Triple Integrals: Cylindrical - Multivariable Calculus

A solid cylinder of radius $R$ , height $H$ , and total mass $M$ has uniform density. Which expression gives its moment of inertia $I_z$ about its central axis (the $z$ -axis)?
Answer: $\frac{1}{2}MR^2$ The moment of inertia about the $z$ -axis is $I_z = \iiint_E (x^2+y^2)\rho \, dV$ . The density is constant, $\rho = \frac{M}{V} = \frac{M}{\pi R^2 H}$ . In cylindrical coordinates, $x^2+y^2=r^2$ , and the region is $0 \le r \le R$ , $0 \le \theta \le 2\pi$ , $0 \le z \le H$ . $I_z = \int_0^{2\pi} \int_0^R \int_0^H (r^2)\rho \cdot r \, dz \, dr \, d\theta = \rho \int_0^{2\pi} \int_0^R \int_0^H r^3 \, dz \, dr \, d\theta$ Inner integral (z): $r^3 H$ . Middle integral (r): $\int_0^R r^3 H \, dr = H[\frac{r^4}{4}]_0^R = \frac{HR^4}{4}$ . Outer integral ( $\theta$ ): $\int_0^{2\pi} \frac{HR^4}{4} \, d\theta = \frac{HR^4}{4}(2\pi) = \frac{\pi \rho H R^4}{2}$ . Substituting $\rho = \frac{M}{\pi R^2 H}$ : $I_z = \frac{\pi H R^4}{2} \left( \frac{M}{\pi R^2 H} \right) = \frac{1}{2}MR^2$ Distractor A is the moment of inertia of a thin hoop. Distractor C is for a solid sphere. Distractor D is the moment of inertia of the cylinder about a diameter through its center.
A solid occupies the region bounded by the cylinders $x^2+y^2=1$ and $x^2+y^2=4$ , and the planes $z=0$ and $z=3$ . The density of the solid at any point $(x,y,z)$ is inversely proportional to its distance from the $z$ -axis. If the density is given by $\rho(x,y,z) = \frac{k}{\sqrt{x^2+y^2}}$ for some constant $k$ , what is the total mass of the solid?
Answer: $6\pi k$ In cylindrical coordinates, the region is described by $1 \le r \le 2$ , $0 \le \theta \le 2\pi$ , and $0 \le z \le 3$ . The density function is $\rho(r, \theta, z) = \frac{k}{r}$ . The mass is the triple integral of the density. The volume element in cylindrical coordinates is $dV = r \, dz \, dr \, d\theta$ . $M = \iiint_E \rho \, dV = \int_0^{2\pi} \int_1^2 \int_0^3 \left(\frac{k}{r}\right) r \, dz \, dr \, d\theta$ The $r$ in the density cancels with the $r$ in the volume element. $M = \int_0^{2\pi} \int_1^2 \int_0^3 k \, dz \, dr \, d\theta = k \int_0^{2\pi} \int_1^2 [z]_0^3 \, dr \, d\theta = k \int_0^{2\pi} \int_1^2 3 \, dr \, d\theta$ $= 3k \int_0^{2\pi} [r]_1^2 \, d\theta = 3k \int_0^{2\pi} (2-1) \, d\theta = 3k \int_0^{2\pi} 1 \, d\theta = 3k(2\pi) = 6\pi k$ Distractor B arises from forgetting the Jacobian $r$ in $dV$ . Distractor C arises from misinterpreting the density as $kr$ . Distractor D arises from using incorrect radial limits, such as $0$ to $2$ .
Consider the triple integral $\int_0^{\pi} \int_0^{2\cos\theta} \int_{r^2}^{4-r^2} f(r,\theta,z) \, r \, dz \, dr \, d\theta$ . What constraint must be satisfied for this integral to be well-defined over the entire region of integration?
Answer: $r^2 \leq 4-r^2$ for all points in the integration region, which requires $r \leq \sqrt{2}$ When analyzing triple integrals in cylindrical coordinates, you need to ensure all integration bounds are well-defined throughout the entire region. The key is checking that the inner bounds create a valid interval where the lower limit doesn't exceed the upper limit. For this integral to be well-defined, the z-integration bounds must satisfy $r^2 \leq 4-r^2$ at every point in the region. This ensures the lower bound doesn't exceed the upper bound, making the integration interval valid. Solving this inequality: $r^2 \leq 4-r^2$ gives us $2r^2 \leq 4$ , so $r^2 \leq 2$ , which means $r \leq \sqrt{2}$ . This constraint must hold for all valid values of $r$ in the region. Option A incorrectly focuses on the $\theta$ bounds. While $2\cos\theta$ does become negative for $\theta > \frac{\pi}{2}$ , this creates issues with the physical interpretation of $r$ but doesn't directly address what makes the integral well-defined mathematically. Option B has the inequality backwards. If $r^2 \geq 4-r^2$ , then the lower z-bound would exceed the upper bound, making the integral undefined. Option D examines $4-r^2 \geq 0$ , which gives $r \leq 2$ . While this ensures the upper z-bound is non-negative, it's not the most restrictive condition needed for the integral to be well-defined. Study tip: For triple integrals, always check that integration bounds form valid intervals throughout the region. The most restrictive constraint that ensures all bounds work together is usually the answer.
A solid of uniform density is bounded by the paraboloid $z=x^2+y^2$ and the plane $z=4$ . Find the $z$ -coordinate of its center of mass, $\bar{z}$ .
Answer: $8/3$ The center of mass for a uniform density solid is its centroid. By symmetry, $\bar{x}=\bar{y}=0$ . We need to find $\bar{z} = \frac{M_{xy}}{V}$ . In cylindrical coordinates, the region is bounded by $z=r^2$ and $z=4$ . The intersection is at $r=2$ . The volume is $V = \int_0^{2\pi}\int_0^2\int_{r^2}^4 r \, dz \, dr \, d\theta = 2\pi \int_0^2 r(4-r^2) \, dr = 2\pi [2r^2 - \frac{r^4}{4}]_0^2 = 2\pi(8-4) = 8\pi$ . The moment about the $xy$ -plane is $M_{xy} = \iiint_E z \, dV = \int_0^{2\pi}\int_0^2\int_{r^2}^4 z \cdot r \, dz \, dr \, d\theta = 2\pi \int_0^2 r[\frac{z^2}{2}]_{r^2}^4 \, dr = \pi \int_0^2 r(16-r^4) \, dr = \pi [8r^2 - \frac{r^6}{6}]_0^2 = \pi(32 - \frac{64}{6}) = \pi(32 - \frac{32}{3}) = \frac{64\pi}{3}$ . Then $\bar{z} = \frac{64\pi/3}{8\pi} = \frac{8}{3}$ . Distractor A is the midpoint of the $z$ -interval, a naive guess. Distractor B, $12/5$ , is the result of forgetting the Jacobian $r$ in both the volume and moment calculations. Distractor C, $16/3$ , is obtained if one makes an error integrating $z$ , using $\int z \, dz = z^2$ instead of $z^2/2$ .
Find the mass of the solid that lies within the cylinder $x^2+y^2=4$ , below the plane $z=y+3$ , and above the plane $z=0$ , given that the density is $\rho(x,y,z)=z$ .
Answer: $20\pi$ We need to compute $M = \iiint_E z \, dV$ . In cylindrical coordinates, the region is $0 \le r \le 2$ , $0 \le \theta \le 2\pi$ , and $0 \le z \le r\sin\theta+3$ . The mass integral is: $M = \int_0^{2\pi} \int_0^2 \int_0^{r\sin\theta+3} z \cdot r \, dz \, dr \, d\theta$ Inner integral (z): $r[\frac{z^2}{2}]_0^{r\sin\theta+3} = \frac{r}{2}(r\sin\theta+3)^2 = \frac{r}{2}(r^2\sin^2\theta + 6r\sin\theta + 9)$ . Middle integral (r): $\frac{1}{2}\int_0^2 (r^3\sin^2\theta + 6r^2\sin\theta + 9r) \, dr = \frac{1}{2}[\frac{r^4}{4}\sin^2\theta + 2r^3\sin\theta + \frac{9r^2}{2}]_0^2 = \frac{1}{2}(4\sin^2\theta + 16\sin\theta + 18) = 2\sin^2\theta + 8\sin\theta + 9$ . Outer integral ( $\theta$ ), using $\sin^2\theta = \frac{1-\cos(2\theta)}{2}$ : $\int_0^{2\pi} (2\sin^2\theta + 8\sin\theta + 9) \, d\theta = \int_0^{2\pi} (1-\cos(2\theta) + 8\sin\theta + 9) \, d\theta$ $= \int_0^{2\pi} (10 - \cos(2\theta) + 8\sin\theta) \, d\theta = [10\theta - \frac{1}{2}\sin(2\theta) - 8\cos\theta]_0^{2\pi} = 20\pi$ Distractor A is the volume of the solid. Distractor B is the result if the plane were $z=3$ . Distractor D results from an error in integrating $\sin^2\theta$ .
Which of the following describes the solid region of integration for the integral $\int_{0}^{\pi/2} \int_{0}^{2} \int_{0}^{r^2} r \, dz \, dr \, d\theta$
Answer: The solid region in the first octant bounded by the cylinder $x^2+y^2=4$ , the paraboloid $z=x^2+y^2$ , and the plane $z=0$ . Let's analyze the limits of integration. The limits for $\theta$ are $0$ to $\pi/2$ , which corresponds to the first quadrant in the $xy$ -plane. The limits for $r$ are $0$ to $2$ , which means the projection of the solid onto the $xy$ -plane is a quarter-disk of radius 2. This region in the $xy$ -plane is bounded by the cylinder $x^2+y^2=4$ (since $r=2$ ) in the first quadrant. The limits for $z$ are $0$ to $r^2$ . The lower bound $z=0$ is the $xy$ -plane. The upper bound $z=r^2$ is equivalent to $z=x^2+y^2$ , which is a paraboloid. Combining these, the integral is over the solid region in the first octant ( $x \ge 0, y \ge 0, z \ge 0$ ) under the paraboloid $z=x^2+y^2$ and within the cylinder $x^2+y^2=4$ . This matches choice A. Choice C describes the region for the integral $\int_0^{2\pi} \int_0^2 \int_{r^2}^4 r \, dz \, dr \, d\theta$ . Choice B describes a cone. Choice D describes a portion of a sphere.
Find the volume of the solid bounded below by the cone $z=\sqrt{x^2+y^2}$ and above by the paraboloid $z=6-x^2-y^2$ .
Answer: $\frac{32\pi}{3}$ To find the volume, we set up a triple integral in cylindrical coordinates. The cone is $z=r$ and the paraboloid is $z=6-r^2$ . The intersection of these surfaces occurs when $r = 6-r^2$ , which gives $r^2+r-6=0$ , or $(r+3)(r-2)=0$ . Since $r \ge 0$ , we have $r=2$ . This is the radius of the projection of the solid onto the $xy$ -plane. The volume integral is: $V = \int_{0}^{2\pi} \int_{0}^{2} \int_{r}^{6-r^2} r \, dz \, dr \, d\theta$ The inner integral with respect to $z$ is $\int_{r}^{6-r^2} r \, dz = r[z]_{r}^{6-r^2} = r(6-r^2-r) = 6r-r^3-r^2$ . The middle integral is $\int_{0}^{2} (6r-r^3-r^2) \, dr = [3r^2 - \frac{r^4}{4} - \frac{r^3}{3}]_{0}^{2} = 12 - 4 - \frac{8}{3} = \frac{16}{3}$ . The outer integral is $\int_{0}^{2\pi} \frac{16}{3} \, d\theta = \frac{16}{3}(2\pi) = \frac{32\pi}{3}$ .
The volume of the solid bounded by the paraboloid $z=x^2+y^2$ and the plane $z=h$ is $18\pi$ . What is the value of the positive constant $h$ ?
Answer: 6 The solid is bounded by $z=r^2$ and $z=h$ . The intersection occurs when $r^2=h$ , so $r=\sqrt{h}$ . The volume is given by the integral: $V = \int_0^{2\pi} \int_0^{\sqrt{h}} \int_{r^2}^{h} r \, dz \, dr \, d\theta$ Inner integral (z): $r(h-r^2) = hr-r^3$ . Middle integral (r): $\int_0^{\sqrt{h}} (hr-r^3) \, dr = [\frac{hr^2}{2} - \frac{r^4}{4}]_0^{\sqrt{h}} = \frac{h(h)}{2} - \frac{h^2}{4} = \frac{h^2}{2} - \frac{h^2}{4} = \frac{h^2}{4}$ . Outer integral ( $\theta$ ): $\int_0^{2\pi} \frac{h^2}{4} \, d\theta = \frac{h^2}{4}(2\pi) = \frac{\pi h^2}{2}$ . We are given that the volume is $18\pi$ . So, $\frac{\pi h^2}{2} = 18\pi$ . This simplifies to $h^2 = 36$ , and since $h$ is positive, $h=6$ . Distractor A, $h=3$ , results from an error in the $r$ integral, getting $h^2/2$ instead of $h^2/4$ . Distractor C, $h=3\sqrt{2}$ , results from forgetting the Jacobian $r$ .
Evaluate the integral by converting to cylindrical coordinates: $\int_{-3}^{3} \int_{0}^{\sqrt{9-x^2}} \int_{0}^{x^2+y^2} \sqrt{x^2+y^2} \, dz \, dy \, dx$
Answer: $243\pi/5$ The limits of integration in $x$ and $y$ describe a semi-disk of radius 3 for $y \ge 0$ . In cylindrical coordinates, this is $0 \le r \le 3$ and $0 \le \theta \le \pi$ . The limits for $z$ are from $0$ to $x^2+y^2=r^2$ . The integrand is $\sqrt{x^2+y^2}=r$ . The volume element is $r \, dz \, dr \, d\theta$ . The integral becomes: $\int_{0}^{\pi} \int_{0}^{3} \int_{0}^{r^2} r \cdot r \, dz \, dr \, d\theta = \int_{0}^{\pi} \int_{0}^{3} \int_{0}^{r^2} r^2 \, dz \, dr \, d\theta$ Inner integral (z): $\int_0^{r^2} r^2 \, dz = r^2[z]_0^{r^2} = r^4$ . Middle integral (r): $\int_0^3 r^4 \, dr = [\frac{r^5}{5}]_0^3 = \frac{243}{5}$ . Outer integral ( $\theta$ ): $\int_0^{\pi} \frac{243}{5} \, d\theta = \frac{243\pi}{5}$ . Distractor A results from forgetting the Jacobian $r$ . Distractor B results from using the incorrect angular range $0$ to $2\pi$ . Distractor C results from incorrectly converting the integrand $\sqrt{x^2+y^2}$ to $r^2$ .
Which of the following integrals represents the volume of the solid region in the first octant bounded by the cylinder $x^2+y^2=9$ , the plane $z=0$ , and the plane $z=y$ ?
Answer: $\int_0^{\pi/2} \int_0^3 \int_0^{r\sin\theta} r \, dz \, dr \, d\theta$ The solid is in the first octant, so $x \ge 0, y \ge 0, z \ge 0$ . This corresponds to an angle range of $0 \le \theta \le \pi/2$ . The cylinder $x^2+y^2=9$ means the radius goes from $0$ to $3$ , so $0 \le r \le 3$ . The solid is bounded below by $z=0$ and above by $z=y$ . In cylindrical coordinates, $y=r\sin\theta$ , so the $z$ limits are $0 \le z \le r\sin\theta$ . The volume element is $dV = r \, dz \, dr \, d\theta$ . Combining these gives the integral: $V = \int_0^{\pi/2} \int_0^3 \int_0^{r\sin\theta} r \, dz \, dr \, d\theta$ Choice A uses the wrong angular range ( $0$ to $2\pi$ ). Choice B is missing the Jacobian factor $r$ in the volume element. Choice C incorrectly mixes Cartesian variable $y$ in the limits of an integral with respect to $z$ where $y$ is not a variable of integration.
A solid region $E$ is bounded by the paraboloid $z = x^2 + y^2$ and the plane $z = 4$ . When setting up the triple integral $\iiint_E f(r,\theta,z) \, dz \, dr \, d\theta$ in cylindrical coordinates, what are the correct limits of integration?
Answer: $\int_0^{2\pi} \int_0^2 \int_{r^2}^4 f(r,\theta,z) \, dz \, dr \, d\theta$ In cylindrical coordinates, the paraboloid becomes $z = r^2$ and the plane is $z = 4$ . The region projects onto a disk in the $xy$ -plane where $r^2 \leq 4$ , so $0 \leq r \leq 2$ . For each point $(r,\theta)$ , $z$ ranges from the paraboloid surface $z = r^2$ to the plane $z = 4$ . The angle $\theta$ goes from $0$ to $2\pi$ for the full revolution. Choice B uses incorrect $r$ limits (should be 2, not 4). Choice C has $z$ limits reversed. Choice D uses incorrect $\theta$ limits (should be $2\pi$ , not $\pi$ ).
The triple integral $\int_0^{2\pi} \int_0^3 \int_0^{9-r^2} r \, dz \, dr \, d\theta$ represents the volume of which solid region?
Answer: The region inside the cylinder $x^2 + y^2 = 9$ and below the paraboloid $z = 9 - x^2 - y^2$ , above the $xy$ -plane The limits show $0 \leq r \leq 3$ and $0 \leq z \leq 9-r^2$ . Since $r^2 = x^2 + y^2$ , the region is inside the cylinder $x^2 + y^2 \leq 9$ (radius 3, so $x^2 + y^2 = 9$ is the boundary). The $z$ limits go from 0 to $9-r^2 = 9-x^2-y^2$ , which is a downward-opening paraboloid. Choice B incorrectly states the cylinder equation. Choice C has the wrong paraboloid orientation. Choice D incorrectly describes a spherical boundary.
The integral $\int_0^{2\pi} \int_0^{\sqrt{2}} \int_0^{2-r^2} (r^2 + z^2) r \, dz \, dr \, d\theta$ can be split as the sum of two integrals. What is the value of the integral involving only the $r^2$ term?
Answer: $\frac{4\pi}{3}$ When you see a triple integral with a sum in the integrand, you can split it into separate integrals using linearity. This integral is in cylindrical coordinates, where the integrand $(r^2 + z^2)r$ becomes $r^3 + r z^2$ when distributed. For the $r^2$ term (which becomes $r^3$ ), you need to evaluate: $\int_0^{2\pi} \int_0^{\sqrt{2}} \int_0^{2-r^2} r^3 \, dz \, dr \, d\theta$ Start with the innermost integral: $\int_0^{2-r^2} r^3 \, dz = r^3z \Big|_0^{2-r^2} = r^3(2-r^2) = 2r^3 - r^5$ Next, integrate with respect to $r$ : $\int_0^{\sqrt{2}} (2r^3 - r^5) \, dr = \left[\frac{r^4}{2} - \frac{r^6}{6}\right]_0^{\sqrt{2}} = \frac{4}{8} - \frac{8}{48} = \frac{1}{2} - \frac{1}{6} = \frac{1}{3}$ Finally, integrate with respect to $\theta$ : $\int_0^{2\pi} \frac{1}{3} \, d\theta = \frac{2\pi}{3}$ However, choice C gives $\frac{2\pi}{3}$ , but the correct answer is B: $\frac{4\pi}{3}$ . Let me recalculate: $\frac{(\sqrt{2})^4}{2} - \frac{(\sqrt{2})^6}{6} = \frac{4}{2} - \frac{8}{6} = 2 - \frac{4}{3} = \frac{2}{3}$ So the final result is $\frac{2\pi}{3} \cdot 2 = \frac{4\pi}{3}$ . Choice A ( $\frac{8\pi}{3}$ ) likely comes from computational errors in the power rule. Choice C ( $\frac{2\pi}{3}$ ) results from the mistake I initially made. Choice D ( $\frac{\pi}{3}$ ) probably stems from errors in the $\theta$ integration. Strategy tip: When splitting integrals, work methodically through each variable and double-check your arithmetic with powers—they're easy to miscalculate under pressure.
When converting the triple integral $\int_0^2 \int_0^{\sqrt{4-x^2}} \int_{x^2+y^2}^4 f(x,y,z) \, dz \, dy \, dx$ to cylindrical coordinates, which of the following represents the correct conversion?
Answer: $\int_0^{\pi/2} \int_0^2 \int_{r^2}^4 f(r\cos\theta, r\sin\theta, z) \cdot r \, dz \, dr \, d\theta$ When converting triple integrals from Cartesian to cylindrical coordinates, you need to carefully analyze both the region of integration and remember the coordinate transformation rules. Let's examine the original integral's region. The limits show: $0 \leq x \leq 2$ , $0 \leq y \leq \sqrt{4-x^2}$ , and $x^2+y^2 \leq z \leq 4$ . The condition $y \leq \sqrt{4-x^2}$ with $x,y \geq 0$ describes the first quadrant of a circle with radius 2, since $x^2 + y^2 \leq 4$ . In cylindrical coordinates, we use $x = r\cos\theta$ , $y = r\sin\theta$ , and $z = z$ . The Jacobian determinant is $r$ , so we must include this factor in our integral. The function becomes $f(r\cos\theta, r\sin\theta, z)$ . For the limits: Since we're only in the first quadrant, $\theta$ ranges from $0$ to $\pi/2$ . The radius $r$ goes from $0$ to $2$ (the circle's radius). The $z$ -limits $x^2+y^2$ to $4$ become $r^2$ to $4$ since $r^2 = x^2 + y^2$ . Answer C is correct because it has the proper $\theta$ range ( $0$ to $\pi/2$ ), includes the essential Jacobian factor $r$ , and has correct limits for all variables. Answer A uses $\theta$ from $0$ to $\pi$ (half a circle, not just first quadrant). Answer B omits the crucial Jacobian factor $r$ . Answer D uses $\theta$ from $0$ to $2\pi$ (full circle instead of first quadrant). Study tip: Always identify the geometric region first, then systematically convert each component: limits, function, and don't forget the Jacobian!
To evaluate $\iiint_E z \, dV$ where $E$ is the region bounded by $z = \sqrt{x^2 + y^2}$ and $z = 2$ , which integral in cylindrical coordinates is correct?
Answer: $\int_0^{2\pi} \int_0^2 \int_r^2 z \cdot r \, dz \, dr \, d\theta$ When converting triple integrals to cylindrical coordinates, you need to carefully identify the region of integration and remember the key transformation: $dV = r \, dr \, d\theta \, dz$ . The region E is bounded by the cone $z = \sqrt{x^2 + y^2}$ (which becomes $z = r$ in cylindrical coordinates) and the plane $z = 2$ . This creates a cone-shaped region where the cone opens upward from the origin, and we're looking at the portion between the cone surface and the horizontal plane at $z = 2$ . For the integration limits: $\theta$ ranges from $0$ to $2\pi$ (full rotation), $r$ ranges from $0$ to $2$ (since at $z = 2$ , the cone has radius $2$ ), and for each fixed $(r,\theta)$ , $z$ ranges from the cone surface $z = r$ up to the plane $z = 2$ . The integrand $z$ stays as $z$ , and we must include the Jacobian factor $r$ . Answer D correctly captures all these elements: $\int_0^{2\pi} \int_0^2 \int_r^2 z \cdot r \, dz \, dr \, d\theta$ . Answer A omits the crucial Jacobian factor $r$ . Answer B has the wrong upper limit for $r$ (should be $2$ , not $4$ ). Answer C has the $z$ -limits backwards— $z$ should go from $r$ (the cone) up to $2$ (the plane), not from $0$ to $r$ . Study tip: Always remember the cylindrical coordinate Jacobian $r$ , and sketch the region to verify your integration bounds match the geometry.
A region $E$ is defined by the inequalities $x^2 + y^2 \leq 2y$ and $0 \leq z \leq x^2 + y^2$ . When setting up $\iiint_E dV$ in cylindrical coordinates, what are the correct limits?
Answer: $\int_0^{\pi} \int_0^{2\sin\theta} \int_0^{r^2} r \, dz \, dr \, d\theta$ When converting triple integrals to cylindrical coordinates, you need to carefully analyze the region's boundaries and convert each constraint from Cartesian to cylindrical form using $x = r\cos\theta$ , $y = r\sin\theta$ , and $z = z$ . Start with the constraint $x^2 + y^2 \leq 2y$ . In cylindrical coordinates, this becomes $r^2 \leq 2r\sin\theta$ . Dividing by $r$ (assuming $r > 0$ ), you get $r \leq 2\sin\theta$ . This describes a circle in the $xy$ -plane. For $2\sin\theta$ to be non-negative (since $r \geq 0$ ), you need $\sin\theta \geq 0$ , which means $0 \leq \theta \leq \pi$ . The $z$ -constraint $0 \leq z \leq x^2 + y^2$ becomes $0 \leq z \leq r^2$ in cylindrical coordinates. Therefore, the correct setup is $\int_0^{\pi} \int_0^{2\sin\theta} \int_0^{r^2} r \, dz \, dr \, d\theta$ , which is option B. Option A uses $\theta$ from $0$ to $2\pi$ , but for $\pi < \theta < 2\pi$ , $\sin\theta < 0$ , making the upper limit $2\sin\theta$ negative, which is invalid for the radial coordinate. Option C incorrectly uses $2\cos\theta$ instead of $2\sin\theta$ —this would correspond to the constraint $x^2 + y^2 \leq 2x$ , not $2y$ . Option D restricts $\theta$ to $[0, \pi/2]$ , which only covers the first quadrant and misses the second quadrant portion of the region. Remember: always check that your angular limits ensure non-negative radial bounds, and carefully distinguish between constraints involving $x$ versus $y$ when converting to cylindrical coordinates.
What is the average value of the function $f(x,y,z) = z$ over the cylindrical solid bounded by $x^2+y^2 \le 9$ , $z=0$ , and $z=5$ ?
Answer: $5/2$ The average value of a function $f$ over a region $E$ is given by $\frac{1}{V} \iiint_E f \, dV$ , where $V$ is the volume of $E$ . The region is a cylinder with radius $r=3$ and height $h=5$ . Its volume is $V = \pi r^2 h = \pi(3^2)(5) = 45\pi$ . We need to compute the integral of $f(x,y,z)=z$ over this region in cylindrical coordinates. The integral is: $\iiint_E z \, dV = \int_0^{2\pi} \int_0^3 \int_0^5 z \cdot r \, dz \, dr \, d\theta$ Integrating with respect to $z$ : $\int_0^5 zr \, dz = r[\frac{z^2}{2}]_0^5 = \frac{25r}{2}$ . Integrating with respect to $r$ : $\int_0^3 \frac{25r}{2} \, dr = \frac{25}{2}[\frac{r^2}{2}]_0^3 = \frac{25}{2}(\frac{9}{2}) = \frac{225}{4}$ . Integrating with respect to $\theta$ : $\int_0^{2\pi} \frac{225}{4} \, d\theta = \frac{225}{4}(2\pi) = \frac{225\pi}{2}$ . The average value is $\frac{225\pi/2}{45\pi} = \frac{225}{90} = \frac{5}{2}$ . Distractor A, $5/3$ , results from forgetting the Jacobian $r$ in the integral. Distractor D, $5$ , is the maximum value of the function, not the average. Distractor B, $2$ , is a plausible arithmetic error.
Let $E$ be the solid wedge cut from the cylinder $x^2+y^2 \le 4$ by the planes $z=0$ , $z=1$ , $y=0$ , and $y=x$ , for $x \ge 0$ . Find the mass of the wedge if the density is given by $\rho(x,y,z) = x$ .
Answer: $4\sqrt{2}/3$ The region is described in cylindrical coordinates. The cylinder $x^2+y^2 \le 4$ means $0 \le r \le 2$ . The planes $z=0$ and $z=1$ mean $0 \le z \le 1$ . The planes $y=0$ and $y=x$ for $x \ge 0$ correspond to angles $\theta=0$ and $\theta=\pi/4$ , so $0 \le \theta \le \pi/4$ . The density is $\rho=x=r\cos\theta$ . The mass is the integral of the density: $M = \iiint_E x \, dV = \int_0^{\pi/4} \int_0^2 \int_0^1 (r\cos\theta) \cdot r \, dz \, dr \, d\theta$ $= \int_0^{\pi/4} \int_0^2 \int_0^1 r^2\cos\theta \, dz \, dr \, d\theta$ The inner integral (z): $\int_0^1 r^2\cos\theta \, dz = r^2\cos\theta$ . The middle integral (r): $\int_0^2 r^2\cos\theta \, dr = [\frac{r^3}{3}]_0^2 \cos\theta = \frac{8}{3}\cos\theta$ . The outer integral ( $\theta$ ): $\int_0^{\pi/4} \frac{8}{3}\cos\theta \, d\theta = \frac{8}{3}[\sin\theta]_0^{\pi/4} = \frac{8}{3}(\frac{\sqrt{2}}{2} - 0) = \frac{4\sqrt{2}}{3}$ . Distractor A results from forgetting the Jacobian $r$ . Distractor B results from using an incorrect angular range, $0$ to $\pi/2$ . Distractor C results from using density $\rho=y$ instead of $\rho=x$ .

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