Calculus 2 Quiz: Shell Method

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Shell MethodQuestion 1 of 20

Let R be the region bounded by $y=1/x^2$ , $y=0$ , $x=1$ , and $x=2$ . Let $V_1$ be the volume when R is revolved about the y-axis and $V_2$ be the volume when R is revolved about the x-axis. Which statement is true?

V_1 = 2\pi \ln 2

and

V_2 = 7\pi/3

V_1 = \pi \ln 2

and

V_2 = 7\pi/24

V_1 = 2\pi \ln 2

and

V_2 = 7\pi/24

V_1 = \pi/2

and

V_2 = 7\pi/3

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Calculus 2 Quiz

Calculus 2 Quiz: Shell Method

Practice Shell Method in Calculus 2 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 Shell Method, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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

$V_1 = 2\pi \ln 2$ and $V_2 = 7\pi/3$
$V_1 = \pi \ln 2$ and $V_2 = 7\pi/24$
$V_1 = 2\pi \ln 2$ and $V_2 = 7\pi/24$ (correct answer)
$V_1 = \pi/2$ and $V_2 = 7\pi/3$

Explanation:

V_1

(revolve about y-axis): Use the shell method.

V_1 = \int_1^2 2\pi x (1/x^2) \,dx = \int_1^2 2\pi (1/x) \,dx = 2\pi [\ln|x|]_1^2 = 2\pi(\ln 2 - \ln 1) = 2\pi \ln 2

V_2

(revolve about x-axis): Use the disk method.

V_2 = \int_1^2 \pi (1/x^2)^2 \,dx = \int_1^2 \pi x^{-4} \,dx = \pi [x^{-3}/(-3)]_1^2 = -\frac{\pi}{3} [1/x^3]_1^2 = -\frac{\pi}{3} (1/8 - 1) = -\frac{\pi}{3} (-7/8) = 7\pi/24

. Thus,

V_1 = 2\pi \ln 2

and

V_2 = 7\pi/24

Question 2

A region in the first quadrant is bounded by $x = y^2$ and $x = 8 - y^2$ . When revolved around the line $x = -1$ , the volume using the shell method requires integration with respect to which variable and over what interval?

Integration with respect to $y$ over $[0, 2]$ (correct answer)
Integration with respect to $x$ over $[0, 4]$
Integration with respect to $y$ over $[0, 4]$
Integration with respect to $x$ over $[0, 8]$

Explanation: The curves intersect where

y^2 = 8 - y^2

, so

2y^2 = 8

and

y = 2

(in the first quadrant). For shell method around

x = -1

, we can use horizontal shells. Each shell has radius

r = y - (-1) = y + 1

and height

h = (8 - y^2) - y^2 = 8 - 2y^2

. We integrate with respect to

y

from

0

2

. Choice B would give vertical shells, but that's more complex since we'd need to solve for

y

in terms of

x

. Choice C uses wrong limits (should be 2, not 4). Choice D also uses wrong variable and limits.

Question 3

The region bounded by $y=x^2$ and $y=2x$ is revolved about the line $x=3$ . The volume of the resulting solid is given by $V = \int_0^2 2\pi (3-x)h(x) \,dx$ . What is the correct function for $h(x)$ ?

$h(x) = x^2 - 2x$
$h(x) = 2x - x^2$ (correct answer)
$h(x) = (2x-x^2)(3-x)$
$h(x) = y$

Explanation: The problem uses the shell method, as indicated by the form of the integral. The term

h(x)

represents the height of the cylindrical shells. The region is bounded by

y=2x

(the upper curve) and

y=x^2

(the lower curve) on the interval where they intersect, which is

[0, 2]

. The height of a vertical strip at a given

x

is the difference between the upper function and the lower function. Therefore,

h(x) = 2x - x^2

Question 4

Let R be the region bounded by $y=e^{x^2}$ , $y=0$ , $x=0$ , and $x=1$ . When revolving R about the y-axis, the shell method is preferable. If the volume is approximated by $n$ cylindrical shells of equal thickness, what is the volume of the $i$ -th shell (where $x_i$ is the right endpoint of the $i$ -th subinterval)?

$V_i = \pi (x_i^2 - x_{i-1}^2) e^{x_i^2}$
$V_i = \frac{2\pi}{n} x_i e^{x_i^2}$ (correct answer)
$V_i = \frac{2\pi}{n^2} i e^{(i/n)^2}$
$V_i = \frac{\pi}{n^2} i^2 e^{(i/n)^2}$

Explanation: The interval

[0,1]

is divided into

n

subintervals of thickness

\Delta x = 1/n

. The right endpoint of the

i

-th interval is

x_i = i/n

. For the shell method, the volume of the

i

-th shell is approximated by

2\pi r_i h_i \Delta x

. Here, the radius is

r_i = x_i

and the height is

h_i = e^{x_i^2}

. The thickness is

\Delta x = 1/n

. So, the volume of the

i

-th shell is

V_i = 2\pi x_i e^{x_i^2} \Delta x = 2\pi (i/n) e^{(i/n)^2} (1/n) = \frac{2\pi i}{n^2} e^{(i/n)^2}

. Option B uses

x_i

instead of

i/n

, which is a more general representation:

V_i = 2\pi x_i e^{x_i^2} (1/n) = \frac{2\pi}{n} x_i e^{x_i^2}

. Let's re-examine C. It's

\frac{2\pi i}{n^2} e^{(i/n)^2}

. This matches my derivation. Let's check B again.

\frac{2\pi}{n} x_i e^{x_i^2} = \frac{2\pi}{n} (i/n) e^{(i/n)^2} = \frac{2\pi i}{n^2} e^{(i/n)^2}

. So B and C are equivalent if we substitute

x_i=i/n

. B is a slightly better form. I will choose B. The key is

2\pi r h \Delta x

Question 5

The integral $V = \int_1^3 2\pi (4-x)(x^2-1) \,dx$ represents the volume of a solid of revolution. What is the region being revolved and the axis of revolution?

The region bounded by $y=x^2-1$ , $y=0$ , $x=1$ , and $x=3$ is revolved about the line $x=4$ . (correct answer)
The region bounded by $y=x^2-1$ , $y=0$ , $x=1$ , and $x=3$ is revolved about the y-axis.
The region bounded by $y=x^2$ and $y=1$ from $x=1$ to $x=3$ is revolved about the line $x=-4$ .
The region bounded by $x=y^2-1$ and $x=0$ is revolved about the line $y=4$ .

Explanation: The integral is in the form of the shell method,

V = \int_a^b 2\pi r(x)h(x) \,dx

. From the integral, we can identify the components: the limits of integration are from

x=1

x=3

. The radius of a shell is

r(x) = 4-x

, which represents the distance from a vertical line

x=4

. The height of a shell is

h(x) = x^2-1

. This corresponds to the region bounded above by

y=x^2-1

and below by

y=0

Question 6

A solid is formed by revolving the region bounded by the curve $y=x-x^4$ and the x-axis about the y-axis. Setting up the volume integral using the washer method would be difficult because it requires solving for $x$ in terms of $y$ . Which integral correctly sets up the volume using the more convenient shell method?

$\int_0^1 \pi (x-x^4)^2 \,dx$
$\int_0^1 2\pi x(x-x^4) \,dx$ (correct answer)
$\int_0^1 2\pi (1-x)(x-x^4) \,dx$
$\int_0^{0.75} 2\pi y (\sqrt[4]{y}) \,dy$

Explanation: To use the shell method for revolution about the y-axis (a vertical axis), we integrate with respect to

x

. The intersection points of

y=x-x^4

and

y=0

are

x(1-x^3)=0

, so

x=0

and

x=1

. These are the limits of integration. The radius of a cylindrical shell is the distance from the y-axis, so

r(x)=x

. The height of the shell is the function value,

h(x) = x-x^4

. The volume is

V = \int_0^1 2\pi r(x)h(x) \,dx = \int_0^1 2\pi x(x-x^4) \,dx

Question 7

Consider the solid formed by revolving the region bounded by $y=x^2$ and $y=4$ about the line $x=3$ . Let $V_S$ be the integral for the volume using the shell method and $V_W$ be the integral for the volume using the washer method. Which pair of integrals is correct?

$V_S = \int_{-2}^2 2\pi (3-x)(4-x^2) dx$ and $V_W = \int_0^4 \pi ((3+\sqrt{y})^2 - (3-\sqrt{y})^2) dy$ (correct answer)
$V_S = \int_{-2}^2 2\pi x(4-x^2) dx$ and $V_W = \int_0^4 \pi (4^2 - (\sqrt{y})^2) dy$
$V_S = \int_{0}^4 2\pi (3-y)(2\sqrt{y}) dy$ and $V_W = \int_{-2}^2 \pi ((4-3)^2 - (x^2-3)^2) dx$
$V_S = \int_{-2}^2 2\pi (x-3)(4-x^2) dx$ and $V_W = \int_0^4 \pi ((\sqrt{y}-3)^2 - (-\sqrt{y}-3)^2) dy$

Explanation: Shell Method (integrate wrt

x

): Axis is

x=3

. The region spans

x \in [-2, 2]

. Radius

r(x)=3-x

. Height

h(x)=4-x^2

. So,

V_S = \int_{-2}^2 2\pi (3-x)(4-x^2) dx

. Washer Method (integrate wrt

y

): Axis is

x=3

. The region spans

y \in [0, 4]

. The right boundary is

x=\sqrt{y}

and left is

x=-\sqrt{y}

. The outer radius is the distance from

x=3

to the far curve

x=-\sqrt{y}

, so

R(y) = 3-(-\sqrt{y}) = 3+\sqrt{y}

. The inner radius is the distance from

x=3

to the near curve

x=\sqrt{y}

, so

r(y) = 3-\sqrt{y}

. Thus,

V_W = \int_0^4 \pi (R(y)^2 - r(y)^2) dy = \int_0^4 \pi ((3+\sqrt{y})^2 - (3-\sqrt{y})^2) dy

Question 8

Let R be the region in the first quadrant bounded by $y=4-x^2$ . Let $V_x$ be the volume when R is revolved about the x-axis, and $V_y$ be the volume when R is revolved about the y-axis. Which method is most efficient for calculating $V_y$ , and what is the ratio $V_y/V_x$ ?

Shell method is most efficient; ratio is $5/16$ .
Washer method is most efficient; ratio is $5/16$ .
Shell method is most efficient; ratio is $15/32$ . (correct answer)
Washer method is most efficient; ratio is $15/32$ .

Explanation:

V_x

(revolve about x-axis): Use disk method. Limits are

x=0

x=2

V_x = \int_0^2 \pi (4-x^2)^2 \,dx = \pi \int_0^2 (16-8x^2+x^4) \,dx = \pi[16x - \frac{8}{3}x^3 + \frac{1}{5}x^5]_0^2 = \pi(32 - 64/3 + 32/5) = 256\pi/15

V_y

(revolve about y-axis): Shell method is efficient. Limits

x=0

x=2

. Radius

r(x)=x

, height

h(x)=4-x^2

V_y = \int_0^2 2\pi x(4-x^2) \,dx = 2\pi[2x^2 - x^4/4]_0^2 = 2\pi(8-4) = 8\pi

. The washer method for

V_y

is also easy:

V_y = \int_0^4 \pi (\sqrt{4-y})^2 \,dy = \pi[4y-y^2/2]_0^4 = \pi(16-8)=8\pi

. So both methods are efficient. The ratio is

V_y/V_x = (8\pi) / (256\pi/15) = 8 \cdot 15 / 256 = 120/256 = 15/32

. Since both methods are efficient for

V_y

, we check the options. Both C and D have the correct ratio. But the shell method setup for

V_y

is arguably more direct from the function's given form

y=f(x)

Question 9

Let R be the region in the first quadrant bounded by the graphs of $y=x^3$ , the x-axis, and the line $x=1$ . Which integral represents the volume of the solid generated by revolving R about the line $x=2$ using the shell method?

$V = \int_0^1 2\pi x(x^3) \,dx$
$V = \int_0^1 2\pi (x-2)(x^3) \,dx$
$V = \int_0^1 2\pi (2-x)(x^3) \,dx$ (correct answer)
$V = \int_0^1 \pi (2-x)^2 (x^3) \,dx$

Explanation: Using the shell method for a vertical axis of revolution, the integral is with respect to

x

. The region is bounded by

x=0

and

x=1

. For a shell at a given

x

, the height

h(x)

is given by the function

y=x^3

, so

h(x)=x^3

. The radius

r(x)

is the distance from the axis of revolution

x=2

to the shell at

x

, which is

r(x)=2-x

. The volume of a shell is

2\pi r(x)h(x)dx

. Therefore, the total volume is given by the integral

V = \int_0^1 2\pi (2-x)(x^3) \,dx

Question 10

The integral $V = \int_0^1 2\pi(y+1)(\sqrt{y} - y^2) \,dy$ represents the volume of a solid generated by revolving a region R about an axis. Which of the following correctly describes the region and the axis?

Region R bounded by $x=\sqrt{y}$ and $x=y^2$ is revolved about the line $y=-1$ . (correct answer)
Region R bounded by $x=\sqrt{y}$ and $x=y^2$ is revolved about the line $y=1$ .
Region R bounded by $y=\sqrt{x}$ and $y=x^2$ is revolved about the line $x=-1$ .
Region R bounded by $y=\sqrt{x}-x^2$ and $y=0$ is revolved about the line $y=-1$ .

Explanation: The integral has the form of the shell method with respect to

y

V = \int_c^d 2\pi r(y)h(y) \,dy

. The radius is

r(y) = y+1 = y - (-1)

, which corresponds to revolution about the horizontal line

y=-1

. The height is

h(y) = \sqrt{y} - y^2

, which represents the horizontal distance between a right curve

x_R = \sqrt{y}

and a left curve

x_L = y^2

. The limits of integration, 0 to 1, are the y-values where these curves intersect. Thus, the region is bounded by

x=\sqrt{y}

and

x=y^2

and is revolved about

y=-1

Question 11

A solid is generated by revolving the region bounded by $y=f(x)$ and $y=g(x)$ , where $f(x) \ge g(x)$ on the interval $[a,b]$ , about the vertical line $x=c$ where $c < a$ . Which integral correctly represents the volume of this solid using the shell method?

$\int_a^b 2\pi (c-x)(f(x)-g(x)) \,dx$
$\int_a^b 2\pi (x-c)(f(x)-g(x)) \,dx$ (correct answer)
$\int_a^b 2\pi x(f(x)-g(x)) \,dx$
$\int_a^b \pi ((f(x)-c)^2 - (g(x)-c)^2) \,dx$

Explanation: For the shell method with a vertical axis of revolution, we integrate with respect to

x

. The height of a cylindrical shell at position

x

is the difference between the upper and lower curves,

h(x) = f(x)-g(x)

. The radius

r(x)

is the distance from the axis of revolution

x=c

to the shell's position

x

. Since

c < a \le x

, the distance is

r(x) = x-c

. The volume integral is

V = \int_a^b 2\pi r(x)h(x) \,dx = \int_a^b 2\pi (x-c)(f(x)-g(x)) \,dx

Question 12

Consider the region bounded by $y = e^x$ , $y = 1$ , and $x = \ln(3)$ . When this region is revolved around the $y$ -axis using the shell method, which integral expression gives the correct volume?

$2\pi \int_1^3 (\ln(y))^2 dy$
$2\pi \int_1^3 y(\ln(y) - 0) dy$
$2\pi \int_0^{\ln(3)} x(1 - e^x) dx$
$2\pi \int_0^{\ln(3)} x(e^x - 1) dx$ (correct answer)

Explanation: When using the shell method to find volumes of revolution around the y-axis, you're essentially summing up cylindrical shells. Each shell has radius x, height given by the difference between functions, and thickness dx. First, let's visualize this region. The curve

y = e^x

intersects

y = 1

when

x = 0

(since

e^0 = 1

). The vertical line

x = \ln(3)

gives us our right boundary, where

y = e^{\ln(3)} = 3

. So we have a region from

x = 0

x = \ln(3)

, bounded above by

y = e^x

and below by

y = 1

. For the shell method around the y-axis, each vertical strip at position x becomes a cylindrical shell with:

Radius: x (distance from y-axis)
Height: $e^x - 1$ (top function minus bottom function)
Volume element: $2\pi x(e^x - 1)dx$

Integrating from

x = 0

x = \ln(3)

gives us

2\pi \int_0^{\ln(3)} x(e^x - 1) dx

, which is answer D. Answer A uses the washer method setup but incorrectly squares the radius term. Answer B also uses washer method thinking but with the wrong integrand. Answer C has the shell method format but incorrectly subtracts

e^x

from 1, giving a negative height—remember that the exponential function is always above the horizontal line in this region. Study tip: Always sketch the region first and identify which function is on top. For shell method, the integrand should be radius × (outer function - inner function).

Question 13

The region bounded by $y = \ln(x)$ , $y = 0$ , $x = 1$ , and $x = e^2$ is revolved around the $y$ -axis. A student sets up the shell method integral as $V = 2\pi \int_1^{e^2} x \ln(x) dx$ but gets the wrong answer. What is most likely the error?

The limits of integration should be from $0$ to $2$
The height function should be $2 - \ln(x)$ instead of $\ln(x)$ (correct answer)
The integral is correct; the error must be in the computation
The height function should be $\ln(x) - 0 = \ln(x)$ , but only for $x \geq 1$

Explanation: Looking at the region,

y = \ln(x)

is negative for

1 < x < e

and positive for

x > e

. For

x \in [1, e]

, the curve

y = \ln(x)

lies below the x-axis, so the height should be

0 - \ln(x) = -\ln(x)

. For

x \in [e, e^2]

, the height is

\ln(x) - 0 = \ln(x)

. The student's setup

x \ln(x)

ignores this sign issue for the first part of the interval. The correct setup requires splitting the integral or using

|\ln(x)|

. Choice A has wrong limits. Choice C is incorrect since there's a conceptual error. Choice D doesn't address the sign issue for

x < e

Question 14

A region is bounded by $y = 2x$ , $y = 6$ , and $x = 0$ . When revolved around the $y$ -axis, both the shell method and disk/washer method can be used. If the shell method gives volume $V_s$ and requires evaluating $\int_0^3 f(x) dx$ , what is $f(x)$ ?

$f(x) = x(6 - 2x)$
$f(x) = \pi x^2(6 - 2x)$
$f(x) = 2\pi x(6 - 2x)$ (correct answer)
$f(x) = 2\pi x^2(6 - 2x)$

Explanation: When using the shell method to find volumes of revolution around the y-axis, you're essentially summing up cylindrical shells with radius

x

, height determined by the function, and thickness

dx

. The general formula is

V = 2\pi \int x \cdot (\text{height}) \, dx

. First, let's understand this region. The boundaries are

y = 2x

y = 6

, and

x = 0

. These intersect at

(0,0)

(3,6)

, and

(0,6)

, forming a triangle. When you slice this region with vertical strips (for the shell method), each strip at position

x

extends from

y = 2x

up to

y = 6

, giving a height of

6 - 2x

. For the shell method around the y-axis, the volume is

V_s = 2\pi \int_0^3 x(6 - 2x) \, dx

. This means

f(x) = 2\pi x(6 - 2x)

, which is answer C. Let's examine why the other options are wrong. Option A,

f(x) = x(6 - 2x)

, gives you just the area element without the

2\pi

factor required for the shell method volume formula. Option B,

f(x) = \pi x^2(6 - 2x)

, incorrectly uses

\pi

instead of

2\pi

and has an extra factor of

x

, suggesting confusion with the disk method. Option D,

f(x) = 2\pi x^2(6 - 2x)

, has the correct

2\pi

but includes an extra

x

factor. Remember: shell method integrands always have the form

2\pi \times \text{radius} \times \text{height}

, where radius is the distance from the axis of rotation.

Question 15

A solid is formed by revolving the region bounded by $y = x^2$ , $y = 0$ , and $x = 2$ around the line $x = 3$ . If we set up the shell method integral as $V = 2\pi \int_a^b R(x) \cdot H(x) dx$ , what are the correct values of $R(x)$ and $H(x)$ ?

$R(x) = 3 + x$ and $H(x) = x^2$
$R(x) = x - 3$ and $H(x) = x^2$
$R(x) = 3 - x$ and $H(x) = x^2$ (correct answer)
$R(x) = 3 - x$ and $H(x) = 2 - x^2$

Explanation: When setting up the shell method for a solid of revolution, you need to identify two key components: the radius of each cylindrical shell and the height of each shell. For this problem, you're revolving the region bounded by

y = x^2

y = 0

, and

x = 2

around the vertical line

x = 3

. Since you're revolving around a vertical line, you'll integrate with respect to

x

from

x = 0

x = 2

. The radius

R(x)

is the horizontal distance from any point

x

in your region to the axis of revolution

x = 3

. Since your region spans from

x = 0

x = 2

(all to the left of

x = 3

), this distance is

3 - x

. The height

H(x)

is the vertical extent of the region at any given

x

-value, which runs from

y = 0

up to

y = x^2

, giving us

H(x) = x^2

. Looking at the wrong answers: Choice A incorrectly uses

R(x) = 3 + x

, which would be the distance if you were measuring from

x = -3

instead of to

x = 3

. Choice B uses

R(x) = x - 3

, which gives negative values since

x < 3

in our region—radius must be positive. Choice D correctly identifies the radius but mistakenly uses

H(x) = 2 - x^2

, which doesn't represent the height of our bounded region. Study tip: Always sketch the region and axis of revolution. For shell method radius, ask yourself: "How far is this strip from the axis?" The distance formula will guide you to the correct expression.

Question 16

A solid is generated by rotating the region R bounded by $y=x^2$ and $y=\sqrt{x}$ about the line $x=-1$ . What is the volume of the solid?

$29\pi/30$ (correct answer)
$13\pi/15$
$49\pi/30$
$\pi/3$

Explanation: We use the shell method. The curves intersect when

x^2 = \sqrt{x}

, giving

x = 0

and

x = 1

. For

0 \leq x \leq 1

, we have

\sqrt{x} \geq x^2

. The radius of each shell is

r(x) = x - (-1) = x + 1

and the height is

h(x) = \sqrt{x} - x^2

. The volume is

V = \int_0^1 2\pi(x+1)(\sqrt{x}-x^2)dx = 2\pi \int_0^1 (x^{3/2} - x^3 + x^{1/2} - x^2)dx = 2\pi[\frac{2}{5}x^{5/2} - \frac{x^4}{4} + \frac{2}{3}x^{3/2} - \frac{x^3}{3}]_0^1 = 2\pi(\frac{2}{5} - \frac{1}{4} + \frac{2}{3} - \frac{1}{3}) = 2\pi(\frac{2}{5} - \frac{1}{4} + \frac{1}{3}) = 2\pi \cdot \frac{24 - 15 + 20}{60} = \frac{29\pi}{30}

Question 17

Which integral represents the volume of the solid generated by revolving the region bounded by $y=x^2$ and $y=x$ about the line $y=2$ , using the shell method?

$\int_0^1 2\pi(2-y)(\sqrt{y}-y) \,dy$ (correct answer)
$\int_0^1 \pi((2-x^2)^2 - (2-x)^2) \,dx$
$\int_0^1 2\pi y(\sqrt{y}-y) \,dy$
$\int_0^1 2\pi(2-x)(x-x^2) \,dx$

Explanation: To use the shell method for revolution about a horizontal line (

y=2

), we must integrate with respect to

y

. First, express

x

in terms of

y

x=y

and

x=\sqrt{y}

. The region is bounded by

y=0

and

y=1

. For a given

y

, the right boundary is

x=\sqrt{y}

and the left is

x=y

. The height of a horizontal shell is

h(y) = \sqrt{y}-y

. The radius is the distance from the axis

y=2

to the shell, so

r(y) = 2-y

. The volume is

V = \int_0^1 2\pi r(y)h(y) \,dy = \int_0^1 2\pi(2-y)(\sqrt{y}-y) \,dy

Question 18

The region bounded by $y=\sin(x)$ and the x-axis from $x=0$ to $x=\pi$ is revolved about the line $x=-1$ . What is the volume of the resulting solid?

$2\pi^2$
$2\pi(\pi-2)$
$2\pi(\pi+2)$ (correct answer)
$4\pi$

Explanation: We use the shell method with respect to

x

. The axis of revolution is

x=-1

. The radius of a shell at

x

is the distance from

x

-1

, which is

r(x) = x - (-1) = x+1

. The height of the shell is

h(x) = \sin(x)

. The limits are from

0

\pi

. The volume is

V = \int_0^\pi 2\pi(x+1)\sin(x) \,dx = 2\pi \int_0^\pi (x\sin(x) + \sin(x)) \,dx

. Using integration by parts,

\int x\sin(x) \,dx = -x\cos(x) + \sin(x)

. The integral becomes

2\pi [-x\cos(x) + \sin(x) - \cos(x)]_0^\pi = 2\pi[(-\pi(-1) + 0 - (-1)) - (0+0-1)] = 2\pi[(\pi+1) - (-1)] = 2\pi(\pi+2)

Question 19

In the formula for the volume of a solid of revolution using the shell method, $V = \int_a^b 2\pi r(x) h(x) \,dx$ , what does the expression $2\pi r(x) h(x) \Delta x$ represent geometrically?

The surface area of a cylinder with radius $r(x)$ and height $h(x)$ .
The volume of a thin cylindrical shell with radius $r(x)$ , height $h(x)$ , and thickness $\Delta x$ . (correct answer)
The volume of a disk with radius $r(x)$ and thickness $\Delta x$ .
The cross-sectional area of the solid at a position $x$ .

Explanation: The shell method approximates the volume by summing the volumes of infinitesimally thin cylindrical shells. The term

2\pi r(x)

is the circumference of the shell,

h(x)

is its height, and

\Delta x

is its thickness. The product of circumference and height,

2\pi r(x) h(x)

, gives the lateral surface area of the cylinder. Multiplying this area by the thickness

\Delta x

gives the approximate volume of that thin shell.

Question 20

The region bounded by $y=1/x$ , the x-axis, $x=1$ , and $x=3$ is revolved about the line $x=-1$ . What is the volume of the solid generated?

$4\pi$
$2\pi(2-\ln 3)$
$2\pi(2+\ln 3)$ (correct answer)
$2\pi/3$

Explanation: Using the shell method, we integrate with respect to

x

from 1 to 3. The axis of revolution is the vertical line

x=-1

. The radius of a shell at

x

r(x) = x - (-1) = x+1

. The height of the shell is

h(x) = 1/x

. The volume integral is

V = \int_1^3 2\pi (x+1) (1/x) \,dx = \int_1^3 2\pi (1 + 1/x) \,dx

. Evaluating the integral gives

2\pi [x + \ln|x|]_1^3 = 2\pi [(3 + \ln 3) - (1 + \ln 1)] = 2\pi(2 + \ln 3)