Calculus 1 Quiz: Washer Method Other Axes

20 questions · exam conditions

0:00

Washer Method Other AxesQuestion 1 of 20

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 formed by revolving $R$ about the line $y=3$ ?

\pi \int_{0}^{4} [3^2 - (3-\sqrt{x})^2] \,dx

\pi \int_{0}^{4} [(3-\sqrt{x})^2 - 3^2] \,dx

\pi \int_{0}^{4} [3^2 - (\sqrt{x})^2] \,dx

\pi \int_{0}^{4} [(3+\sqrt{x})^2 - 3^2] \,dx

← Back to quizzes

Calculus 1 Quiz

Calculus 1 Quiz: Washer Method Other Axes

Practice Washer 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.

What this quiz covers

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

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

$\pi \int_{0}^{4} [3^2 - (3-\sqrt{x})^2] \,dx$ (correct answer)
$\pi \int_{0}^{4} [(3-\sqrt{x})^2 - 3^2] \,dx$
$\pi \int_{0}^{4} [3^2 - (\sqrt{x})^2] \,dx$
$\pi \int_{0}^{4} [(3+\sqrt{x})^2 - 3^2] \,dx$

Explanation: The region is bounded by

y=\sqrt{x}

and

y=0

from

x=0

x=4

. The axis of revolution is the horizontal line

y=3

, which is above the region. Using the washer method, we integrate with respect to

x

. The outer radius

R(x)

is the distance from the axis of revolution

y=3

to the farther function, which is

y=0

. So,

R(x) = 3 - 0 = 3

. The inner radius

r(x)

is the distance from the axis of revolution

y=3

to the closer function, which is

y=\sqrt{x}

. So,

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

. The volume is given by

V = \pi \int_{0}^{4} [R(x)^2 - r(x)^2] \,dx = \pi \int_{0}^{4} [3^2 - (3-\sqrt{x})^2] \,dx

Question 2

Let $R$ be the region in the first quadrant enclosed by the graphs of $y=x$ and $y=\sqrt[3]{x}$ . Which definite integral represents the volume of the solid generated when $R$ is revolved about the line $y=-2$ ?

$\pi \int_{0}^{1} (\sqrt[3]{x}-x)^2 \,dx$
$\pi \int_{0}^{1} [(x+2)^2 - (\sqrt[3]{x}+2)^2] \,dx$
$\pi \int_{0}^{1} [(\sqrt[3]{x})^2 - x^2] \,dx$
$\pi \int_{0}^{1} [(\sqrt[3]{x}+2)^2 - (x+2)^2] \,dx$ (correct answer)

Explanation: When finding the volume of a solid of revolution about a horizontal line, you need to use the washer method if the region has both an upper and lower boundary. The key insight is measuring distances from each boundary curve to the axis of rotation. First, let's identify the region R. The curves

y = x

and

y = \sqrt[3]{x}

intersect when

x = \sqrt[3]{x}

, which occurs at

x = 0

and

x = 1

. Between these points,

\sqrt[3]{x} > x

(since the cube root function grows faster than the linear function for

0 < x < 1

), so

y = \sqrt[3]{x}

is the upper boundary. When revolving about

y = -2

, each cross-section perpendicular to the x-axis forms a washer. The outer radius extends from

y = -2

to the upper curve

y = \sqrt[3]{x}

, giving distance

\sqrt[3]{x} - (-2) = \sqrt[3]{x} + 2

. The inner radius extends from

y = -2

to the lower curve

y = x

, giving distance

x - (-2) = x + 2

. The washer method formula is

V = \pi \int_a^b [R_{outer}^2 - R_{inner}^2] \, dx

, so we get

\pi \int_0^1 [(\sqrt[3]{x} + 2)^2 - (x + 2)^2] \, dx

, which is choice D. Choice A uses the difference of functions, not the washer method. Choice B incorrectly swaps which radius is outer versus inner. Choice C ignores the axis of rotation entirely, treating it as if revolving about the x-axis. Strategy tip: Always sketch the region and identify which curve is farther from the axis of rotation—that determines your outer radius in the washer method.

Question 3

The volume of a solid is given by the definite integral $\pi \int_{1}^{5} [ (6-y)^2 - (6-(y^2-4y+5))^2 ] \,dy$ . This solid is formed by revolving a region $R$ about an axis. Which of the following correctly describes the region $R$ and the axis of revolution?

Region $R$ is bounded by $x=y$ and $x=y^2-4y+5$ , revolved about the line $x=y^2-4y+5$ .
Region $R$ is bounded by $y=x$ and $y=x^2-4x+5$ , revolved about the line $y=6$ .
Region $R$ is bounded by $x=y$ and $x=y^2-4y+5$ , revolved about the y-axis.
Region $R$ is bounded by $x=y$ and $x=y^2-4y+5$ , revolved about the line $x=6$ . (correct answer)

Explanation: When you encounter a volume integral in the form

\pi \int_a^b [R(y)^2 - r(y)^2] dy

, you're looking at the washer method for solids of revolution. The key insight is recognizing what each component represents: the outer radius

R(y)

, inner radius

r(y)

, and axis of revolution. Looking at this integral, we have outer radius

(6-y)

and inner radius

(6-(y^2-4y+5))

. Notice both radii are measured as distances from the line

x = 6

. The outer radius goes from

x = 6

x = y

, while the inner radius goes from

x = 6

x = y^2-4y+5

. Since we're integrating with respect to

y

from 1 to 5, the region is bounded by these two curves between those y-values. Answer D correctly identifies that region

R

is bounded by

x = y

and

x = y^2-4y+5

, revolved about the line

x = 6

. Answer A incorrectly suggests revolving about

x = y^2-4y+5

, which would produce different radius expressions. Answer B switches to functions of

x

and revolution about

y = 6

, but our integral uses

dy

, indicating functions of

y

with horizontal axis of revolution. Answer C suggests revolution about the y-axis, which would give radii of

y

and

y^2-4y+5

, not the

(6-\text{something})

form we see. Remember: in washer method problems, the radius expressions tell you the axis of revolution. When you see

(c - f(y))

, the axis is the vertical line

x = c

Question 4

Let $R$ be the region bounded by the graphs of $y=\sin(\pi x)$ and $y=x-x^2$ . Which integral gives the volume of the solid generated when $R$ is revolved about the line $y=2$ ?

$\pi \int_{0}^{1} [\sin^2(\pi x) - (x-x^2)^2] \,dx$
$\pi \int_{0}^{1} [(2-\sin(\pi x))^2 - (2-x+x^2)^2] \,dx$
$\pi \int_{0}^{1} [(2-x+x^2)^2 - (2-\sin(\pi x))^2] \,dx$ (correct answer)
$\pi \int_{0}^{1} (\sin(\pi x) - x + x^2)^2 \,dx$

Explanation: When you encounter a volume of revolution problem, you need to identify which function is the outer radius and which is the inner radius when rotating about a horizontal line. The washer method formula is

V = \pi \int_a^b [R_{outer}^2 - R_{inner}^2] \,dx

, where the radii are measured as distances from the axis of rotation. First, determine which function is farther from the line

y = 2

. Since

y = \sin(\pi x)

oscillates between 0 and 1, and

y = x - x^2

forms a parabola opening downward with maximum at

x = \frac{1}{2}

, you need to check their relative positions. The parabola

y = x - x^2

is actually above

y = \sin(\pi x)

throughout most of the interval

[0,1]

, making it closer to the line

y = 2

. The outer radius is the distance from

y = 2

y = x - x^2

R_{outer} = 2 - (x - x^2) = 2 - x + x^2

. The inner radius is the distance from

y = 2

y = \sin(\pi x)

R_{inner} = 2 - \sin(\pi x)

. This gives us

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

, which is answer C. Answer A uses the disk method incorrectly without accounting for the axis shift. Answer B reverses the outer and inner radii. Answer D attempts to use the difference of functions rather than the washer method. Remember: always sketch the region and identify which function is farther from the axis of rotation to determine your outer radius correctly.

Question 5

The region bounded by the parabola $y=x^2$ and the line $y=4$ is revolved about the line $y=k$ , where $k$ is a constant and $k > 4$ . The volume of the resulting solid is $V(k)$ . Which of the following expressions represents $V(k)$ ?

$\pi \int_{-2}^{2} [ (k-4)^2 - (k-x^2)^2 ] \,dx$
$\pi \int_{-2}^{2} [ (k-x^2)^2 - (k-4)^2 ] \,dx$ (correct answer)
$\pi \int_{-2}^{2} (4-x^2)^2 \,dx$
$\pi \int_{-2}^{2} [ (k-x^2)^2 - 16 ] \,dx$

Explanation: When you encounter a solid of revolution problem, you need to carefully set up the washer method formula, paying close attention to which function is farther from and closer to the axis of rotation. Since we're revolving about the line

y = k

where

k > 4

, we're rotating above the entire region. The region is bounded by

y = x^2

(parabola) and

y = 4

(horizontal line), with intersection points at

x = -2

and

x = 2

. For the washer method, the volume formula is

V = \pi \int_a^b [R^2 - r^2] \, dx

, where

R

is the outer radius and

r

is the inner radius. Since

k > 4

, the distance from the axis

y = k

to any point on the parabola

y = x^2

k - x^2

, and the distance to the line

y = 4

k - 4

. Because

x^2 \leq 4

in our region, we have

k - x^2 \geq k - 4

, making

k - x^2

the outer radius and

k - 4

the inner radius. Therefore,

V(k) = \pi \int_{-2}^{2} [(k-x^2)^2 - (k-4)^2] \, dx

, which is choice B. Choice A incorrectly swaps the outer and inner radii. Choice C ignores the axis of rotation entirely, using a disk method setup as if rotating about the x-axis. Choice D uses the wrong inner radius, substituting

16

instead of

(k-4)^2

. Strategy tip: Always identify which boundary is farther from the rotation axis to determine your outer radius correctly. Draw a quick sketch to visualize the setup before writing your integral.

Question 6

Let $R$ be the region enclosed by the graphs of the line $y=x-1$ and the parabola $y^2=2x+6$ . Which of the following definite integrals represents the volume of the solid generated when $R$ is revolved about the line $x=10$ ?

$\pi \int_{-2}^{4} \left[ \left(13 - \frac{1}{2}y^2\right)^2 - (9-y)^2 \right] \,dy$ (correct answer)
$\pi \int_{-2}^{4} \left[ (y+1)^2 - \left(\frac{1}{2}y^2-3\right)^2 \right] \,dy$
$\pi \int_{-2}^{4} \left[ (9-y)^2 - \left(13 - \frac{1}{2}y^2\right)^2 \right] \,dy$
$\pi \int_{-3}^{5} \left[ (10-(x-1))^2 - (10-\sqrt{2x+6})^2 \right] \,dx$

Explanation: To revolve around a vertical line

x=10

, we must integrate with respect to

y

. First, express

x

in terms of

y

x=y+1

and

x=\frac{1}{2}y^2-3

. Find intersection points:

y+1 = \frac{1}{2}y^2-3 \implies y^2-2y-8=0 \implies (y-4)(y+2)=0

. So, the limits of integration are

y=-2

y=4

. The axis of revolution

x=10

is to the right of the region. The outer radius

R(y)

is the distance from

x=10

to the farther curve (the left curve), which is

x=\frac{1}{2}y^2-3

. So,

R(y) = 10 - (\frac{1}{2}y^2-3) = 13 - \frac{1}{2}y^2

. The inner radius

r(y)

is the distance from

x=10

to the closer curve (the right curve), which is

x=y+1

. So,

r(y) = 10 - (y+1) = 9-y

. The volume is

V = \pi \int_{-2}^{4} [R(y)^2 - r(y)^2] \,dy = \pi \int_{-2}^{4} [ (13 - \frac{1}{2}y^2)^2 - (9-y)^2 ] \,dy

Question 7

Let R be the region in the first quadrant enclosed by the graphs of $x = y^2$ and $x = 2y$ . Which integral gives the volume of the solid generated when R is revolved about the line $x = -1$ ?

$\pi \int_{0}^{2} [(2y+1)^2 - (y^2+1)^2] \,dy$ (correct answer)
$\pi \int_{0}^{2} [(y^2+1)^2 - (2y+1)^2] \,dy$
$\pi \int_{0}^{2} [(2y - y^2)^2] \,dy$
$\pi \int_{0}^{4} [(\sqrt{x}+1)^2 - (x/2+1)^2] \,dx$

Explanation: The curves intersect when

y^2 = 2y

, which gives

y=0

and

y=2

. Since the revolution is about a vertical line, we integrate with respect to

y

. For

y \in [0, 2]

2y \ge y^2

, so

x=2y

is the right curve and

x=y^2

is the left curve. The axis is

x=-1

. The outer radius

R(y)

is the distance from

x=-1

to the right curve

x=2y

, so

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

. The inner radius

r(y)

is the distance from

x=-1

to the left curve

x=y^2

, so

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

. The volume is

\pi \int_{0}^{2} [(2y+1)^2 - (y^2+1)^2] \,dy

Question 8

Let R be the region in the first quadrant bounded by $y = x$ and $y = x^2/4$ . Which integral gives the volume when R is rotated about the line $x=5$ ?

$\pi \int_{0}^{4} [(5-y)^2 - (5-2\sqrt{y})^2] \,dy$ (correct answer)
$\pi \int_{0}^{4} [(5-2\sqrt{y})^2 - (5-y)^2] \,dy$
$\pi \int_{0}^{4} [(x-5)^2 - (x^2/4-5)^2] \,dx$
$\pi \int_{0}^{4} [(2\sqrt{y})^2 - y^2] \,dy$

Explanation: To rotate around the vertical line

x=5

, we integrate with respect to

y

. First, find intersections:

x = x^2/4 \implies 4x = x^2 \implies x(x-4)=0

, so

x=0, 4

. The corresponding y-values are

y=0, 4

. We rewrite the functions as

x=y

and

x=2\sqrt{y}

. In the interval

y \in [0,4]

2\sqrt{y} \ge y

, so

x=2\sqrt{y}

is the right boundary and

x=y

is the left boundary. The axis

x=5

is to the right of the region. The outer radius is the distance from the axis to the farther (left) curve:

R(y) = 5-y

. The inner radius is the distance to the closer (right) curve:

r(y) = 5-2\sqrt{y}

. The volume is

\pi \int_{0}^{4} [(5-y)^2 - (5-2\sqrt{y})^2] \,dy

Question 9

Let R be the region bounded by $y=x^2$ and $y=2x$ . Which integral represents the volume of the solid obtained by revolving R about the line $y=5$ ?

$\pi \int_{0}^{2} [(5-2x)^2 - (5-x^2)^2] \,dx$
$\pi \int_{0}^{2} [(5-x^2)^2 - (5-2x)^2] \,dx$ (correct answer)
$\pi \int_{0}^{4} [(5-\sqrt{y})^2 - (5-y/2)^2] \,dy$
$\pi \int_{0}^{2} [(2x-x^2)^2] \,dx$

Explanation: When finding volumes of revolution using the washer method, you need to identify the outer and inner radii from the axis of rotation to each bounding curve, then apply the formula

V = \pi \int [R_{outer}^2 - R_{inner}^2] \, dx

. First, find where the curves intersect by solving

x^2 = 2x

, giving

x = 0

and

x = 2

. Between these points,

y = 2x

lies above

y = x^2

. When revolving about the horizontal line

y = 5

, the distance from this axis to each curve determines the radii. For any

x

[0,2]

Distance to $y = x^2$ : $5 - x^2$ (this is the outer radius since $x^2 < 5$ )
Distance to $y = 2x$ : $5 - 2x$ (this is the inner radius since $2x < 5$ and $2x > x^2$ )

The volume integral becomes

\pi \int_0^2 [(5-x^2)^2 - (5-2x)^2] \, dx

, which is choice B. Choice A incorrectly swaps the outer and inner radii, treating

5-2x

as outer and

5-x^2

as inner. This happens when you don't carefully determine which curve is farther from the axis of rotation. Choice C uses integration with respect to

y

, which requires different bounds and curve expressions, but the setup shown doesn't correctly represent this region. Choice D attempts to use just the difference between curves without accounting for revolution about

y = 5

, missing the washer method entirely. Key strategy: Always sketch the region and axis of rotation. Identify which curve is farther from the axis (outer radius) and closer (inner radius) before setting up your integral.

Question 10

Let R be the region enclosed by the parabola $x = (y-1)^2$ and the line $x=1$ . Which integral gives the volume of the solid generated by revolving R about the line $x=2$ ?

$\pi \int_{0}^{2} [(2+(y-1)^2)^2 - 3^2] \,dy$
$\pi \int_{0}^{2} [1^2 - (2-(y-1)^2)^2] \,dy$
$\pi \int_{0}^{1} [(2-(1+\sqrt{x}))^2 - (2-(1-\sqrt{x}))^2] \,dx$
$\pi \int_{0}^{2} [(2-(y-1)^2)^2 - 1^2] \,dy$ (correct answer)

Explanation: When finding volumes of revolution using the washer method, you need to identify the outer and inner radii of the washers formed when slicing perpendicular to the axis of rotation. First, let's understand the region R. The parabola

x = (y-1)^2

opens rightward with vertex at (0,1), and the line

x = 1

is vertical. These curves intersect when

(y-1)^2 = 1

, giving

y-1 = ±1

, so

y = 0

and

y = 2

. The region extends from

y = 0

y = 2

. When revolving about the vertical line

x = 2

, we use horizontal slicing (integrating with respect to

y

). At any height

y

, the distance from the axis

x = 2

to the parabola

x = (y-1)^2

2 - (y-1)^2

(outer radius), and the distance to the line

x = 1

2 - 1 = 1

(inner radius). The washer method formula gives us:

V = \pi \int_{0}^{2} [(2-(y-1)^2)^2 - 1^2] \,dy

Option A incorrectly adds the parabola's

x

-coordinate to 2 instead of subtracting, and uses 3 as the inner radius. Option B has the outer and inner radii backwards and gets a negative volume. Option C attempts to integrate with respect to

x

, but the bounds and setup are incorrect for this geometry. Study tip: Always sketch the region and identify which boundary is farther from the axis of rotation—that determines your outer radius. The washer method is outer radius squared minus inner radius squared.

Question 11

Let $R$ be the region enclosed by the graphs of $y=x^2+1$ and $y=x+3$ . Which of the following integrals gives the volume of the solid generated when $R$ is revolved about the line $y=-1$ ?

$\pi \int_{-1}^{2} [(x+4)^2 - (x^2+2)^2] \,dx$ (correct answer)
$\pi \int_{-1}^{2} [(x+3)^2 - (x^2+1)^2] \,dx$
$\pi \int_{-1}^{2} [(x+3)-(x^2+1)]^2 \,dx$
$\pi \int_{-1}^{2} [(x^2+2)^2 - (x+4)^2] \,dx$

Explanation: First, find the points of intersection by setting the functions equal:

x^2+1 = x+3 \implies x^2-x-2=0 \implies (x-2)(x+1)=0

. The points of intersection are

x=-1

and

x=2

. In the interval

[-1, 2]

, the line

y=x+3

is above the parabola

y=x^2+1

. The solid is generated by revolving the region about the horizontal line

y=-1

. The method of washers is appropriate. The outer radius

R(x)

is the distance from the axis of revolution

y=-1

to the outer curve

y=x+3

, so

R(x) = (x+3) - (-1) = x+4

. The inner radius

r(x)

is the distance from the axis of revolution to the inner curve

y=x^2+1

, so

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

. The volume

V

is given by the integral

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

. Substituting the radii and limits gives

V = \pi \int_{-1}^{2} [(x+4)^2 - (x^2+2)^2] \,dx

Question 12

Let $R$ be the region in the first quadrant bounded by the graphs of $y=x^2$ , $y=4$ , and $x=0$ . What is the volume of the solid generated by revolving $R$ about the line $y=5$ ?

$\pi \int_{0}^{2} [1 - (5-x^2)^2] \,dx$
$\pi \int_{0}^{2} [(5-x^2)^2 - 1] \,dx$ (correct answer)
$\pi \int_{0}^{2} [25 - (5-x^2)^2] \,dx$
$\pi \int_{0}^{4} [5^2 - (5-\sqrt{y})^2] \,dy$

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

R

is bounded by

y = x^2

y = 4

, and

x = 0

in the first quadrant, and we're rotating about the horizontal line

y = 5

. Since we're rotating about a horizontal line and our region is naturally described with vertical boundaries, we should use the washer method with vertical slices. For any

x

-value from 0 to 2, we have a vertical strip extending from

y = x^2

up to

y = 4

. When this strip rotates about

y = 5

, it creates a washer. The outer radius is the distance from

y = 5

down to the bottom of the strip:

R(x) = 5 - x^2

. The inner radius is the distance from

y = 5

down to the top of the strip:

r(x) = 5 - 4 = 1

. Using the washer method formula

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

, we get:

V = \pi \int_0^2 [(5-x^2)^2 - 1^2] dx = \pi \int_0^2 [(5-x^2)^2 - 1] dx

This matches answer choice B. Answer choice A incorrectly subtracts the outer radius squared from the inner radius squared—backwards from the washer formula. Answer choice C uses

25

instead of

1

for the inner radius squared, suggesting confusion about which distance to measure. Answer choice D attempts integration with respect to

y

, but sets up the radii incorrectly for that approach. Study tip: Always sketch the region and identify both radii clearly before setting up your integral. The washer method is always outer radius squared minus inner radius squared.

Question 13

Let $R$ be the region bounded by $y=x^3$ , $x=2$ , and the x-axis. Which integral represents the volume of the solid formed by revolving $R$ about the line $x=3$ ?

$2\pi \int_{0}^{2} (3-x)x^3 \,dx$
$\pi \int_{0}^{8} [(3-\sqrt[3]{y})^2 - 1] \,dy$ (correct answer)
$\pi \int_{0}^{8} [1 - (3-\sqrt[3]{y})^2] \,dy$
$\pi \int_{0}^{2} (8^2 - (x^3)^2) \,dx$

Explanation: When you're finding volumes of revolution, you need to choose the right method based on the axis of rotation. Since we're revolving around the vertical line

x=3

, the washer method with horizontal slices is most efficient. First, let's set up the region. You have

y=x^3

from

x=0

x=2

, bounded below by the x-axis. When using horizontal slices, you need to express everything in terms of

y

. Since

y=x^3

, we get

x=\sqrt[3]{y}

, and

y

ranges from

0

8

(since when

x=2

y=2^3=8

). For each horizontal slice at height

y

, you're creating a washer. The outer radius is the distance from the axis

x=3

to the right boundary

x=2

, which is

3-2=1

. The inner radius is the distance from

x=3

to the curve

x=\sqrt[3]{y}

, which is

3-\sqrt[3]{y}

. The washer method gives us volume =

\pi \int [(\text{outer radius})^2 - (\text{inner radius})^2] \,dy = \pi \int_{0}^{8} [1^2 - (3-\sqrt[3]{y})^2] \,dy

. This matches choice B. Choice A uses the shell method incorrectly—it's missing the

\pi

factor and has the wrong integrand structure. Choice C has the radii reversed (subtracting the larger from the smaller). Choice D attempts to use vertical slices but sets up the radii incorrectly, using

8

instead of the proper distance calculations. Study tip: Always sketch the region and identify which radius is larger when using the washer method—the formula is always

\pi \int [R^2 - r^2]

where

R > r

Question 14

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

$\pi \int_{-1}^{2} [(y+2)^2 - (y^2)^2] \,dy$
$\pi \int_{-1}^{2} [(y^2+1)^2 - (y+3)^2] \,dy$
$\pi \int_{-1}^{2} [(y+3)^2 - (y^2+1)^2] \,dy$ (correct answer)
$\pi \int_{-1}^{2} (-y^2+y+2)^2 \,dy$

Explanation: When finding volumes of revolution about vertical or horizontal lines that aren't the coordinate axes, you need to use the washer method with careful attention to the radii from each curve to the axis of revolution. First, find where the curves intersect by solving

y^2 = y + 2

, which gives

y^2 - y - 2 = 0

, so

(y-2)(y+1) = 0

. The curves intersect at

y = -1

and

y = 2

. Since we're revolving about the vertical line

x = -1

, we integrate with respect to

y

. For any horizontal slice at height

y

, we need the distances from each curve to the line

x = -1

Distance from parabola $x = y^2$ to $x = -1$ : $y^2 - (-1) = y^2 + 1$
Distance from line $x = y + 2$ to $x = -1$ : $(y + 2) - (-1) = y + 3$

The line

x = y + 2

is farther from the axis of revolution (forms the outer radius), while the parabola forms the inner radius. Using the washer method:

V = \pi \int_{-1}^{2} [R^2 - r^2] \,dy = \pi \int_{-1}^{2} [(y+3)^2 - (y^2+1)^2] \,dy

. Answer A incorrectly uses

(y^2)^2

instead of

(y^2+1)^2

for the inner radius. Answer B switches the outer and inner radii, which would give a negative volume under the integral. Answer D attempts to use a single function approach, missing that this is a washer (not disk) problem. Strategy tip: Always sketch the region and identify which curve is farther from the axis of revolution—that becomes your outer radius in the washer formula.

Question 15

Let $R$ be the triangular region enclosed by the lines $y=x$ , $y=2-x$ , and $y=0$ . Find the volume of the solid generated by revolving $R$ about the vertical line $x=3$ .

$\pi \int_{0}^{1} [(2-y)^2 - y^2] \,dy$
$\pi \int_{0}^{1} [(1+y)^2 - (3-y)^2] \,dy$
$\pi \int_{0}^{1} [(3-y)^2 - (1+y)^2] \,dy$ (correct answer)
$\pi \int_{0}^{1} (2-2y)^2 \,dy$

Explanation: When finding volumes of revolution about vertical lines, you need to carefully set up the washer method by identifying the outer and inner radii from the axis of revolution. First, sketch the triangular region R. The lines

y=x

y=2-x

, and

y=0

intersect at points

(0,0)

(2,0)

, and

(1,1)

, forming a triangle. Since you're revolving about the vertical line

x=3

, it's most efficient to integrate with respect to

y

from

y=0

y=1

. For any horizontal slice at height

y

, you need the distances from

x=3

to the boundaries of region R. From the line

y=x

, we get

x=y

, so the distance from

x=3

3-y

. From the line

y=2-x

, we get

x=2-y

, so the distance from

x=3

3-(2-y) = 1+y

. Since we're revolving about

x=3

(which lies to the right of region R), the outer radius is the farther distance

3-y

, and the inner radius is the closer distance

1+y

. Using the washer method:

V = \pi \int_{0}^{1} [(3-y)^2 - (1+y)^2] \,dy

. Answer A incorrectly uses

x

-coordinates as radii instead of distances from

x=3

. Answer B reverses the outer and inner radii, which would give a negative volume under the integral. Answer D uses the disk method with width

2-2y

, ignoring that this creates a washer, not a solid disk. Study tip: Always identify which boundary is farther from the axis of revolution—that becomes your outer radius in the washer method.

Question 16

Let R be the region bounded by $y = \sin(x)$ and $y = \cos(x)$ over the interval $[0, \pi/4]$ . What is the volume of the solid generated by revolving R about the line $y = 2$ ?

$\pi \int_{0}^{\pi/4} [(2-\sin(x))^2 - (2-\cos(x))^2] \,dx$ (correct answer)
$\pi \int_{0}^{\pi/4} [(2-\cos(x))^2 - (2-\sin(x))^2] \,dx$
$\pi \int_{0}^{\pi/4} [(\cos(x)-2)^2 - (\sin(x)-2)^2] \,dx$
$\pi \int_{0}^{\pi/4} [\cos^2(x) - \sin^2(x)] \,dx$

Explanation: On the interval

[0, \pi/4]

\cos(x) \ge \sin(x)

. The axis of revolution

y=2

is above the region. The outer radius

R(x)

is the distance from the axis

y=2

to the farther curve, which is

y=\sin(x)

. Thus,

R(x) = 2 - \sin(x)

. The inner radius

r(x)

is the distance from the axis

y=2

to the closer curve, which is

y=\cos(x)

. Thus,

r(x) = 2 - \cos(x)

. The volume is

V = \pi \int_{0}^{\pi/4} [R(x)^2 - r(x)^2] \,dx = \pi \int_{0}^{\pi/4} [(2-\sin(x))^2 - (2-\cos(x))^2] \,dx

Question 17

The region R is enclosed by the parabola $y = 4 - x^2$ and the x-axis. Find the volume of the solid generated by revolving R about the line $y = 5$ .

$\pi \int_{-2}^{2} [5^2 - (1+x^2)^2] \,dx$ (correct answer)
$\pi \int_{-2}^{2} [(1+x^2)^2 - 5^2] \,dx$
$\pi \int_{-2}^{2} [(5 - (4-x^2))^2] \,dx$
$\pi \int_{-2}^{2} [(4-x^2)-5]^2 \,dx$

Explanation: The region is bounded by

y=4-x^2

and

y=0

, which intersect at

x = \pm 2

. The axis of revolution

y=5

is above the region. The outer radius

R(x)

is the distance from

y=5

to the farther boundary

y=0

, so

R(x) = 5 - 0 = 5

. The inner radius

r(x)

is the distance from

y=5

to the closer boundary

y=4-x^2

, so

r(x) = 5 - (4-x^2) = 1+x^2

. The volume is

\pi \int_{-2}^{2} [R(x)^2 - r(x)^2] \,dx = \pi \int_{-2}^{2} [5^2 - (1+x^2)^2] \,dx

Question 18

Let R be the region in the first quadrant bounded by $y = x^3$ , $y = 8$ , and the y-axis. Which integral represents the volume of the solid obtained by rotating R about the vertical line $x = 3$ ?

$\pi \int_{0}^{8} [3^2 - (3-\sqrt[3]{y})^2] \,dy$ (correct answer)
$\pi \int_{0}^{8} [(3-\sqrt[3]{y})^2 - 3^2] \,dy$
$\pi \int_{0}^{2} [8^2 - (x^3)^2] \,dx$
$\pi \int_{0}^{8} [(3+\sqrt[3]{y})^2 - 3^2] \,dy$

Explanation: The revolution is about a vertical line, so we integrate with respect to

y

. The region is bounded by

x = \sqrt[3]{y}

on the right and

x=0

on the left, from

y=0

y=8

. The axis of revolution

x=3

is to the right of the region. The outer radius

R(y)

is the distance from

x=3

to the farther boundary

x=0

, so

R(y) = 3 - 0 = 3

. The inner radius

r(y)

is the distance from

x=3

to the closer boundary

x=\sqrt[3]{y}

, so

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

. The volume is

\pi \int_{0}^{8} [R(y)^2 - r(y)^2] \,dy = \pi \int_{0}^{8} [3^2 - (3-\sqrt[3]{y})^2] \,dy

Question 19

The region R is bounded by $y=x$ , $y=2-x$ , and the x-axis. Which expression gives the volume of the solid formed by revolving R about the line $y=3$ ?

$\pi \int_{0}^{1} [3^2 - (3-x)^2] \,dx + \pi \int_{1}^{2} [3^2 - (3-(2-x))^2] \,dx$ (correct answer)
$\pi \int_{0}^{2} [3^2 - (3-x)^2] \,dx$
$\pi \int_{0}^{1} [(3-x)^2 - 3^2] \,dx + \pi \int_{1}^{2} [(3-(2-x))^2 - 3^2] \,dx$
$\pi \int_{0}^{2} (x - (2-x))^2 \,dx$

Explanation: The region is a triangle with vertices at (0,0), (2,0), and (1,1). The upper boundary is defined by

y=x

for

x \in [0,1]

and

y=2-x

for

x \in [1,2]

. The axis of revolution is

y=3

. The outer boundary of the solid is formed by revolving the line

y=0

(the x-axis) around

y=3

, so the outer radius is constant:

R(x) = 3-0=3

. The inner radius changes based on the upper boundary of the region. For

x \in [0,1]

r(x) = 3-x

. For

x \in [1,2]

r(x) = 3-(2-x) = 1+x

. The total volume requires two integrals:

V = \pi \int_{0}^{1} [3^2 - (3-x)^2] \,dx + \pi \int_{1}^{2} [3^2 - (1+x)^2] \,dx

. The second part of the correct answer simplifies

3-(2-x)

1+x

. The expression in choice A is equivalent.

Question 20

Let R be the region bounded by $y=x^2$ and $y=2x$ . Which integral represents the volume of the solid obtained by revolving R about the line $x=-1$ ?

$\pi \int_{0}^{4} [(y/2+1)^2 - (\sqrt{y}+1)^2] \,dy$
$\pi \int_{0}^{4} [(\sqrt{y}+1)^2 - (y/2+1)^2] \,dy$ (correct answer)
$\pi \int_{0}^{2} [(2x+1)^2 - (x^2+1)^2] \,dx$
$\pi \int_{0}^{4} [(\sqrt{y})^2 - (y/2)^2] \,dy$

Explanation: When finding volumes of revolution using the washer method, you need to identify the outer and inner radii from the axis of rotation to each bounding curve, then apply the formula

V = \pi \int [R_{outer}^2 - R_{inner}^2] \, dx

. First, find where the curves intersect by solving

x^2 = 2x

, giving

x = 0

and

x = 2

. Between these points,

y = 2x

lies above

y = x^2

. When revolving about the horizontal line

y = 5

, the distance from this axis to each curve determines the radii. For any

x

[0,2]

Distance to $y = x^2$ : $5 - x^2$ (this is the outer radius since $x^2 < 5$ )
Distance to $y = 2x$ : $5 - 2x$ (this is the inner radius since $2x < 5$ and $2x > x^2$ )

The volume integral becomes

\pi \int_0^2 [(5-x^2)^2 - (5-2x)^2] \, dx

, which is choice B. Choice A incorrectly swaps the outer and inner radii, treating

5-2x

as outer and

5-x^2

as inner. This happens when you don't carefully determine which curve is farther from the axis of rotation. Choice C uses integration with respect to

y

y = 5