Calculus 1 Quiz: Cross Sections Squares And Rectangles

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Cross Sections Squares And RectanglesQuestion 1 of 20

The region bounded by $y = x^2 + 1$ , $y=0$ , $x=0$ , and $x=2$ is the base of a solid. Cross-sections perpendicular to the x-axis are squares. Find the volume.

\frac{14}{3}

\frac{56}{3}

\frac{32}{5}

\frac{206}{15}

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Calculus 1 Quiz

Calculus 1 Quiz: Cross Sections Squares And Rectangles

Practice Cross Sections Squares And Rectangles 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 Cross Sections Squares And Rectangles, 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

The region bounded by $y = x^2 + 1$ , $y=0$ , $x=0$ , and $x=2$ is the base of a solid. Cross-sections perpendicular to the x-axis are squares. Find the volume.

$\frac{14}{3}$
$\frac{56}{3}$
$\frac{32}{5}$
$\frac{206}{15}$ (correct answer)

Explanation: This problem tests your understanding of volumes using cross-sections, a key application of integration in Calculus 1. When you see "cross-sections perpendicular to an axis," think about slicing the solid and finding the area of each slice. First, you need to understand the base region. It's bounded by

y = x^2 + 1

(top),

y = 0

(bottom),

x = 0

(left), and

x = 2

(right). At any x-value, the height of this region is

(x^2 + 1) - 0 = x^2 + 1

. Since cross-sections perpendicular to the x-axis are squares, and the side length of each square equals the height of the base region, each square has side length

x^2 + 1

. The area of each square cross-section is therefore

(x^2 + 1)^2

. To find the volume, integrate these cross-sectional areas from

x = 0

x = 2

V = \int_0^2 (x^2 + 1)^2 \, dx = \int_0^2 (x^4 + 2x^2 + 1) \, dx

= \left[\frac{x^5}{5} + \frac{2x^3}{3} + x\right]_0^2 = \frac{32}{5} + \frac{16}{3} + 2 = \frac{96 + 80 + 30}{15} = \frac{206}{15}

Answer choice A (

\frac{14}{3}

) likely comes from integrating just

x^2 + 1

instead of squaring it. Choice B (

\frac{56}{3}

) might result from incorrect expansion or integration errors. Choice C (

\frac{32}{5}

) represents only the

\frac{x^5}{5}

term, missing the other components. Key strategy: Always square the side length when dealing with square cross-sections, and carefully expand

(x^2 + 1)^2

before integrating.

Question 2

A solid's base is the triangle with vertices (0,0), (6,0), and (0,3). Cross-sections perpendicular to the x-axis are squares. Find the volume.

9
18 (correct answer)
27
54

Explanation: The vertices define a triangular region bounded by the x-axis (

y=0

), the y-axis (

x=0

), and the line connecting (6,0) and (0,3). The equation of this line is

y - 0 = \frac{3-0}{0-6}(x-6) \implies y = -\frac{1}{2}(x-6) = -\frac{1}{2}x + 3

. The integration is from

x=0

x=6

. The side length of a square cross-section is

s(x) = (-\frac{1}{2}x + 3) - 0 = 3 - \frac{1}{2}x

. The area is

A(x) = (3 - \frac{1}{2}x)^2 = 9 - 3x + \frac{1}{4}x^2

. The volume is

V = \int_0^6 (9 - 3x + \frac{1}{4}x^2) \,dx = [9x - \frac{3}{2}x^2 + \frac{1}{12}x^3]_0^6 = 9(6) - \frac{3}{2}(36) + \frac{1}{12}(216) = 54 - 54 + 18 = 18

Question 3

A solid has a base bounded by $y=\sin(x)$ and $y=\cos(x)$ between $x=0$ and $x=\pi/4$ . If cross-sections perpendicular to the x-axis are squares, what is the volume?

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

Explanation: When you encounter a volume problem with cross-sections, you need to set up an integral where each slice has a known area. Here, the cross-sections are squares perpendicular to the x-axis, so you need to find the side length of each square at any x-value. First, determine which function is on top in the given interval. At

x = 0

\sin(0) = 0

and

\cos(0) = 1

, so

\cos(x) > \sin(x)

initially. The functions intersect when

\sin(x) = \cos(x)

, which occurs at

x = \pi/4

(where both equal

\sqrt{2}/2

). Throughout

[0, \pi/4]

, cosine remains above sine. The side length of each square cross-section is the distance between the curves:

\cos(x) - \sin(x)

. Since the area of a square is side length squared, each cross-section has area

[\cos(x) - \sin(x)]^2

. The volume is:

V = \int_0^{\pi/4} [\cos(x) - \sin(x)]^2 \, dx

Expanding:

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

Therefore:

V = \int_0^{\pi/4} [1 - \sin(2x)] \, dx = [x + \frac{1}{2}\cos(2x)]_0^{\pi/4} = [\frac{\pi}{4} + 0] - [0 + \frac{1}{2}] = \frac{\pi}{4} - \frac{1}{2}

Answer D is correct. Options A and B likely come from integration errors or incorrect setup. Option C represents

\cos(\pi/4) - \sin(0) = \sqrt{2}/2 - 0

, suggesting confusion about what to integrate. Strategy tip: Always expand squared expressions completely and use trigonometric identities to simplify before integrating.

Question 4

Let a solid have a base defined by the region between $y = \sqrt{x}$ and the x-axis from $x=0$ to $x=4$ . Cross-sections perpendicular to the x-axis are rectangles whose height is half of their base. What is the volume of the solid?

2
4 (correct answer)
8
16

Explanation: The limits of integration are given as

x=0

x=4

. The base of a rectangular cross-section at a given

x

s(x) = \sqrt{x} - 0 = \sqrt{x}

. The height is half the base, so

h(x) = \frac{1}{2}s(x) = \frac{1}{2}\sqrt{x}

. The area of the cross-section is

A(x) = s(x) \cdot h(x) = \sqrt{x} \cdot \frac{1}{2}\sqrt{x} = \frac{1}{2}x

. The volume is the integral of the area:

V = \int_0^4 \frac{1}{2}x \,dx = \frac{1}{2}[\frac{x^2}{2}]_0^4 = \frac{1}{4}[x^2]_0^4 = \frac{1}{4}(4^2 - 0^2) = \frac{16}{4} = 4

Question 5

The base of a solid is the region bounded by $y = x^3$ , the line $x = 2$ , and the x-axis. The cross-sections perpendicular to the y-axis are rectangles with a height equal to one-fourth of their base. What is the volume of the solid?

$24/5$
$16/5$
$32/7$
$4/5$ (correct answer)

Explanation: This problem tests your ability to set up volume integrals when cross-sections are perpendicular to the y-axis rather than the x-axis. When you see "perpendicular to the y-axis," you need to think of thin horizontal slices and integrate with respect to y. First, identify the region. The solid's base is bounded by

y = x^3

x = 2

, and the x-axis, so it extends from

x = 0

x = 2

and from

y = 0

y = 8

(since when

x = 2

y = 2^3 = 8

). For integration with respect to y, express x in terms of y: from

y = x^3

, we get

x = y^{1/3}

. At height y, the rectangular cross-section has a base extending from

x = 0

x = y^{1/3}

, so the base length is

y^{1/3}

. The height is one-fourth of the base, which is

\frac{1}{4}y^{1/3}

. The area of each rectangular cross-section is:

A(y) = y^{1/3} \cdot \frac{1}{4}y^{1/3} = \frac{1}{4}y^{2/3}

The volume is:

V = \int_0^8 \frac{1}{4}y^{2/3} dy = \frac{1}{4} \cdot \frac{3}{5}y^{5/3}\Big|_0^8 = \frac{3}{20}(8^{5/3}) = \frac{3}{20}(32) = \frac{4}{5}

Answer D is correct. Answer A (24/5) likely comes from incorrect integration bounds or forgetting the 1/4 factor. Answer B (16/5) suggests integrating

y^{2/3}

without the 1/4 coefficient. Answer C (32/7) indicates using the wrong power in integration. Strategy tip: When cross-sections are perpendicular to the y-axis, always solve for x in terms of y and integrate with respect to y.

Question 6

The base of a solid is a triangle in the xy-plane with vertices at $(0,0)$ , $(4,0)$ , and $(0,2)$ . The cross-sections of the solid perpendicular to the x-axis are squares. What is the volume of the solid?

$8$
$32/3$
$4$
$16/3$ (correct answer)

Explanation: This problem involves finding the volume of a solid using cross-sections, a key technique in Calculus 1. When you see "cross-sections perpendicular to an axis," you're integrating the area of those cross-sections along that axis. First, you need to understand the triangular base. The vertices

(0,0)

(4,0)

, and

(0,2)

form a right triangle. The hypotenuse connects

(4,0)

(0,2)

, with equation

y = 2 - \frac{x}{2}

. Since cross-sections perpendicular to the x-axis are squares, at any x-value between 0 and 4, the square's side length equals the height of the triangle at that x-coordinate. From the line equation, this height is

2 - \frac{x}{2}

. The area of each square cross-section is therefore

A(x) = \left(2 - \frac{x}{2}\right)^2 = 4 - 2x + \frac{x^2}{4}

. To find the volume, integrate from

x = 0

x = 4

V = \int_0^4 \left(4 - 2x + \frac{x^2}{4}\right) dx = \left[4x - x^2 + \frac{x^3}{12}\right]_0^4

V = 16 - 16 + \frac{64}{12} = \frac{16}{3}

Answer choice A (

8

) likely comes from using just the side length times the base. Choice B (

32/3

) might result from incorrectly squaring the integral bounds. Choice C (

4

) could come from confusing this with a simpler geometric calculation. Remember: for cross-section problems, always identify the shape's dimensions at each point, square or cube as needed for area/volume, then integrate along the specified axis.

Question 7

The base of a solid is the region enclosed by the graphs of $y = \sec(x)$ , $y = \tan(x)$ , $x = 0$ , and $x = \pi/4$ . Cross-sections perpendicular to the x-axis are squares. Which integral represents the volume of the solid?

$\int_{0}^{\pi/4} \sec^2(x) \,dx$
$\int_{0}^{\pi/4} (\sec^2(x) - \tan^2(x)) \,dx$
$\pi \int_{0}^{\pi/4} (\sec(x) - \tan(x))^2 \,dx$
$\int_{0}^{\pi/4} (\sec(x) - \tan(x))^2 \,dx$ (correct answer)

Explanation: When finding volumes of solids with known cross-sections, you need to identify the area of each cross-section as a function of x, then integrate along the axis of cross-sections. First, visualize the region. Between

x = 0

and

x = \pi/4

\sec(x)

lies above

\tan(x)

(since

\sec^2(x) - \tan^2(x) = 1 > 0

, so

\sec(x) > \tan(x)

in this interval where both are positive). The region is bounded by these curves and the vertical lines

x = 0

and

x = \pi/4

. For any value of x in this interval, the cross-section is a square whose side length equals the vertical distance between the curves:

\sec(x) - \tan(x)

. Since the area of a square equals (side length)², the cross-sectional area is

(\sec(x) - \tan(x))^2

. The volume is therefore

\int_{0}^{\pi/4} (\sec(x) - \tan(x))^2 \,dx

, which is answer D. Answer A gives

\int_{0}^{\pi/4} \sec^2(x) \,dx

, which would be correct if the side length were

\sec(x)

instead of the difference between the functions. Answer B gives

\int_{0}^{\pi/4} (\sec^2(x) - \tan^2(x)) \,dx = \int_{0}^{\pi/4} 1 \,dx

, which treats the integrand as if it were already squared. Answer C incorrectly includes the factor

\pi

, suggesting confusion with circular cross-sections rather than squares. Remember: for volume problems with known cross-sections, the key steps are (1) find the dimension(s) of each cross-section as functions of the integration variable, (2) calculate the area using the appropriate geometric formula, then (3) integrate.

Question 8

The volume of a solid is given by the integral $\int_{0}^{2} (3-x)^2 \,dx$ . Which of the following could describe the solid?

A solid generated by rotating the region bounded by $y=3-x$ , $x=0$ , $x=2$ , and the x-axis about the x-axis.
A solid with a base bounded by $y=3-x$ , $x=0$ , $x=2$ , and the x-axis, and square cross-sections perpendicular to the y-axis.
A solid with a base bounded by $y=3-x$ , $x=0$ , $x=2$ , and the x-axis, and square cross-sections perpendicular to the x-axis. (correct answer)
A solid with a base bounded by $y=\sqrt{3-x}$ , $x=0$ , $x=2$ , and the x-axis, and square cross-sections perpendicular to the x-axis.

Explanation: When you encounter a volume integral, you need to identify which method of volume calculation produced that specific integrand. The key is recognizing the relationship between the integrand and the geometric setup. The given integral

\int_{0}^{2} (3-x)^2 \,dx

has the form of a cross-sectional area method, where you integrate area functions over an interval. Since the integrand is

(3-x)^2

, this represents the area of square cross-sections with side length

(3-x)

. The limits of integration tell you that

x

ranges from 0 to 2, meaning the cross-sections are taken perpendicular to the x-axis. Answer C correctly describes this setup: a solid with square cross-sections perpendicular to the x-axis, where the side length of each square equals the height of the function

y = 3-x

at that x-value. Answer A is incorrect because rotation about the x-axis would produce the integral

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

, which includes the factor

\pi

from the disk method formula. Answer B is wrong because cross-sections perpendicular to the y-axis would require integrating with respect to

y

, not

x

, and you'd need to express everything in terms of

y

. Answer D fails because if the boundary were

y = \sqrt{3-x}

, then the side length of the square cross-sections would be

\sqrt{3-x}

, making the area

(\sqrt{3-x})^2 = 3-x

, not

(3-x)^2

. Study tip: Always match the integrand to the geometry—look for missing

\pi

factors (indicating rotation) and check whether the variable of integration matches the direction of the cross-sections.

Question 9

The base of a solid is a semicircle of radius 3 with its diameter along the x-axis. The cross-sections perpendicular to the x-axis are squares with one side on the base. Find the volume of the solid.

$72$
$36$ (correct answer)
$18\pi$
$144$

Explanation: This is a volumes of solids with known cross-sections problem, where you need to set up an integral based on the geometry described. When you see "cross-sections perpendicular to an axis," think about slicing the solid and finding the area of each slice. Start by visualizing the setup: a semicircle of radius 3 centered at the origin with diameter from (-3, 3) on the x-axis. The equation of this semicircle is

y = \sqrt{9 - x^2}

for

-3 \leq x \leq 3

. At any x-value, the cross-section is a square with one side lying on the diameter (along the x-axis) and extending upward to the semicircle. The side length of each square cross-section equals the height of the semicircle at that x-value, which is

\sqrt{9 - x^2}

. Since the area of a square is side length squared, each cross-sectional area is

(\sqrt{9 - x^2})^2 = 9 - x^2

. The volume is:

V = \int_{-3}^{3} (9 - x^2) dx = \left[9x - \frac{x^3}{3}\right]_{-3}^{3} = (27 - 9) - (-27 + 9) = 18 - (-18) = 36

Choice A (72) likely comes from doubling the correct answer or making an error in the limits. Choice C (18π) incorrectly uses the area formula for the semicircle base rather than integrating cross-sections. Choice D (144) represents a significant computational error, possibly squaring an intermediate result incorrectly. Remember: for cross-section problems, always identify what determines the dimensions of each slice, then integrate those areas over the appropriate interval.

Question 10

Let $R$ be a region in the xy-plane that forms the base of two different solids. Solid $S_1$ has cross-sections perpendicular to the x-axis that are squares. Solid $S_2$ has cross-sections perpendicular to the x-axis that are rectangles with height three times their base. If the volume of $S_1$ is 15, what is the volume of $S_2$ ?

$135$
$45$ (correct answer)
$5$
$15\sqrt{3}$

Explanation: When you encounter problems involving solids with known cross-sections, you're working with the method of cross-sectional areas. The key insight is that both solids share the same base region

R

, so at any x-value, their cross-sections have the same width. Let's say the width of the region at position

x

w(x)

. For solid

S_1

, each cross-section is a square with side length

w(x)

, giving area

[w(x)]^2

. For solid

S_2

, each cross-section is a rectangle where the base is

w(x)

and the height is

3w(x)

(three times the base), giving area

w(x) \cdot 3w(x) = 3[w(x)]^2

. Notice that at every x-value, the area of

S_2

's cross-section is exactly 3 times the area of

S_1

's cross-section. When you integrate to find volumes:

V_{S_1} = \int [w(x)]^2 dx = 15

and

V_{S_2} = \int 3[w(x)]^2 dx = 3\int [w(x)]^2 dx = 3 \cdot 15 = 45

The answer is B) 45. Choice A) 135 likely comes from incorrectly multiplying by 9 (perhaps squaring the factor of 3). Choice C) 5 suggests dividing by 3 instead of multiplying. Choice D)

15\sqrt{3}

might result from confusion about how the geometric relationship translates to volume. Strategy tip: In cross-section problems, always identify the relationship between the areas (not just the linear dimensions) of corresponding cross-sections. The volume ratio equals the area ratio since both solids extend over the same interval.

Question 11

The base of a solid is the region in the first quadrant bounded by $y = x^3$ , $y = 8$ , and the y-axis ( $x=0$ ). Cross-sections perpendicular to the y-axis are rectangles whose height is half the length of their base. What is the volume of the solid?

$96/5$
$48/5$ (correct answer)
$288/7$
$6$

Explanation: When you encounter a volume problem with cross-sections, you're working with the method of integrating cross-sectional areas. The key is to set up your integral in terms of the variable perpendicular to the cross-sections—here, that's the y-axis. First, identify the region. The base is bounded by

y = x^3

y = 8

, and

x = 0

in the first quadrant. From

y = x^3

, we get

x = y^{1/3}

, so the region extends from

y = 0

y = 8

, with width

x = y^{1/3}

at each height

y

. Each cross-section is a rectangle with base length equal to the width of the region at that y-value, which is

y^{1/3}

. The height is half the base length, so height =

\frac{1}{2}y^{1/3}

. Therefore, the cross-sectional area is:

A(y) = \text{base} \times \text{height} = y^{1/3} \cdot \frac{1}{2}y^{1/3} = \frac{1}{2}y^{2/3}

The volume is:

V = \int_0^8 \frac{1}{2}y^{2/3} \, dy = \frac{1}{2} \cdot \frac{3}{5}y^{5/3}\Big|_0^8 = \frac{3}{10}(8^{5/3}) = \frac{3}{10}(32) = \frac{96}{10} = \frac{48}{5}

This confirms answer (B). (A) 96/5 likely comes from forgetting the factor of

\frac{1}{2}

in the height condition. (C) 288/7 suggests using the wrong exponent in integration. (D) 6 probably results from computational errors in evaluating

8^{5/3}

. Always carefully read the relationship between base and height in cross-section problems, and double-check your integration limits and exponent arithmetic.

Question 12

The base of a solid is the region in the first quadrant enclosed by the graphs of $y = \sqrt{x}$ and $y = x^2$ . The cross-sections of the solid perpendicular to the x-axis are squares. What is the volume of the solid?

$9/70$ (correct answer)
$3/10$
$1/3$
$7/10$

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

\sqrt{x} = x^2 \implies x = x^4 \implies x^4 - x = 0 \implies x(x^3 - 1) = 0

. The intersection points are at

x=0

and

x=1

. In the interval

(0, 1)

\sqrt{x} > x^2

. The side length

s(x)

of a square cross-section at

x

is the vertical distance between the curves,

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

. The area of the square cross-section is

A(x) = s(x)^2 = (\sqrt{x} - x^2)^2 = x - 2x^{5/2} + x^4

. The volume is the integral of the area from

x=0

x=1

V = \int_{0}^{1} (x - 2x^{5/2} + x^4) \,dx = \left[ \frac{1}{2}x^2 - 2\left(\frac{2}{7}x^{7/2}\right) + \frac{1}{5}x^5 \right]_{0}^{1} = \frac{1}{2} - \frac{4}{7} + \frac{1}{5} = \frac{35 - 40 + 14}{70} = \frac{9}{70}

Question 13

The base of a solid is the region in the first quadrant bounded by the graph of $y = \ln(x)$ , the line $x = e$ , and the x-axis. For this solid, each cross-section perpendicular to the x-axis is a rectangle of constant height 3. What is the volume of the solid?

$3e - 6$
$e-2$
$3$ (correct answer)
$1$

Explanation: This problem tests your ability to set up and evaluate volume integrals using the cross-sectional method. When you see "cross-sections perpendicular to an axis," think about slicing the solid and finding the area of each slice, then integrating. First, you need to identify the region. The base is bounded by

y = \ln(x)

x = e

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

\ln(x)

intersects the x-axis when

x = 1

(because

\ln(1) = 0

), your region extends from

x = 1

x = e

. For any vertical slice at position

x

, the cross-section is a rectangle. The width of this rectangle is the distance from the x-axis up to the curve

y = \ln(x)

, which is simply

\ln(x)

. The height is given as the constant 3. Therefore, the area of each cross-section is

A(x) = 3 \ln(x)

. The volume is:

V = \int_1^e 3\ln(x) \, dx

Using integration by parts with

u = \ln(x)

and

dv = dx

\int \ln(x) \, dx = x\ln(x) - x + C

So:

V = 3[x\ln(x) - x]_1^e = 3[(e \cdot 1 - e) - (1 \cdot 0 - 1)] = 3[0 - (-1)] = 3

Answer choice A)

3e - 6

likely comes from incorrectly applying integration by parts. Answer choice B)

e - 2

results from forgetting the factor of 3 for the rectangle height. Answer choice D)

1

occurs if you forget the coefficient 3 entirely. Remember: always identify your integration bounds carefully and don't forget constant factors like the given height.

Question 14

Let $R$ be the region enclosed by the graphs of $y = x^2$ and $y = 4$ . A solid has $R$ as its base. For this solid, each cross-section perpendicular to the x-axis is a rectangle whose height is twice the length of its base. Find the volume of the solid.

$1024/15$ (correct answer)
$512/15$
$64/3$
$64$

Explanation: The region

R

is bounded by

y=4

(top) and

y=x^2

(bottom). The intersection points are at

x^2 = 4

, so

x = \pm 2

. The cross-sections are perpendicular to the x-axis. At a given

x

, the base of the rectangle is the vertical distance

b(x) = 4 - x^2

. The height is

h(x) = 2 \cdot b(x) = 2(4 - x^2)

. The area of the cross-section is

A(x) = b(x)h(x) = (4 - x^2) \cdot 2(4 - x^2) = 2(4 - x^2)^2 = 2(16 - 8x^2 + x^4)

. The volume is

V = \int_{-2}^{2} 2(16 - 8x^2 + x^4) \,dx = 2 \cdot 2 \int_{0}^{2} (16 - 8x^2 + x^4) \,dx = 4 \left[ 16x - \frac{8}{3}x^3 + \frac{1}{5}x^5 \right]_{0}^{2} = 4 \left( 32 - \frac{64}{3} + \frac{32}{5} \right) = 4 \left( \frac{256}{15} \right) = \frac{1024}{15}

Question 15

Let $R$ be the region bounded by the graph of $y = e^x$ , the line $y = 1$ , and the line $x = 2$ . A solid has base $R$ and square cross-sections perpendicular to the x-axis. Find the volume of the solid.

$\frac{1}{2}e^4 - \frac{5}{2}$
$\frac{1}{2}e^4 - \frac{1}{2}$
$\frac{1}{2}e^4 - 2e^2 + \frac{7}{2}$ (correct answer)
$e^2 - 3$

Explanation: When you encounter a problem about solids with cross-sections, you're dealing with volume by integration. The key insight is that each cross-section contributes a small "slice" to the total volume, and you integrate these slices along the axis perpendicular to the cross-sections. First, you need to understand the region R. The curve

y = e^x

intersects the line

y = 1

when

e^x = 1

, so

x = 0

. Therefore, region R is bounded by

y = e^x

(top),

y = 1

(bottom),

x = 0

(left), and

x = 2

(right). Since the cross-sections are squares perpendicular to the x-axis, at each x-value, the side length of the square equals the height of region R at that point:

e^x - 1

. The area of each square cross-section is therefore

(e^x - 1)^2

. To find the volume, integrate these areas from

x = 0

x = 2

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

Evaluating:

V = \left[\frac{1}{2}e^{2x} - 2e^x + x\right]_0^2 = \left(\frac{1}{2}e^4 - 2e^2 + 2\right) - \left(\frac{1}{2} - 2 + 0\right) = \frac{1}{2}e^4 - 2e^2 + \frac{7}{2}

Answer A forgot to expand

(e^x - 1)^2

correctly. Answer B made an error in the definite integral evaluation. Answer D likely confused this with a different integration setup entirely. Strategy tip: Always sketch the region first and clearly identify what determines the cross-sectional dimensions before setting up your integral.

Question 16

The base of a solid is the circle $x^2 + y^2 = 9$ . The cross-sections perpendicular to the x-axis are squares. What is the volume of the solid?

36
72
144 (correct answer)
216

Explanation: The limits of integration along the x-axis are from

x=-3

x=3

. For a given

x

, the vertical chord of the circle has endpoints on

y = \sqrt{9-x^2}

and

y = -\sqrt{9-x^2}

. The length of this chord, which is the side length of the square cross-section, is

s(x) = \sqrt{9-x^2} - (-\sqrt{9-x^2}) = 2\sqrt{9-x^2}

. The area of the square is

A(x) = [s(x)]^2 = (2\sqrt{9-x^2})^2 = 4(9-x^2) = 36 - 4x^2

. The volume is

V = \int_{-3}^3 (36 - 4x^2) \,dx

. Since the integrand is even,

V = 2 \int_0^3 (36 - 4x^2) \,dx = 2[36x - \frac{4}{3}x^3]_0^3 = 2(36(3) - \frac{4}{3}(27)) = 2(108 - 36) = 2(72) = 144

Question 17

The base of a solid is the region bounded by $y=x$ , $y=-x$ , and $x=2$ . Cross-sections perpendicular to the x-axis are squares. The volume is:

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

Explanation: The region is a triangle with vertices (0,0), (2,2), and (2,-2). The integration is with respect to

x

from 0 to 2. The top boundary is

y=x

and the bottom boundary is

y=-x

. The side length of a square cross-section is

s(x) = x - (-x) = 2x

. The area is

A(x) = (2x)^2 = 4x^2

. The volume is

V = \int_0^2 4x^2 \,dx = 4[\frac{x^3}{3}]_0^2 = 4(\frac{8}{3}) = \frac{32}{3}

Question 18

Let $R$ be the region enclosed by the graph of $y = \frac{1}{x}$ , the x-axis, and the lines $x=1$ and $x=5$ . A solid is formed with base $R$ and cross-sections perpendicular to the x-axis that are rectangles of constant height 3. What is the volume of the solid?

$3 \ln(5)$ (correct answer)
$\frac{12}{5}$
$\frac{4}{5}$
$15$

Explanation: The integration is with respect to

x

from

1

5

. For a given

x

, the base of the rectangular cross-section is the height of the region

R

, which is

s(x) = \frac{1}{x} - 0 = \frac{1}{x}

. The height of the rectangle is given as a constant,

h=3

. The area of a cross-section is

A(x) = s(x) \cdot h = \frac{1}{x} \cdot 3 = \frac{3}{x}

. The volume is

V = \int_1^5 \frac{3}{x} \,dx = 3[\ln|x|]_1^5 = 3(\ln(5) - \ln(1)) = 3\ln(5)

Question 19

The base of a solid is the region in the first quadrant bounded by the graph of $y = 4 - x^2$ and the coordinate axes. Cross-sections perpendicular to the x-axis are squares. What is the volume of the solid?

$\frac{32}{3}$
$\frac{128}{5}$
$\frac{256}{15}$ (correct answer)
$\frac{512}{15}$

Explanation: The region is bounded by

y=4-x^2

y=0

, and

x=0

. The x-intercept of the parabola is at

x=2

, so the integration is from

x=0

x=2

. The side length of a square cross-section at a given

x

s(x) = (4-x^2) - 0 = 4-x^2

. The area of the cross-section is

A(x) = [s(x)]^2 = (4-x^2)^2 = 16 - 8x^2 + x^4

. The volume is the integral of the area:

V = \int_0^2 (16 - 8x^2 + x^4) \,dx = [16x - \frac{8}{3}x^3 + \frac{1}{5}x^5]_0^2 = (16(2) - \frac{8}{3}(8) + \frac{1}{5}(32)) - 0 = 32 - \frac{64}{3} + \frac{32}{5} = \frac{480 - 320 + 96}{15} = \frac{256}{15}

Question 20

A solid has a base in the xy-plane bounded by the graphs of $y=x$ and $y=x^2$ . If the cross-sections perpendicular to the x-axis are rectangles with height three times the base, what is the volume of the solid?

$\frac{1}{30}$
$\frac{1}{10}$ (correct answer)
$\frac{1}{6}$
$\frac{3}{10}$

Explanation: First, find the intersection points by setting

x=x^2

, which gives

x=0

and

x=1

. These are the limits of integration. In the interval

[0,1]

y=x

is the upper curve. The base of a rectangular cross-section is

s(x) = x - x^2

. The height is given as three times the base, so

h(x) = 3(x-x^2)

. The area of a cross-section is

A(x) = \text{base} \times \text{height} = (x-x^2) \cdot 3(x-x^2) = 3(x-x^2)^2 = 3(x^2 - 2x^3 + x^4)

. The volume is

V = \int_0^1 3(x^2 - 2x^3 + x^4) \,dx = 3[\frac{x^3}{3} - \frac{2x^4}{4} + \frac{x^5}{5}]_0^1 = 3(\frac{1}{3} - \frac{1}{2} + \frac{1}{5}) = 3(\frac{10-15+6}{30}) = 3(\frac{1}{30}) = \frac{1}{10}