Calculus 1 Quiz: Disc Method Other Axes

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Disc Method Other AxesQuestion 1 of 20

Let R be the region bounded by $y=1/x$ , $x=1$ , $x=3$ , and $y=0$ . Which integral gives the volume of the solid generated by revolving R about the line $y=2$ ?

\pi \int_1^3 (2 - 1/x)^2 dx

\pi \int_1^3 (4 - (1/x)^2) dx

\pi \int_1^3 (2^2 - (2 - 1/x)^2) dx

\pi \int_1^3 ((2-1/x)^2 - 4) dx

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Calculus 1 Quiz

Calculus 1 Quiz: Disc Method Other Axes

Practice Disc Method Other Axes in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Disc Method Other Axes, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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Question 1

Let R be the region bounded by $y=1/x$ , $x=1$ , $x=3$ , and $y=0$ . Which integral gives the volume of the solid generated by revolving R about the line $y=2$ ?

$\pi \int_1^3 (2 - 1/x)^2 dx$
$\pi \int_1^3 (4 - (1/x)^2) dx$
$\pi \int_1^3 (2^2 - (2 - 1/x)^2) dx$ (correct answer)
$\pi \int_1^3 ((2-1/x)^2 - 4) dx$

Explanation: The axis of revolution is

y=2

, which is above the region. This is a washer problem. The outer radius

R(x)

is the distance from

y=2

to the farther boundary

y=0

, so

R(x) = 2 - 0 = 2

. The inner radius

r(x)

is the distance from

y=2

to the closer boundary

y=1/x

, so

r(x) = 2 - 1/x

. The volume is

V = \pi \int_1^3 (R(x)^2 - r(x)^2) dx = \pi \int_1^3 (2^2 - (2 - 1/x)^2) dx

Question 2

A solid is formed by revolving the region bounded by $y=x^4$ , $y=1$ , and $x=0$ about the line $y=2$ . Which integral represents the volume?

$\pi \int_0^1 (2-1)^2 dx$
$\pi \int_0^1 (1-(2-x^4)^2) dx$
$\pi \int_0^1 (2-x^4)^2 dx$
$\pi \int_0^1 ((2-x^4)^2-1) dx$ (correct answer)

Explanation: When finding the volume of a solid of revolution using the disk/washer method, you need to carefully identify the outer and inner radii from the axis of rotation to the boundaries of your region. First, sketch the region bounded by

y = x^4,y = 1

, and

x = 0

. This creates a region from

x = 0

x = 1

(where

x^4 = 1

), with

y = 1

as the top boundary and

y = x^4

as the bottom boundary. Since we're rotating about

y = 2

, this creates washers (rings with holes). For any

x

-value, the outer radius extends from the axis

y = 2

down to the bottom curve

y = x^4

, giving

R = 2 - x^4

. The inner radius extends from

y = 2

down to the top boundary

y = 1

, giving

r = 2 - 1 = 1

. Using the washer formula

V = \pi \int_a^b (R^2 - r^2) dx

, we get:

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

Choice A uses only the inner radius and ignores the outer radius entirely. Choice B incorrectly subtracts the outer radius squared from the inner radius, which reverses the washer formula. Choice C uses only the outer radius and forgets about the hole created by the inner radius. Key strategy: Always identify both radii when rotating around an external axis, and remember the washer formula subtracts the smaller area from the larger:

\pi(R^2 - r^2)

Question 3

The region enclosed by $y=x$ and $y=x^2/4$ is rotated about the line $x=4$ . What is the resulting volume?

$\frac{16\pi}{3}$
$\frac{32\pi}{3}$ (correct answer)
$\frac{64\pi}{3}$
$8\pi$

Explanation: The curves intersect at

x=0

and

x=4

, which correspond to

y=0

and

y=4

. Since the axis of revolution

x=4

is vertical, we integrate with respect to

y

. The curves are

x=y

and

x=2\sqrt{y}

. On the interval

y \in [0,4]

2\sqrt{y} \ge y

, so

x=2\sqrt{y}

is the right curve and

x=y

is the left curve. The axis of revolution

x=4

is to the right of the region. The outer radius is the distance from

x=4

to the left curve,

R(y) = 4-y

. The inner radius is the distance from

x=4

to the right curve,

r(y) = 4-2\sqrt{y}

. The volume is

V = \pi \int_0^4 ((4-y)^2 - (4-2\sqrt{y})^2) dy = \pi \int_0^4 ((16-8y+y^2) - (16-16\sqrt{y}+4y)) dy = \pi \int_0^4 (y^2 - 12y + 16y^{1/2}) dy = \pi [\frac{y^3}{3} - 6y^2 + \frac{32}{3}y^{3/2}]_0^4 = \pi (\frac{64}{3} - 96 + \frac{32}{3}(8)) = \pi (\frac{64-288+256}{3}) = \frac{32\pi}{3}

Question 4

The region in the first quadrant bounded by $y=x^2$ , $y=1$ , and the y-axis is revolved about the line $y=3$ . The volume of the solid is given by:

$\pi \int_0^1 (3-x^2)^2 dx$
$\pi \int_0^1 ((3-1)^2 - (3-x^2)^2) dx$
$\pi \int_0^1 ((3-x^2)^2 - (3-1)^2) dx$ (correct answer)
$\pi \int_0^1 (x^4 - 1) dx$

Explanation: The region is bounded by

y=x^2

(bottom),

y=1

(top), and

x=0

(left). The intersection of

y=x^2

and

y=1

is at

x=1

. So

x

ranges from 0 to 1. The axis of revolution is

y=3

, which is above the region. The outer radius

R(x)

is the distance from

y=3

to the bottom curve

y=x^2

, so

R(x) = 3 - x^2

. The inner radius

r(x)

is the distance from

y=3

to the top curve

y=1

, so

r(x) = 3 - 1 = 2

. The volume is

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

Question 5

What is the volume of the solid generated by revolving the region bounded by $y=|x-2|$ and $y=1$ about the line $y=1$ ?

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

Explanation: When you encounter a solid of revolution problem, you need to identify the region being revolved and the axis of revolution. Here, you're revolving the region between

y=|x-2|

and

y=1

about the line

y=1

. First, find where these curves intersect. Setting

|x-2|=1

gives us

x-2=±1

, so

x=1

x=3

. The absolute value function

y=|x-2|

has its vertex at

(2,0)

and forms a V-shape. Between

x=1

and

x=3

, this curve lies below the horizontal line

y=1

. Since you're revolving about

y=1

, use the washer method. The radius of each circular cross-section is the distance from

y=1

down to the curve

y=|x-2|

. This radius is

r(x)=1-|x-2|

. The volume is:

V = \pi\int_1^3 [1-|x-2|]^2 \, dx

Since

|x-2| = 2-x

for

1 \leq x \leq 3

, we have

1-|x-2| = 1-(2-x) = x-1.V = \pi\int_1^3 (x-1)^2 \, dx = \pi\int_1^3 (x^2-2x+1) \, dx

= \pi\left[\frac{x^3}{3}-x^2+x\right]_1^3 = \pi\left[(9-9+3)-\left(\frac{1}{3}-1+1\right)\right] = \pi\left[3-\frac{1}{3}\right] = \frac{2\pi}{3}

This confirms answer D. Answer A (

\frac{\pi}{3}

) likely comes from calculation errors, B (

\frac{4\pi}{3}

) might result from doubling the correct answer, and C (

\pi

) could come from forgetting to account for the absolute value properly. Strategy tip: When revolving about a horizontal line other than the x-axis, always measure your radius as the distance from that line to your curve, not from the x-axis.

Question 6

Let $R$ be the region bounded by $y = \cos(x) + 3$ and $y=2$ on the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ . Which integral represents the volume of the solid generated by revolving $R$ about the line $y=2$ ?

$\pi \int_{-\pi/2}^{\pi/2} ((\cos(x)+3)^2 - 4) dx$
$\pi \int_{-\pi/2}^{\pi/2} (\cos(x)+1)^2 dx$ (correct answer)
$\pi \int_{-\pi/2}^{\pi/2} (\cos(x)+3)^2 dx$
$\pi \int_{-\pi/2}^{\pi/2} (\cos(x)+1) dx$

Explanation: The solid is generated by revolving a region about the horizontal line

y=2

. The disc method applies because one of the region's boundaries is the axis of revolution. The radius of a representative disc at a value

x

is the distance from the axis of revolution (

y=2

) to the outer curve (

y=\cos(x)+3

). Therefore, the radius is

R(x) = (\cos(x)+3) - 2 = \cos(x)+1

. The volume is given by

V = \pi \int_{a}^{b} [R(x)]^2 dx

. The interval is given as

[-\frac{\pi}{2}, \frac{\pi}{2}]

. So, the integral is

V = \pi \int_{-\pi/2}^{\pi/2} (\cos(x)+1)^2 dx

A incorrectly uses the washer method formula $V = \pi \int (R^2 - r^2) dx$ , treating $R(x) = \cos(x)+3$ and $r=2$ . This would be the volume if the region was revolved around the x-axis, which is not the case.
C uses an incorrect radius, $R(x)=\cos(x)+3$ . This would be the radius if the region were revolved around the x-axis ( $y=0$ ).
D is incorrect because the radius, $R(x)=\cos(x)+1$ , has not been squared in the volume formula.

Question 7

Let $S$ be the region in the first quadrant bounded by $y=x^3$ , $y=8$ , and the y-axis. Let $V_1$ be the volume of the solid generated by revolving $S$ about the line $y=8$ , and let $V_2$ be the volume of the solid generated by revolving $S$ about the y-axis. Which statement is true?

$V_1 < V_2$
The relationship between $V_1$ and $V_2$ cannot be determined from the information given.
$V_1 = V_2$
$V_1 > V_2$ (correct answer)

Explanation: When you encounter volume problems involving revolution around different axes, you need to set up and evaluate separate integrals for each case, then compare the results. First, identify the region S: it's bounded by

y = x^3

y = 8

, and the y-axis in the first quadrant. Since

y = x^3

intersects

y = 8

when

x = 2

, the region extends from

x = 0

x = 2

. For

V_1

(revolution about

y = 8

), use the washer method. The outer radius is

8 - x^3

and inner radius is

8 - 8 = 0

, so:

V_1 = \pi \int_0^2 (8 - x^3)^2 dx = \pi \int_0^2 (64 - 16x^3 + x^6) dx = \pi[64x - 4x^4 + \frac{x^7}{7}]_0^2 = \frac{512\pi}{7}

For

V_2

(revolution about the y-axis), use the disk method with

x = y^{1/3}

V_2 = \pi \int_0^8 (y^{1/3})^2 dy = \pi \int_0^8 y^{2/3} dy = \pi[\frac{3y^{5/3}}{5}]_0^8 = \frac{96\pi}{5}

Comparing:

\frac{512\pi}{7} \approx 228.6

while

\frac{96\pi}{5} \approx 60.3

, so

V_1 > V_2

. Choice A incorrectly reverses the relationship. Choice B suggests the volumes can't be compared, but direct calculation shows they can be. Choice C claims they're equal, which contradicts our calculations. Choice D correctly identifies that

V_1 > V_2

. Study tip: When comparing volumes of revolution, always calculate both explicitly rather than trying to guess from the geometry alone—the mathematics often reveals surprising relationships.

Question 8

Let $k$ be a positive constant. The region bounded by the graphs of $y=kx^2$ and $y=k$ for $x \ge 0$ is revolved about the line $y=k$ . If the volume of the resulting solid is $\frac{16\pi}{15}$ , what is the value of $k$ ?

$\frac{1}{2}$
$4$
$2$
$\sqrt{2}$ (correct answer)

Explanation: When you encounter a volume of revolution problem, you need to identify the region being revolved and choose the appropriate method. Here, the region is bounded by

y = kx^2

and

y = k

for

x \geq 0

, and it's revolved about the horizontal line

y = k

. First, find where these curves intersect. Setting

kx^2 = k

gives

x^2 = 1

, so

x = 1

(since

x \geq 0

). The region extends from

x = 0

x = 1

. Since we're revolving about

y = k

, use the washer method. The outer radius is the distance from

y = k

y = k

, which is

0

. Wait - this suggests we should think differently. The outer radius is actually the distance from the axis of revolution (

y = k

) to the line

y = k

, but we need to consider that the parabola

y = kx^2

is below the line

y = k

in our region. The radius at any

x

r(x) = k - kx^2 = k(1 - x^2)

. Using the disk method:

V = \pi \int_0^1 [k(1-x^2)]^2 \, dx = \pi k^2 \int_0^1 (1-x^2)^2 \, dx

Expanding:

(1-x^2)^2 = 1 - 2x^2 + x^4

V = \pi k^2 \left[x - \frac{2x^3}{3} + \frac{x^5}{5}\right]_0^1 = \pi k^2 \left(1 - \frac{2}{3} + \frac{1}{5}\right) = \pi k^2 \cdot \frac{8}{15} = \frac{8\pi k^2}{15}

Setting this equal to

\frac{16\pi}{15}

\frac{8\pi k^2}{15} = \frac{16\pi}{15}

Solving:

8k^2 = 16

, so

k^2 = 2

, giving

k = \sqrt{2}

. Answer D is correct. Answer A (

\frac{1}{2}

) gives

k^2 = \frac{1}{4}

, making the volume too small. Answer B (

4

) gives

k^2 = 16

, making the volume too large. Answer C (

2

) gives

k^2 = 4

, also too large. Remember: carefully set up your radius function by measuring distance from the axis of revolution, and always expand polynomials completely before integrating.

Question 9

The region bounded by $y=e^x$ , $y=e$ , and $x=0$ is revolved about the line $y=e$ . What is the volume of the resulting solid?

$\frac{\pi}{2}(e^2 - 1)$
$\pi(e-1)$
$\frac{\pi}{2}(4e - e^2 - 1)$ (correct answer)
$\pi(2e - \frac{e^2}{2})$

Explanation: The revolution is about the horizontal line

y=e

. The region is bounded by

y=e^x

y=e

, and

x=0

. The intersection of

y=e^x

and

y=e

is at

x=1

. So, the integration is from

x=0

x=1

. Since the axis of revolution

y=e

is a boundary of the region, we use the disc method. The radius is

R(x) = e - e^x

. The volume is

V = \pi \int_{0}^{1} (e - e^x)^2 dx = \pi \int_{0}^{1} (e^2 - 2e \cdot e^x + e^{2x}) dx

. The antiderivative is

\pi [e^2x - 2e \cdot e^x + \frac{1}{2}e^{2x}]_{0}^{1}

. Evaluating at the limits gives

\pi [ (e^2(1) - 2e(e^1) + \frac{1}{2}e^2) - (e^2(0) - 2e(e^0) + \frac{1}{2}e^0) ] = \pi [ (e^2 - 2e^2 + \frac{1}{2}e^2) - (0 - 2e + \frac{1}{2}) ] = \pi [ -\frac{1}{2}e^2 - (-2e + \frac{1}{2}) ] = \pi ( -\frac{e^2}{2} + 2e - \frac{1}{2}) = \frac{\pi}{2}(4e - e^2 - 1)

A is the result of incorrectly setting up the integral as $\pi \int (e^2 - (e^x)^2) dx$ , a common washer method mistake.
B is the result of forgetting to square the radius, calculating $\pi \int (e-e^x)dx$ .
D results from a calculation error, specifically forgetting to subtract the value of the antiderivative at $x=0$ .

Question 10

The interior of a decorative bowl is formed by rotating the curve $x=2\sqrt{y-1}$ for $1 \le y \le 5$ about the line $x=6$ . The solid material of the bowl is the region between the curve and the line $x=6$ . Which integral gives the volume of the bowl's material?

$2\pi \int_{0}^{4} x ((\frac{x^2}{4}+1)-6) dx$
$\pi \int_{1}^{5} (36 - 4(y-1)) dy$
$\pi \int_{1}^{5} (6 - 2\sqrt{y-1})^2 dy$ (correct answer)
$\pi \int_{1}^{5} (2\sqrt{y-1})^2 dy$

Explanation: When you encounter a volume problem involving rotation about a vertical or horizontal line, you need to carefully identify what solid is being created and choose the appropriate method. Here, the bowl's material is the region between the curve

x = 2\sqrt{y-1}

and the line

x = 6

, rotated about

x = 6

. Since we're rotating about a vertical line and have the function in terms of

y

, the washer method with horizontal slices is most natural. For any fixed

y

between 1 and 5, the horizontal slice extends from the curve at

x = 2\sqrt{y-1}

to the axis of rotation at

x = 6

. The radius of each circular slice is the distance from the curve to the axis:

R(y) = 6 - 2\sqrt{y-1}

. Using the disk method, the volume is

\pi \int_{1}^{5} [R(y)]^2 \, dy = \pi \int_{1}^{5} (6 - 2\sqrt{y-1})^2 \, dy

, which is answer C. Answer A uses the shell method incorrectly—it has the wrong integrand and bounds. Answer B attempts the washer method but fails to square the radius function, giving area instead of volume. Answer D only accounts for the "hole" in a washer (if there were one) but ignores the outer radius entirely. Study tip: When setting up rotation problems, always sketch the region first, identify your axis of rotation, then determine whether each cross-section is a disk, washer, or shell. Remember that areas get squared in the disk/washer method, but not in the shell method.

Question 11

Let $R$ be the region enclosed by the graph of $x = y^2 + 1$ and the line $x=5$ . Which of the following definite integrals represents the volume of the solid generated when $R$ is revolved about the line $x=5$ ?

$\pi \int_{-2}^{2} (5 - (y^2+1))^2 dy$ (correct answer)
$\pi \int_{-2}^{2} (25 - (y^2+1)^2) dy$
$2\pi \int_{1}^{5} x \sqrt{x-1} dx$
$\pi \int_{1}^{5} (\sqrt{x-1})^2 dx$

Explanation: The revolution is about the vertical line

x=5

. This requires integration with respect to

y

. The region is bounded by

x=y^2+1

and

x=5

. The intersections occur when

y^2+1=5

, so

y^2=4

, which means

y=\pm 2

. These are the limits of integration. The axis of revolution,

x=5

, is a boundary of the region, so we use the disc method. The radius

R(y)

is the horizontal distance from the axis of revolution (

x=5

) to the curve (

x=y^2+1

). Thus,

R(y) = 5 - (y^2+1)

. The volume is

V = \pi \int_{-2}^{2} [R(y)]^2 dy = \pi \int_{-2}^{2} (5 - (y^2+1))^2 dy

B represents a common error, using the washer formula form $\pi(R^2 - r^2)$ with $R=5$ and $r=y^2+1$ . This is incorrect; the radius itself, $(R-r)$ , must be squared.
C represents the volume using the shell method for revolving a different region about the y-axis.
D represents the volume of the solid formed by revolving the region about the x-axis, using integration with respect to $x$ . This is the wrong axis and wrong variable of integration.

Question 12

Let $R$ be the region bounded by $y=f(x)$ , $y=c$ , $x=a$ , and $x=b$ , where $f(x) \le c$ for all $x$ in $[a,b]$ . The volume of the solid formed by revolving $R$ about the line $y=c$ is $V$ . A new region, $R'$ , is formed by translating $R$ up by $k$ units, where $k>0$ . This new region $R'$ is then revolved about the line $y=c+k$ . What is the volume of the resulting solid?

$V + \pi k^2 (b-a)$
$V + k$
$V$ (correct answer)
$k V$

Explanation: When you encounter problems involving solids of revolution with translations, the key insight is understanding how the radius function changes when both the region and axis of rotation are shifted together. Originally, region

R

is revolved about

y = c

. Since

f(x) \leq c

, the radius at any point

x

r(x) = c - f(x)

. Using the disk method, the volume is

V = \pi \int_a^b [c - f(x)]^2 \, dx

When

R

is translated up by

k

units to form

R'

, the new upper boundary becomes

y = f(x) + k

, and the axis of rotation shifts to

y = c + k

. The crucial observation is that the radius function remains unchanged: the distance from

y = f(x) + k

to the axis

y = c + k

is still

(c + k) - (f(x) + k) = c - f(x)

. Since the radius function is identical in both cases, the volume integral remains exactly the same, giving us the same volume

V

. This makes choice C correct. Choice A incorrectly assumes you add the volume of a cylinder with radius

k

, which would apply if only the region moved but not the axis. Choice B suggests adding a constant

k

, which has no geometric meaning in this context. Choice D implies the volume scales by factor

k

, which would only occur if the radius function itself were multiplied by

k

. Remember: when both a region and its axis of rotation translate by the same amount in the same direction, the volume of revolution stays constant because the radius function doesn't change.

Question 13

The region R is bounded by the graph of $y=\sin x$ and the x-axis from $x=0$ to $x=\pi$ . Which integral represents the volume of the solid obtained by rotating R about the line $y=-1$ ?

$\pi \int_0^\pi (\sin x - 1)^2 dx$
$\pi \int_0^\pi (\sin^2 x - 1) dx$
$\pi \int_0^\pi ((\sin x + 1)^2 - 1^2) dx$ (correct answer)
$\pi \int_0^\pi (\sin x + 1)^2 dx$

Explanation: The axis of revolution is

y=-1

. The region is bounded by

y=\sin x

(top) and

y=0

(bottom). The outer radius

R(x)

is the distance from

y=-1

y=\sin x

, which is

\sin x - (-1) = \sin x + 1

. The inner radius

r(x)

is the distance from

y=-1

y=0

, which is

0 - (-1) = 1

. The volume is given by the washer method:

V = \pi \int_0^\pi (R(x)^2 - r(x)^2) dx = \pi \int_0^\pi ((\sin x + 1)^2 - 1^2) dx

Question 14

The region enclosed by the graphs of $x=y^2$ and $x=1$ is revolved about the vertical line $x=2$ . Which integral represents the volume of the resulting solid?

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

Explanation: Since the axis of revolution is vertical (

x=2

), we integrate with respect to

y

. The region is bounded by

x=y^2

and

x=1

, which intersect at

y=\pm 1

. The axis

x=2

is to the right of the region. The outer radius

R(y)

is the distance from

x=2

to the farther curve

x=y^2

, so

R(y) = 2 - y^2

. The inner radius

r(y)

is the distance from

x=2

to the closer curve

x=1

, so

r(y) = 2 - 1 = 1

. The volume is

V = \pi \int_{-1}^1 (R(y)^2 - r(y)^2) dy = \pi \int_{-1}^1 ((2-y^2)^2 - 1^2) dy

Question 15

The region in the first quadrant bounded by $x=y^2+1$ , the y-axis, and the lines $y=0$ and $y=2$ is revolved about the line $x=-2$ . Which integral gives the volume of the solid?

$\pi \int_0^2 (((y^2+1)+2)^2 - 2^2) dy$ (correct answer)
$\pi \int_0^2 ((y^2+1)-2)^2 dy$
$\pi \int_1^5 ((\sqrt{x-1})+2)^2 dx$
$\pi \int_0^2 ((y^2+1)^2 - (-2)^2) dy$

Explanation: The axis of revolution is vertical (

x=-2

), so we integrate with respect to

y

. The limits are given as

y=0

y=2

. The right boundary of the region is

x=y^2+1

and the left boundary is

x=0

(the y-axis). The axis

x=-2

is to the left of the region. The outer radius

R(y)

is the distance from

x=-2

to the right boundary

x=y^2+1

, so

R(y) = (y^2+1) - (-2) = y^2+3

. The inner radius

r(y)

is the distance from

x=-2

to the left boundary

x=0

, so

r(y) = 0 - (-2) = 2

. The volume is

V = \pi \int_0^2 (R(y)^2 - r(y)^2) dy = \pi \int_0^2 ((y^2+3)^2 - 2^2) dy

. Choice A simplifies to this:

\pi \int_0^2 (((y^2+1)+2)^2 - 2^2) dy = \pi \int_0^2 ((y^2+3)^2 - 2^2) dy

Question 16

The region R is enclosed by $y=f(x)$ and $y=g(x)$ on $[a, b]$ , where $k < g(x) < f(x)$ for a constant $k$ . If R is revolved about the line $y=k$ , the volume is $\pi \int_a^b ([f(x)-k]^2 - [g(x)-k]^2) dx$ . What is the volume if the same region R is revolved about the line $y=m$ , where $m > f(x)$ ?

$\pi \int_a^b ([m-f(x)]^2 - [m-g(x)]^2) dx$
$\pi \int_a^b ([f(x)-m]^2 - [g(x)-m]^2) dx$
$\pi \int_a^b ([m-g(x)]^2 - [m-f(x)]^2) dx$ (correct answer)
$\pi \int_a^b ([f(x)+m]^2 - [g(x)+m]^2) dx$

Explanation: When revolving around a line

y=m

that is above the region (

m > f(x) > g(x)

), the outer radius

R(x)

is the distance from

y=m

to the farther curve,

y=g(x)

. So,

R(x) = m - g(x)

. The inner radius

r(x)

is the distance from

y=m

to the closer curve,

y=f(x)

. So,

r(x) = m - f(x)

. The volume is

V = \pi \int_a^b (R(x)^2 - r(x)^2) dx = \pi \int_a^b ([m-g(x)]^2 - [m-f(x)]^2) dx

Question 17

The region bounded by $y=e^{-x}$ , $y=0$ , $x=0$ , and $x=1$ is revolved about the line $y=-2$ . The volume of the solid is:

$\pi \int_0^1 (e^{-x}+2)^2 dx$
$\pi \int_0^1 (e^{-2x}-4) dx$
$\pi \int_0^1 ((e^{-x}+2)^2 - 4) dx$ (correct answer)
$\pi \int_0^1 (e^{-x}-2)^2 dx$

Explanation: The axis of revolution is

y=-2

, which is below the region. The upper boundary is

y=e^{-x}

and the lower boundary is

y=0

. We use the washer method. The outer radius is the distance from the axis to the upper curve:

R(x) = e^{-x} - (-2) = e^{-x}+2

. The inner radius is the distance from the axis to the lower curve:

r(x) = 0 - (-2) = 2

. The volume is

V = \pi \int_0^1 (R(x)^2 - r(x)^2) dx = \pi \int_0^1 ((e^{-x}+2)^2 - 2^2) dx = \pi \int_0^1 ((e^{-x}+2)^2 - 4) dx

Question 18

Let R be the region in the first quadrant bounded by the graph of $y=\sqrt{x}$ , the x-axis, and the line $x=4$ . Which integral represents the volume of the solid generated when R is revolved about the line $y=-2$ ?

$\pi \int_0^4 (\sqrt{x}-2)^2 dx$
$\pi \int_0^4 ((\sqrt{x})^2 - (-2)^2) dx$
$\pi \int_0^4 ((\sqrt{x}+2)^2 - 2^2) dx$ (correct answer)
$\pi \int_0^4 (\sqrt{x}+2)^2 dx$

Explanation: The solid is a washer. The axis of revolution is the horizontal line

y=-2

. The outer radius

R(x)

is the distance from

y=-2

to the upper curve

y=\sqrt{x}

, which is

R(x) = \sqrt{x} - (-2) = \sqrt{x}+2

. The inner radius

r(x)

is the distance from

y=-2

to the lower curve

y=0

, which is

r(x) = 0 - (-2) = 2

. The volume is given by the washer method formula

V = \pi \int_a^b (R(x)^2 - r(x)^2) dx

. The limits of integration are from

x=0

x=4

. Therefore, the integral is

\pi \int_0^4 ((\sqrt{x}+2)^2 - 2^2) dx

Question 19

Let $R$ be the region bounded by the curve $x=y^2$ and the line $x=1$ . What is the volume of the solid formed by revolving $R$ about the line $x=1$ ?

$\frac{8\pi}{5}$
$\frac{16\pi}{15}$ (correct answer)
$\frac{4\pi}{3}$
$\frac{8\pi}{3}$

Explanation: The revolution is around a vertical line (

x=1

), so we integrate with respect to

y

. The radius

R(y)

is the horizontal distance from the axis of revolution (

x=1

) to the curve (

x=y^2

). So,

R(y) = 1 - y^2

. The bounds of integration are the y-values where the region begins and ends. The curves intersect when

y^2=1

, so

y=-1

and

y=1

. The volume is

V = \pi \int_{-1}^{1} [R(y)]^2 dy = \pi \int_{-1}^{1} (1-y^2)^2 dy

. Expanding the integrand gives

\pi \int_{-1}^{1} (1 - 2y^2 + y^4) dy

. Integrating gives

\pi [y - \frac{2}{3}y^3 + \frac{1}{5}y^5]_{-1}^{1}

. Evaluating at the limits:

\pi [ (1 - \frac{2}{3} + \frac{1}{5}) - (-1 + \frac{2}{3} - \frac{1}{5}) ] = \pi [ 2 - \frac{4}{3} + \frac{2}{5} ] = \pi [ \frac{30 - 20 + 6}{15} ] = \frac{16\pi}{15}

A results from incorrectly calculating the integral of $(1-y^4)$ , which would come from a mistaken washer method setup $1^2 - (y^2)^2$ .
C results from forgetting to square the radius, i.e., calculating $\pi \int_{-1}^{1} (1-y^2) dy$ .
D results from a sign error in the radius setup, for example using $R(y)=y^2+1$ or a similar miscalculation.

Question 20

The integral $\pi \int_{0}^{2} (5 - e^x)^2 dx$ represents the volume of a solid of revolution. Which of the following describes the solid?

The region bounded by $y=e^x$ , $y=5$ , $x=0$ , and $x=2$ is revolved about the line $y=5$ . (correct answer)
The region bounded by $y=5-e^x$ , the x-axis, $x=0$ , and $x=2$ is revolved about the x-axis.
The region bounded by $y=e^x$ , $y=0$ , $x=0$ , and $x=2$ is revolved about the line $y=5$ .
The region bounded by $x=\ln(5-y)$ , the y-axis, $y=e^0$ , and $y=e^2$ is revolved about the y-axis.

Explanation: The integral is in the form of the disc method for revolution about a horizontal line:

V = \pi \int_{a}^{b} [R(x)]^2 dx

. From the integral, the limits of integration are

a=0

and

b=2

. The radius is

R(x) = 5 - e^x

. The radius is the distance between the axis of revolution

y=c

and the curve

y=f(x)

, so

R(x) = c - f(x)

R(x) = f(x) - c

. If we identify

R(x)=5-e^x

, this matches the form

c-f(x)

with

c=5

and

f(x)=e^x

. For the disc method to apply, the region must be flush against the axis of revolution. This means the region is bounded by the axis

y=5

and the curve

y=e^x

. The limits

x=0

and

x=2

provide the other boundaries of the region.

B is incorrect. While revolving the region under $y=5-e^x$ about the x-axis would use the radius $R(x) = 5-e^x$ , this is a different region and solid from the one described by option A.
C describes a region that is not flush against the axis of revolution $y=5$ . Revolving this region would require the washer method, with outer radius $R(x)=5-0$ and inner radius $r(x)=5-e^x$ .
D describes a revolution about a vertical axis (the y-axis), which would require an integral with respect to $y$ , not $x$ .