Multivariable Calculus Quiz: Triple Integrals Cylindrical

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Triple Integrals CylindricalQuestion 1 of 18

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?

6\pi k

6\pi k \ln(2)

14\pi k

12\pi k

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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Triple Integrals Cylindrical

Practice Triple Integrals Cylindrical in Multivariable Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Triple Integrals Cylindrical, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$6\pi k$ (correct answer)
$6\pi k \ln(2)$
$14\pi k$
$12\pi k$

Explanation: 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

dV

. Distractor C arises from misinterpreting the density as

kr

. Distractor D arises from using incorrect radial limits, such as

0

2

Question 2

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 $\rho(x,y,z) = x$ .

$\sqrt{2}$
$8/3$
$4(2-\sqrt{2})/3$
$4\sqrt{2}/3$ (correct answer)

Explanation: 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

\pi/2

. Distractor C results from using density

\rho=y

instead of

\rho=x

Question 3

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?

$\int_0^{2\pi} \int_0^2 \int_{r^2}^4 f(r,\theta,z) \, dz \, dr \, d\theta$ (correct answer)
$\int_0^{2\pi} \int_0^4 \int_{r^2}^4 f(r,\theta,z) \, dz \, dr \, d\theta$
$\int_0^{2\pi} \int_0^2 \int_0^{r^2} f(r,\theta,z) \, dz \, dr \, d\theta$
$\int_0^{\pi} \int_0^2 \int_{r^2}^4 f(r,\theta,z) \, dz \, dr \, d\theta$

Explanation: 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

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

Question 4

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?

$\frac{8\pi}{3}$
$\frac{4\pi}{3}$ (correct answer)
$\frac{2\pi}{3}$
$\frac{\pi}{3}$

Explanation: 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.

Question 5

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?

$2\cos\theta \geq 0$ for all $\theta \in [0,\pi]$ , which requires $\theta \leq \frac{\pi}{2}$
$r^2 \geq 4-r^2$ for all points in the integration region, which requires $r \geq \sqrt{2}$
$r^2 \leq 4-r^2$ for all points in the integration region, which requires $r \leq \sqrt{2}$ (correct answer)
$4-r^2 \geq 0$ for all points in the integration region, which requires $r \leq 2$

Explanation: 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.

Question 6

Which of the following describes the solid region of integration for the integral $\int_{0}^{\pi/2} \int_{0}^{2} \int_{0}^{r^2} r \, dz \, dr \, d\theta$

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 $z=0$ . (correct answer)
The solid region bounded by the cone $z^2=x^2+y^2$ and the plane $z=4$ .
The solid region inside the cylinder $x^2+y^2=4$ for $z \ge 0$ , bounded above by the plane $z=4$ and below by the paraboloid $z=x^2+y^2$ .
The solid region in the first octant bounded by the sphere $x^2+y^2+z^2=4$ and the coordinate planes.

Explanation: Let's analyze the limits of integration. The limits for

\theta

are

0

\pi/2

, which corresponds to the first quadrant in the

xy

-plane. The limits for

r

are

0

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

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.

Question 7

Evaluate the integral by converting to cylindrical coordinates: $\int_{-3}^{3} \int_{0}^{\sqrt{9-x^2}} \int_{0}^{x^2+y^2} \sqrt{x^2+y^2} \, dz \, dy \, dx$

$81\pi/4$
$486\pi/5$
$243\pi/2$
$243\pi/5$ (correct answer)

Explanation: 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

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

2\pi

. Distractor C results from incorrectly converting the integrand

\sqrt{x^2+y^2}

r^2

Question 8

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)?

$MR^2$
$\frac{1}{2}MR^2$ (correct answer)
$\frac{2}{5}MR^2$
$\frac{1}{4}MR^2 + \frac{1}{3}MH^2$

Explanation: 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.

Question 9

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 $h$ ?

3
6 (correct answer)
$3\sqrt{2}$
$3\sqrt[3]{2}$

Explanation: 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

Question 10

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 $\rho(x,y,z)=z$ .

$12\pi$
$18\pi$
$20\pi$ (correct answer)
$22\pi$

Explanation: 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

Question 11

Find the volume of the solid bounded below by the cone $z=\sqrt{x^2+y^2}$ and above by the paraboloid $z=6-x^2-y^2$ .

$\frac{16\pi}{3}$
$\frac{32\pi}{3}$ (correct answer)
$16\pi$
$\frac{44\pi}{3}$

Explanation: 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

\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}

Question 12

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 $z=5$ ?

$5/3$
$2$
$5/2$ (correct answer)
$5$

Explanation: 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.

Question 13

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 $z=y$ ?

$\int_0^{2\pi} \int_0^3 \int_0^{r\sin\theta} r \, dz \, dr \, d\theta$
$\int_0^{\pi/2} \int_0^3 \int_0^{r\sin\theta} \, dz \, dr \, d\theta$
$\int_0^{\pi/2} \int_0^3 \int_0^{y} r \, dz \, dr \, d\theta$
$\int_0^{\pi/2} \int_0^3 \int_0^{r\sin\theta} r \, dz \, dr \, d\theta$ (correct answer)

Explanation: 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

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

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.

Question 14

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, $\bar{z}$ .

$2$
$12/5$
$16/3$
$8/3$ (correct answer)

Explanation: 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

Question 15

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?

The region inside the cylinder $x^2 + y^2 = 9$ and below the paraboloid $z = 9 - x^2 - y^2$ , above the $xy$ -plane (correct answer)
The region inside the cylinder $x^2 + y^2 = 3$ and below the paraboloid $z = 9 - x^2 - y^2$ , above the $xy$ -plane
The region inside the cylinder $x^2 + y^2 = 9$ and above the paraboloid $z = x^2 + y^2 - 9$ , below the plane $z = 0$
The region inside the sphere $x^2 + y^2 + z^2 = 9$ and below the paraboloid $z = 9 - x^2 - y^2$

Explanation: 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.

Question 16

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?

$\int_0^{2\pi} \int_0^2 \int_r^2 z \, dz \, dr \, d\theta$
$\int_0^{2\pi} \int_0^4 \int_r^2 z \cdot r \, dz \, dr \, d\theta$
$\int_0^{2\pi} \int_0^2 \int_0^r z \cdot r \, dz \, dr \, d\theta$
$\int_0^{2\pi} \int_0^2 \int_r^2 z \cdot r \, dz \, dr \, d\theta$ (correct answer)

Explanation: 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

2\pi

(full rotation),

r

ranges from

0

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

r

. Study tip: Always remember the cylindrical coordinate Jacobian

r

, and sketch the region to verify your integration bounds match the geometry.

Question 17

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?

$\int_0^{2\pi} \int_0^{2\sin\theta} \int_0^{r^2} r \, dz \, dr \, d\theta$
$\int_0^{\pi} \int_0^{2\sin\theta} \int_0^{r^2} r \, dz \, dr \, d\theta$ (correct answer)
$\int_0^{\pi} \int_0^{2\cos\theta} \int_0^{r^2} r \, dz \, dr \, d\theta$
$\int_0^{\pi/2} \int_0^{2\sin\theta} \int_0^{r^2} r \, dz \, dr \, d\theta$

Explanation: 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

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

[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.

Question 18

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?

$\int_0^{\pi} \int_0^2 \int_{r^2}^4 f(r\cos\theta, r\sin\theta, z) \cdot r \, dz \, dr \, d\theta$
$\int_0^{\pi/2} \int_0^2 \int_{r^2}^4 f(r\cos\theta, r\sin\theta, z) \, dz \, dr \, d\theta$
$\int_0^{\pi/2} \int_0^2 \int_{r^2}^4 f(r\cos\theta, r\sin\theta, z) \cdot r \, dz \, dr \, d\theta$ (correct answer)
$\int_0^{2\pi} \int_0^2 \int_{r^2}^4 f(r\cos\theta, r\sin\theta, z) \cdot r \, dz \, dr \, d\theta$

Explanation: 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

\pi/2

. The radius

r

goes from

0

2

(the circle's radius). The

z

-limits

x^2+y^2

4

become

r^2

4

since

r^2 = x^2 + y^2

. Answer C is correct because it has the proper

\theta

range (

0

\pi/2

), includes the essential Jacobian factor

r

, and has correct limits for all variables. Answer A uses

\theta

from

0

\pi

(half a circle, not just first quadrant). Answer B omits the crucial Jacobian factor

r

. Answer D uses

\theta

from

0

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!