{2}.\]

Quadratic: \(1+x+\dfrac{x^2-y^2}{2}\). Cubic: add \(\dfrac{x^3}{6}-\dfrac{xy^2}{2}\).

Function 2: \(f(x,y)=\ln(2x+y+1)\)

Let \(u=2x+y\). Use \(\ln(1+u)=u-\dfrac{u^2}{2}+\dfrac{u^3}{3}-\cdots\)

\[\ln(1+2x+y) = (2x+y)-\frac{(2x+y)^2}{2}+\frac{(2x+y)^3}{3}-\cdots\]

Quadratic: \((2x+y)-\dfrac{(2x+y)^2}{2} = 2x+y-\dfrac{4x^2+4xy+y^2}{2}.\)

Cubic: add \(+\dfrac{(2x+y)^3}{3} = \dfrac{8x^3+12x^2y+6xy^2+y^3}{3}\).

4.93. Quadratic Approximation and Error Estimate (Exercises, Task 16)

Use Taylor’s formula to find a quadratic approximation of \(f(x,y)=\cos x\cos y\) and \(f(x,y)=e^x\sin y\) at the origin. Estimate the error in the approximation if \(|x|\leq 0.1\) and \(|y|\leq 0.1\).

Click to see the solution

Key Concept: The error in a degree-\(n\) Taylor polynomial near the origin is bounded by the \((n+1)\)-th order terms, which are \(O(|x|^{n+1}+|y|^{n+1})\) for small \(x,y\).

Function 1: \(f(x,y)=\cos x\cos y\)

\[\cos x = 1-\frac{x^2}{2}+\cdots, \quad \cos y = 1-\frac{y^2}{2}+\cdots\]

\[\cos x\cos y = \left(1-\frac{x^2}{2}\right)\!\left(1-\frac{y^2}{2}\right)+\cdots \approx 1-\frac{x^2}{2}-\frac{y^2}{2}.\]

Quadratic approximation: \(Q(x,y) = 1-\dfrac{x^2+y^2}{2}\).

Error: the next terms are of degree 4 (the cross term \(\dfrac{x^2y^2}{4}\) and the degree-4 contributions from each cosine). For \(|x|,|y|\leq 0.1\): \[|R| \lesssim \frac{x^2 y^2}{4}+\frac{x^4}{24}+\frac{y^4}{24} \leq \frac{(0.1)^4}{4}+\frac{2(0.1)^4}{24} \approx 2.5\times10^{-4}+8.3\times10^{-5}\approx 3.3\times10^{-4}.\]

Function 2: \(f(x,y)=e^x\sin y\)

\[e^x\approx 1+x+\frac{x^2}{2},\quad \sin y\approx y-\frac{y^3}{6}.\]

\[e^x\sin y\approx \left(1+x+\frac{x^2}{2}\right)\!y+\cdots = y+xy+\frac{x^2 y}{2}+\cdots\]

Up to degree 2: \(Q(x,y)=y+xy\).

Error: the next terms involve \(y^3/6\) from \(\sin y\) and the product \(x^2y/2\) (degree 3). For \(|x|,|y|\leq 0.1\): \[|R|\lesssim \frac{|y|^3}{6}+\frac{x^2|y|}{2}\leq \frac{(0.1)^3}{6}+\frac{(0.1)^2(0.1)}{2} = \frac{10^{-3}}{6}+\frac{10^{-3}}{2}\approx 6.7\times10^{-4}.\]