Calculus 2 Quiz: Disc Method Other Axes

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Disc Method Other AxesQuestion 1 of 20

A region is bounded by $y = \ln x$ , $x = 1$ , $x = e^2$ , and $y = 0$ . When this region is revolved around the horizontal line $y = -3$ , the volume using the disc method is:

\pi \int_0^2 (e^y + 3)^2 \, dy

\pi \int_1^{e^2} (\ln x + 3)^2 \, dx

\pi \int_0^2 [(e^y + 3)^2 - 9] \, dy

\pi \int_1^{e^2} [(\ln x + 3)^2 - 9] \, dx

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Calculus 2 Quiz

Calculus 2 Quiz: Disc Method Other Axes

Practice Disc Method Other Axes in Calculus 2 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 2.

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Question 1

A region is bounded by $y = \ln x$ , $x = 1$ , $x = e^2$ , and $y = 0$ . When this region is revolved around the horizontal line $y = -3$ , the volume using the disc method is:

$\pi \int_0^2 (e^y + 3)^2 \, dy$
$\pi \int_1^{e^2} (\ln x + 3)^2 \, dx$ (correct answer)
$\pi \int_0^2 [(e^y + 3)^2 - 9] \, dy$
$\pi \int_1^{e^2} [(\ln x + 3)^2 - 9] \, dx$

Explanation: When revolving around

y = -3

, each vertical slice at

x

creates a disc with radius equal to the distance from

y = -3

y = \ln x

, which is

\ln x - (-3) = \ln x + 3

. We integrate with respect to

x

from

x = 1

x = e^2

. The volume is

\pi \int_1^{e^2} (\ln x + 3)^2 \, dx

. Choice A uses the wrong variable (should integrate with respect to

x

, not

y

). Choices C and D incorrectly subtract

9

, applying washer method when there's no inner hole.

Question 2

Let R be the region enclosed by the graphs of $y = x^2$ , $y = 4$ , and $x = 0$ . The region R is revolved about the line $y = 4$ . What is the volume of the solid generated?

$\frac{256\pi}{15}$ (correct answer)
$\frac{128\pi}{15}$
$\frac{64\pi}{5}$
$\frac{32\pi}{3}$

Explanation: The region is revolved around a horizontal line, so we integrate with respect to

x

. The intersection of

y=x^2

and

y=4

is at

x=2

. The bounds of integration are from

x=0

x=2

. The axis of revolution

y=4

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

x

R(x) = 4 - x^2

. The volume is given by the integral

V = \pi \int_{0}^{2} (R(x))^2 dx = \pi \int_{0}^{2} (4 - x^2)^2 dx

. Expanding the integrand gives

V = \pi \int_{0}^{2} (16 - 8x^2 + x^4) dx

. Evaluating the integral gives

V = \pi [16x - \frac{8}{3}x^3 + \frac{1}{5}x^5]_{0}^{2} = \pi (32 - \frac{64}{3} + \frac{32}{5}) = \pi (\frac{480 - 320 + 96}{15}) = \frac{256\pi}{15}

Question 3

The integral $\pi \int_{0}^{4} (2 - \sqrt{y})^2 dy$ represents the volume of a solid of revolution. The solid is generated by revolving a region R about which line?

The line $x=2$ (correct answer)
The line $x=0$ (the y-axis)
The line $y=2$
The line $y=0$ (the x-axis)

Explanation: The integral is with respect to

y

, which means the discs are horizontal and the axis of revolution is vertical. The formula for the disc method is

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

. Here, the radius is

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

. The radius for a vertical axis of revolution

x=h

R(y) = |h - g(y)|

, where

x=g(y)

is the boundary curve. From

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

, we can identify the axis as

x=2

and the curve as

x=\sqrt{y}

. The region R is bounded by

x=\sqrt{y}

and the axis

x=2

, over the y-interval

[0, 4]

. Therefore, the line of revolution is

x=2

Question 4

The region R is bounded by $y = x^2$ , the x-axis, and the line $x = a$ for $a > 0$ . R is revolved about the line $x = a$ . If the volume of the resulting solid is $\frac{8\pi}{3}$ , what is the value of $a$ ?

$\sqrt{2}$
$2$ (correct answer)
$\sqrt[4]{16/5}$
$\sqrt[4]{10}$

Explanation: The axis of revolution is vertical (

x=a

), so we integrate with respect to

y

. We rewrite the function as

x=\sqrt{y}

. The y-bounds are from

y=0

y=a^2

. The axis

x=a

is a boundary of the region. The radius is

R(y) = a - \sqrt{y}

. The volume is

V = \pi \int_{0}^{a^2} (a - \sqrt{y})^2 dy = \pi \int_{0}^{a^2} (a^2 - 2ay^{1/2} + y) dy

. Integrating gives

V = \pi [a^2y - 2a(\frac{2}{3})y^{3/2} + \frac{1}{2}y^2]_{0}^{a^2} = \pi [a^2(a^2) - \frac{4a}{3}(a^2)^{3/2} + \frac{1}{2}(a^2)^2] = \pi [a^4 - \frac{4a}{3}(a^3) + \frac{1}{2}a^4] = \pi [a^4 - \frac{4}{3}a^4 + \frac{1}{2}a^4] = \pi a^4 (1 - \frac{4}{3} + \frac{1}{2}) = \pi a^4 (\frac{6-8+3}{6}) = \frac{\pi a^4}{6}

. We are given

V=\frac{8\pi}{3}

. So,

\frac{\pi a^4}{6} = \frac{8\pi}{3} \implies a^4 = \frac{8 \cdot 6}{3} = 16

. Thus,

a = 2

(since

a>0

Question 5

The region bounded by $y=1$ , $y=1/\sqrt{x}$ , and $x=4$ is revolved about the line $y=1$ . What is the volume of the solid?

$\pi (3 - \ln(4))$
$\pi (1 - \ln(2))$
$\pi (\ln(2) - 1/2)$
$\pi (\ln(4) - 1)$ (correct answer)

Explanation: When you encounter a solid of revolution problem, you need to identify the region being revolved and choose the appropriate method. Since we're revolving around a horizontal line (y = 1), the washer method is ideal. First, visualize the region. The curve

y = \frac{1}{\sqrt{x}}

intersects

y = 1

when

\frac{1}{\sqrt{x}} = 1

, so

x = 1

. The region is bounded by these curves from

x = 1

x = 4

. For the washer method revolving around

y = 1

, you need the outer radius

R(x)

and inner radius

r(x)

. Since

\frac{1}{\sqrt{x}} > 1

when

x < 1

and

\frac{1}{\sqrt{x}} < 1

when

x > 1

, in our interval

[1,4]

, the curve

y = \frac{1}{\sqrt{x}}

lies below

y = 1

. Therefore:

Outer radius: $R(x) = 1 - 1 = 0$ (distance from axis $y = 1$ to line $y = 1$ )
Inner radius: $r(x) = 1 - \frac{1}{\sqrt{x}}$ (distance from axis to curve)

Wait - this setup is incorrect. The outer radius should be

R(x) = 1 - \frac{1}{\sqrt{x}}

and inner radius is 0, giving us:

V = \pi \int_1^4 \left(1 - \frac{1}{\sqrt{x}}\right)^2 dx = \pi \int_1^4 \left(1 - \frac{2}{\sqrt{x}} + \frac{1}{x}\right) dx

= \pi \left[x - 4\sqrt{x} + \ln|x|\right]_1^4 = \pi[(4 - 8 + \ln 4) - (1 - 4 + 0)] = \pi(\ln 4 - 1)

This confirms answer D. Answer A uses

\ln 4

but with wrong arithmetic. Answer B substitutes

\ln 2

incorrectly. Answer C has the wrong sign relationship. Strategy tip: Always sketch the region first and carefully identify which function is farther from the axis of revolution - this determines your outer radius.

Question 6

The region R is bounded by $y=f(x)$ , $y=k$ , $x=a$ , and $x=b$ , where $f(x) \le k$ for $x$ in $[a, b]$ . A solid is formed by revolving R about the line $y=k$ . Which expression gives the volume of this solid?

$\pi \int_{a}^{b} (k^2 - [f(x)]^2) dx$
$\pi \int_{a}^{b} (f(x) + k)^2 dx$
$\pi \int_{a}^{b} (k - f(x))^2 dx$ (correct answer)
$2\pi \int_{a}^{b} x(k - f(x)) dx$

Explanation: The solid is generated by revolving a region about one of its boundaries (

y=k

), so the disc method is applicable. The axis of revolution is horizontal, so the integration is with respect to

x

from

a

b

. The radius

R(x)

of a disc at a given

x

is the distance between the axis

y=k

and the curve

y=f(x)

. Since

f(x) \le k

, the radius is

R(x) = k - f(x)

. The volume of the solid is given by

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

Question 7

The region enclosed by the parabola $x = y^2+1$ and the line $x=2$ is revolved about the line $x=2$ . What is the volume of the resulting solid?

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

Explanation: The axis of revolution is vertical (

x=2

), so we integrate with respect to

y

. First, find the bounds of integration by setting the x-values equal:

y^2+1 = 2 \implies y^2 = 1 \implies y = \pm 1

. The axis of revolution is a boundary of the region, so we use the disc method. The radius of a disc at height

y

is the horizontal distance from the axis to the curve:

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

. The volume is given by

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

. By symmetry, this is

2\pi \int_{0}^{1} (1-2y^2+y^4) dy = 2\pi [y - \frac{2}{3}y^3 + \frac{1}{5}y^5]_{0}^{1} = 2\pi (1 - \frac{2}{3} + \frac{1}{5}) = 2\pi (\frac{15-10+3}{15}) = 2\pi (\frac{8}{15}) = \frac{16\pi}{15}

Question 8

The region enclosed by $y=\cos(x)$ , $y=1$ , $x=0$ , and $x=\pi/2$ is revolved about the line $y=1$ . Which of the following integrals gives the volume of the generated solid?

$\pi \int_{0}^{\pi/2} (1 - \cos(x))^2 dx$ (correct answer)
$\pi \int_{0}^{\pi/2} (1 - \cos^2(x)) dx$
$\pi \int_{0}^{\pi/2} \cos^2(x) dx$
$\pi \int_{0}^{\pi/2} (1 + \cos(x))^2 dx$

Explanation: The axis of revolution is horizontal (

y=1

), so we integrate with respect to

x

. The bounds are given as

x=0

x=\pi/2

. Since the region is bounded by

y=1

and revolved about

y=1

, this is a disc method problem. The radius of a disc at

x

is the distance from the axis of revolution

y=1

to the curve

y=\cos(x)

. This distance is

R(x) = 1 - \cos(x)

, since

1 \ge \cos(x)

on the interval. The volume is given by the formula

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

, which in this case is

V = \pi \int_{0}^{\pi/2} (1 - \cos(x))^2 dx

Question 9

The region bounded by $y=\ln(x)$ , $y=1$ , and $x=1$ is revolved about the line $x=1$ . What is the volume of the solid?

$\pi (\frac{1}{2}e^2 - 2e + \frac{1}{2})$
$\pi (\frac{1}{2}e^2 - 2e + \frac{5}{2})$ (correct answer)
$\pi (\frac{1}{2}e^2 - 2e - \frac{1}{2})$
$\pi (e^2 - 2e + 1)$

Explanation: The axis of revolution is vertical (

x=1

), so we integrate with respect to

y

. We express the curve as

x = e^y

. The region is bounded by

x=1

and

x=e^y

. The y-bounds are determined by the horizontal lines bounding the region. The curve

y=\ln(x)

passes through

(1,0)

, so the lower bound is

y=0

. The upper bound is given as

y=1

. The axis

x=1

is a boundary of the region. The radius is

R(y) = e^y - 1

. The volume is

V = \pi \int_{0}^{1} (e^y - 1)^2 dy = \pi \int_{0}^{1} (e^{2y} - 2e^y + 1) dy

. Evaluating the integral gives

V = \pi [\frac{1}{2}e^{2y} - 2e^y + y]_{0}^{1} = \pi [(\frac{1}{2}e^2 - 2e + 1) - (\frac{1}{2}e^0 - 2e^0 + 0)] = \pi [(\frac{1}{2}e^2 - 2e + 1) - (\frac{1}{2} - 2)] = \pi [\frac{1}{2}e^2 - 2e + 1 + \frac{3}{2}] = \pi [\frac{1}{2}e^2 - 2e + \frac{5}{2}]

Question 10

The region enclosed by the graphs of $y=1/x$ , $y=-1$ , $x=1$ , and $x=3$ is revolved about the line $y=-1$ . Which integral correctly represents the volume of the solid?

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

Explanation: The axis of revolution is the horizontal line

y=-1

, which is a boundary of the specified region. Therefore, the disc method applies. We integrate with respect to

x

from

x=1

x=3

. The radius

R(x)

of a disc at a given

x

is the distance from the axis of revolution

y=-1

to the other boundary curve,

y=1/x

. Since

1/x > -1

for

x>0

, the radius is

R(x) = \frac{1}{x} - (-1) = \frac{1}{x} + 1

. The volume formula for the disc method is

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

. Substituting the radius and bounds gives

V = \pi \int_{1}^{3} (\frac{1}{x} + 1)^2 dx

Question 11

The region in the first quadrant enclosed by $y = x^3$ , $y=8$ , and the y-axis is revolved about the line $y=8$ . Find the volume of the solid generated.

$\frac{128\pi}{7}$
$\frac{256\pi}{7}$
$\frac{576\pi}{7}$ (correct answer)
$\frac{704\pi}{7}$

Explanation: The axis of revolution is horizontal (

y=8

), so we integrate with respect to

x

. The region is bounded by

x=0

(the y-axis) and the intersection of

y=x^3

and

y=8

, which is

x^3=8 \implies x=2

. The bounds are

x=0

x=2

. The axis of revolution

y=8

is a boundary of the region, so we use the disc method. The radius is the vertical distance from the axis to the curve:

R(x) = 8 - x^3

. The volume is

V = \pi \int_{0}^{2} (8-x^3)^2 dx = \pi \int_{0}^{2} (64 - 16x^3 + x^6) dx

. Integrating gives

V = \pi [64x - 4x^4 + \frac{1}{7}x^7]_{0}^{2} = \pi (64(2) - 4(2^4) + \frac{1}{7}(2^7)) = \pi (128 - 64 + \frac{128}{7}) = \pi (64 + \frac{128}{7}) = \pi (\frac{448+128}{7}) = \frac{576\pi}{7}

Question 12

The area bounded by $x = (y-2)^2$ and the line $x=1$ is revolved about the line $x=1$ . What is the resulting volume?

$\frac{16\pi}{15}$ (correct answer)
$\frac{8\pi}{15}$
$\frac{4\pi}{3}$
$\frac{32\pi}{15}$

Explanation: The axis of revolution is vertical (

x=1

), so we integrate with respect to

y

. Find the intersection points:

1 = (y-2)^2 \implies y-2 = \pm 1 \implies y=3

y=1

. These are the bounds of integration. The axis

x=1

is a boundary of the region. The radius is the horizontal distance from the axis to the curve:

R(y) = 1 - x = 1 - (y-2)^2

. The volume is

V = \pi \int_{1}^{3} (1 - (y-2)^2)^2 dy

. Let

u=y-2

, so

du=dy

. The bounds become

u=1-2=-1

and

u=3-2=1

. The integral becomes

V = \pi \int_{-1}^{1} (1-u^2)^2 du = \pi \int_{-1}^{1} (1-2u^2+u^4) du

. This is an even function, so

V = 2\pi \int_{0}^{1} (1-2u^2+u^4) du = 2\pi [u - \frac{2}{3}u^3 + \frac{1}{5}u^5]_0^1 = 2\pi (1 - \frac{2}{3} + \frac{1}{5}) = 2\pi(\frac{8}{15}) = \frac{16\pi}{15}

Question 13

The region bounded by $y = \tan(x)$ , $x = \pi/4$ , and the x-axis is revolved around the line $x = \pi/4$ . Which of the following integrals represents the volume of the resulting solid?

$\pi \int_{0}^{1} (\frac{\pi}{4} - \arctan(y))^2 dy$ (correct answer)
$\pi \int_{0}^{\pi/4} (\tan(x))^2 dx$
$\pi \int_{0}^{1} ((\frac{\pi}{4})^2 - (\arctan(y))^2) dy$
$\pi \int_{0}^{\pi/4} (\frac{\pi}{4} - \tan(x))^2 dx$

Explanation: The axis of revolution is vertical (

x = \pi/4

), so we must integrate with respect to

y

. The function must be expressed as

x

in terms of

y

x = \arctan(y)

. The bounds for

y

correspond to

x=0

and

x=\pi/4

. When

x=0

y=\tan(0)=0

. When

x=\pi/4

y=\tan(\pi/4)=1

. So,

y

ranges from 0 to 1. The axis of revolution

x=\pi/4

is a boundary of the region. The radius of a disc at height

y

R(y) = \frac{\pi}{4} - x = \frac{\pi}{4} - \arctan(y)

. The volume is

V = \pi \int_{0}^{1} [R(y)]^2 dy = \pi \int_{0}^{1} (\frac{\pi}{4} - \arctan(y))^2 dy

Question 14

The region bounded by the curve $x=2\sqrt{y}$ and the line $x=2$ is revolved about the line $x=2$ . What is the volume of the generated solid?

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

Explanation: The axis of revolution is vertical (

x=2

), so we integrate with respect to

y

. The region is bounded by

x=2\sqrt{y}

and

x=2

. They intersect when

2=2\sqrt{y} \implies \sqrt{y}=1 \implies y=1

. The parabola

x=2\sqrt{y}

starts at

(0,0)

, so the y-bounds are from 0 to 1. The axis of revolution

x=2

is a boundary of the region. The radius is

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

. The volume is

V = \pi \int_{0}^{1} (2 - 2\sqrt{y})^2 dy = 4\pi \int_{0}^{1} (1 - \sqrt{y})^2 dy = 4\pi \int_{0}^{1} (1 - 2y^{1/2} + y) dy

. Evaluating the integral:

V = 4\pi [y - 2(\frac{2}{3})y^{3/2} + \frac{1}{2}y^2]_{0}^{1} = 4\pi [1 - \frac{4}{3} + \frac{1}{2}] = 4\pi [\frac{6 - 8 + 3}{6}] = 4\pi [\frac{1}{6}] = \frac{2\pi}{3}

Question 15

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

$3\pi/2$
$3\pi$
$9\pi$ (correct answer)
$18\pi$

Explanation: The axis of revolution is horizontal (

y=1

), so we integrate with respect to

x

. The region is bounded by the y-axis (

x=0

), the line

y=1

, and the line

y=x-2

. The intersection of

y=x-2

and

y=1

occurs at

1=x-2

, which gives

x=3

. So the bounds of integration are from

x=0

x=3

. The axis of revolution

y=1

is a boundary of the region. The radius of a disc is the vertical distance from the axis to the other boundary:

R(x) = 1 - (x-2) = 3-x

. The volume is

V = \pi \int_{0}^{3} (3-x)^2 dx

. Evaluating the integral:

V = \pi [-\frac{(3-x)^3}{3}]_{0}^{3} = \pi [0 - (-\frac{3^3}{3})] = \pi [\frac{27}{3}] = 9\pi

Question 16

The region bounded by $y=x^2-2$ and $y=-2$ from $x=0$ to $x=1$ is revolved about the line $y=-2$ . Find the volume of the solid.

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

Explanation: When you encounter a volume of revolution problem, start by visualizing the region and identifying the axis of rotation. Here, you're rotating the region between

y = x^2 - 2

and

y = -2

around the line

y = -2

. Since the axis of rotation is

y = -2

, which is also the lower boundary of your region, this becomes a disk method problem. The radius of each disk at position

x

is the distance from the axis

y = -2

to the upper curve

y = x^2 - 2

. This radius is:

r(x) = (x^2 - 2) - (-2) = x^2

. Using the disk method formula:

V = \pi \int_0^1 [r(x)]^2 \, dx = \pi \int_0^1 (x^2)^2 \, dx = \pi \int_0^1 x^4 \, dx

Evaluating:

V = \pi \left[\frac{x^5}{5}\right]_0^1 = \pi \cdot \frac{1}{5} = \frac{\pi}{5}

The incorrect answers stem from common mistakes: A)

\frac{2\pi}{3}

likely comes from incorrectly using the washer method or miscalculating the integral. B)

\frac{2\pi}{5}

suggests you found the correct integral but doubled the result, possibly by confusion about the setup. C)

\frac{\pi}{3}

indicates an integration error, perhaps evaluating

\int x^4 dx

\frac{x^4}{3}

instead of

\frac{x^5}{5}

. The correct answer is D)

\frac{\pi}{5}

. Strategy tip: Always identify whether the axis of rotation forms a boundary of your region. If it does, use the disk method; if not, you'll likely need the washer method.

Question 17

The region in the first quadrant bounded by $x=y^2$ and $x=4$ is revolved about the line $x=4$ . What is the volume of the solid?

$\frac{128\pi}{15}$
$\frac{256\pi}{15}$ (correct answer)
$\frac{64\pi}{5}$
$\frac{512\pi}{15}$

Explanation: The axis of revolution is vertical (

x=4

), so we integrate with respect to

y

. The curve is

x=y^2

. The intersection with

x=4

4=y^2 \implies y=2

in the first quadrant. So the y-bounds are from 0 to 2. The axis

x=4

is a boundary of the region. The radius is

R(y) = 4 - x = 4 - y^2

. The volume is

V = \pi \int_{0}^{2} (4-y^2)^2 dy = \pi \int_{0}^{2} (16 - 8y^2 + y^4) dy

. Integrating gives

V = \pi [16y - \frac{8}{3}y^3 + \frac{1}{5}y^5]_{0}^{2} = \pi [16(2) - \frac{8}{3}(2^3) + \frac{1}{5}(2^5)] = \pi [32 - \frac{64}{3} + \frac{32}{5}] = \pi [\frac{480 - 320 + 96}{15}] = \frac{256\pi}{15}

Question 18

A solid is formed by revolving the region bounded by $x=9-y^2$ and $x=5$ about the line $x=5$ . What is the volume?

$\frac{256\pi}{5}$
$\frac{512\pi}{15}$ (correct answer)
$\frac{256\pi}{15}$
$\frac{128\pi}{5}$

Explanation: The axis of revolution is vertical (

x=5

), so we integrate with respect to

y

. Find the intersections:

9-y^2=5 \implies y^2=4 \implies y=\pm 2

. These are the bounds. The axis

x=5

is a boundary of the region. The radius is

R(y) = (9-y^2) - 5 = 4-y^2

. The volume is

V = \pi \int_{-2}^{2} (4-y^2)^2 dy

. This is an even function, so

V = 2\pi \int_{0}^{2} (16 - 8y^2 + y^4) dy = 2\pi [16y - \frac{8}{3}y^3 + \frac{1}{5}y^5]_{0}^{2} = 2\pi [32 - \frac{64}{3} + \frac{32}{5}] = 2\pi [\frac{480-320+96}{15}] = 2\pi [\frac{256}{15}] = \frac{512\pi}{15}

Question 19

The region bounded by $y=2\sqrt{x}$ , $y=4$ , and $x=0$ is revolved about the line $y=4$ . What is the volume of the solid?

$\frac{16\pi}{5}$
$\frac{32\pi}{5}$
$\frac{64\pi}{3}$
$\frac{32\pi}{3}$ (correct answer)

Explanation: The axis of revolution is horizontal (

y=4

), so we integrate with respect to

x

. Find the intersection of

y=2\sqrt{x}

and

y=4

4=2\sqrt{x} \implies \sqrt{x}=2 \implies x=4

. The bounds of integration are from

x=0

x=4

. The axis of revolution,

y=4

, is a boundary of the region. The radius of a disc is

R(x) = 4 - 2\sqrt{x}

. The volume is

V = \pi \int_{0}^{4} (4 - 2\sqrt{x})^2 dx = \pi \int_{0}^{4} (16 - 16x^{1/2} + 4x) dx

. Evaluating the integral:

V = \pi [16x - 16(\frac{2}{3})x^{3/2} + 2x^2]_{0}^{4} = \pi [(16(4) - \frac{32}{3}(4^{3/2}) + 2(4^2)) - 0] = \pi [64 - \frac{32}{3}(8) + 32] = \pi [96 - \frac{256}{3}] = \pi [\frac{288 - 256}{3}] = \frac{32\pi}{3}

Question 20

Let R be the region bounded by $y = e^x$ , $y=1$ , and $x=2$ . R is revolved about the line $y=1$ . The volume of the generated solid is:

$\pi (\frac{1}{2}e^4 - 2e^2 + \frac{7}{2})$ (correct answer)
$\pi (\frac{1}{2}e^4 - 2e^2 + \frac{3}{2})$
$\pi (\frac{1}{2}e^4 - 1)$
$\pi (e^4 - 2e^2 + 1)$

Explanation: The axis of revolution is horizontal (

y=1

), so we integrate with respect to

x

. The region is bounded by

y=e^x

and

y=1

. They intersect when

e^x=1

, which is

x=0

. The other boundary is

x=2

. So the bounds are from 0 to 2. The axis of revolution is a boundary of the region. The radius is

R(x) = e^x - 1

. The volume is

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

. Integrating gives

V = \pi [\frac{1}{2}e^{2x} - 2e^x + x]_{0}^{2} = \pi ((\frac{1}{2}e^4 - 2e^2 + 2) - (\frac{1}{2}e^0 - 2e^0 + 0)) = \pi (\frac{1}{2}e^4 - 2e^2 + 2 - (\frac{1}{2} - 2)) = \pi (\frac{1}{2}e^4 - 2e^2 + 2 + \frac{3}{2}) = \pi (\frac{1}{2}e^4 - 2e^2 + \frac{7}{2})