Calculus 2 Quiz: Cross Sections Triangles And Semicircles

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Cross Sections Triangles And SemicirclesQuestion 1 of 20

A solid has its base in the xy-plane, bounded by the curve $y = \sqrt{x}$ , the x-axis, and the line $x = 4$ . If the cross-sections of the solid perpendicular to the x-axis are equilateral triangles, what is the volume of the solid?

2\sqrt{3}

\sqrt{3}

4

8

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Calculus 2 Quiz

Calculus 2 Quiz: Cross Sections Triangles And Semicircles

Practice Cross Sections Triangles And Semicircles in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Cross Sections Triangles And Semicircles, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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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

$2\sqrt{3}$ (correct answer)
$\sqrt{3}$
$4$
$8$

Explanation: The side length

s

of an equilateral triangle at a given

x

is the height of the region, so

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

. The area of an equilateral triangle with side

s

A = \frac{\sqrt{3}}{4}s^2

. Thus, the area of a cross-section is

A(x) = \frac{\sqrt{3}}{4}(\sqrt{x})^2 = \frac{\sqrt{3}}{4}x

. The volume is found by integrating this area from

x=0

x=4

V = \int_{0}^{4} \frac{\sqrt{3}}{4}x \,dx = \frac{\sqrt{3}}{4} \left[ \frac{x^2}{2} \right]_{0}^{4} = \frac{\sqrt{3}}{4} \left( \frac{16}{2} - 0 \right) = \frac{\sqrt{3}}{4}(8) = 2\sqrt{3}

Question 2

The base of a solid is the region between the curves $y = x^2$ and $y = \sqrt{x}$ . Cross-sections perpendicular to the x-axis are equilateral triangles. Which expression gives the volume of the solid?

$\frac{\sqrt{3}}{4} \int_{0}^{1} (x - x^4) dx$
$\frac{1}{2} \int_{0}^{1} (\sqrt{x} - x^2)^2 dx$
$\frac{\sqrt{3}}{4} \int_{0}^{1} (\sqrt{x} - x^2)^2 dx$ (correct answer)
$\frac{\sqrt{3}}{4} \int_{0}^{1} (x^2 - \sqrt{x})^2 dx$

Explanation: First, find the points of intersection:

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

, so

x=0

and

x=1

. In the interval

[0, 1]

\sqrt{x} \ge x^2

. The side length of the equilateral triangle is

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

. The area of an equilateral triangle is

A = \frac{\sqrt{3}}{4}s^2

. Therefore, the cross-sectional area is

A(x) = \frac{\sqrt{3}}{4}(\sqrt{x} - x^2)^2

. The volume is the integral of this area from 0 to 1:

V = \int_{0}^{1} \frac{\sqrt{3}}{4}(\sqrt{x} - x^2)^2 dx

Question 3

A solid's base is enclosed by $x = y^{1/2}$ , the y-axis, and the line $y = 4$ . If cross-sections taken perpendicular to the y-axis are isosceles right triangles with their hypotenuse on the base, what is the volume?

2 (correct answer)
4
8
16

Explanation: Cross-sections are perpendicular to the y-axis, so we integrate with respect to

y

. The horizontal length of the base at height

y

s(y) = y^{1/2} - 0 = \sqrt{y}

. The cross-sections are isosceles right triangles with the hypotenuse

s(y)

on the base. The area of such a triangle is

A(s) = \frac{1}{4}s^2

. So,

A(y) = \frac{1}{4}(\sqrt{y})^2 = \frac{1}{4}y

. The volume is found by integrating from

y=0

y=4

V = \int_{0}^{4} \frac{1}{4}y \,dy = \frac{1}{4} \left[ \frac{y^2}{2} \right]_{0}^{4} = \frac{1}{4} \left( \frac{16}{2} - 0 \right) = \frac{1}{4}(8) = 2

Question 4

Let $S_1$ be a solid with a base region R and cross-sections perpendicular to the x-axis that are semicircles. Let $S_2$ be a solid with the same base region R and cross-sections perpendicular to the x-axis that are equilateral triangles. What is the ratio of the volume of $S_1$ to the volume of $S_2$ ?

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

Explanation: Let the side length of a cross-section at

x

s(x)

. For the semicircular cross-sections of

S_1

, the area is

A_1(x) = \frac{\pi}{8}(s(x))^2

. For the equilateral triangle cross-sections of

S_2

, the area is

A_2(x) = \frac{\sqrt{3}}{4}(s(x))^2

. The volume of each solid is the integral of its cross-sectional area. The ratio of the volumes is the ratio of the area formulas:

\frac{V(S_1)}{V(S_2)} = \frac{\int A_1(x) dx}{\int A_2(x) dx} = \frac{\int \frac{\pi}{8}(s(x))^2 dx}{\int \frac{\sqrt{3}}{4}(s(x))^2 dx} = \frac{\frac{\pi}{8}}{\frac{\sqrt{3}}{4}} = \frac{\pi}{8} \cdot \frac{4}{\sqrt{3}} = \frac{\pi}{2\sqrt{3}}

Question 5

The base of a solid is the region enclosed by $y = \sin(x)$ and the x-axis for $0 \le x \le \pi$ . The cross-sections perpendicular to the x-axis are isosceles right triangles with one leg lying on the base. Find the volume of the solid.

$\frac{\pi}{8}$
$\frac{\pi}{4}$ (correct answer)
$\frac{\pi}{2}$
$\pi$

Explanation: The length of the leg on the base at a given

x

s(x) = \sin(x) - 0 = \sin(x)

. For an isosceles right triangle with leg

s

, the area is

A = \frac{1}{2}s^2

. So, the area of a cross-section is

A(x) = \frac{1}{2}(\sin(x))^2 = \frac{1}{2}\sin^2(x)

. The volume is

V = \int_{0}^{\pi} \frac{1}{2}\sin^2(x) dx

. Using the identity

\sin^2(x) = \frac{1 - \cos(2x)}{2}

, we get

V = \frac{1}{2} \int_{0}^{\pi} \frac{1 - \cos(2x)}{2} dx = \frac{1}{4} \left[ x - \frac{1}{2}\sin(2x) \right]_{0}^{\pi} = \frac{1}{4} \left( (\pi - 0) - (0 - 0) \right) = \frac{\pi}{4}

Question 6

A solid's base is the region bounded by $y = \sqrt{x}$ and $y=x/2$ . Solid A has cross-sections perpendicular to the x-axis that are semicircles. Solid B has cross-sections perpendicular to the y-axis that are equilateral triangles. Which integral represents the volume of Solid B?

$\frac{\pi}{8} \int_{0}^{4} (\sqrt{x} - x/2)^2 dx$
$\frac{\sqrt{3}}{4} \int_{0}^{4} (\sqrt{x} - x/2)^2 dx$
$\frac{\sqrt{3}}{4} \int_{0}^{2} (2y - y^2)^2 dy$ (correct answer)
$\frac{\sqrt{3}}{4} \int_{0}^{2} (y^2 - 2y)^2 dy$

Explanation: For Solid B, cross-sections are perpendicular to the y-axis, so we must express the boundaries as functions of

y

y=\sqrt{x} \implies x=y^2

y=x/2 \implies x=2y

. Find intersections:

y^2=2y \implies y(y-2)=0

, so

y=0, 2

. For

y \in [0, 2]

2y \ge y^2

, so the right boundary is

x=2y

and the left boundary is

x=y^2

. The side length of the equilateral triangle is

s(y) = 2y - y^2

. The area is

A(y) = \frac{\sqrt{3}}{4}s(y)^2 = \frac{\sqrt{3}}{4}(2y - y^2)^2

. The volume is the integral from

y=0

y=2

\int_{0}^{2} \frac{\sqrt{3}}{4}(2y - y^2)^2 dy

Question 7

The base of a solid is the region bounded by $y = \ln(x)$ , the x-axis, and the line $x = e$ . Cross-sections perpendicular to the x-axis are isosceles right triangles with one leg on the base. Which of the following integrals represents the volume of the solid?

$\frac{1}{4} \int_{1}^{e} (\ln x)^2 dx$
$\frac{1}{2} \int_{1}^{e} (\ln x)^2 dx$ (correct answer)
$\frac{1}{2} \int_{0}^{1} (e - e^y)^2 dy$
$\frac{1}{2} \int_{1}^{e} (\ln x) dx$

Explanation: The region is bounded by

y=\ln(x)

and

y=0

, from

x=1

(where

\ln(x)=0

) to

x=e

. The length of the leg on the base is

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

. The area of an isosceles right triangle with leg

s

A = \frac{1}{2}s^2

. Thus, the cross-sectional area is

A(x) = \frac{1}{2}(\ln x)^2

. The volume is the integral of this area over the interval

[1, e]

, which is

V = \int_{1}^{e} \frac{1}{2}(\ln x)^2 dx

Question 8

A solid has a base in the first quadrant bounded by $y = x^3$ and $y = x$ . If cross-sections perpendicular to the y-axis are equilateral triangles, which integral represents the volume?

$\frac{\sqrt{3}}{4} \int_{0}^{1} (x - x^3)^2 dx$
$\frac{\sqrt{3}}{4} \int_{0}^{1} (\sqrt[3]{y} - y)^2 dy$ (correct answer)
$\frac{1}{2} \int_{0}^{1} (\sqrt[3]{y} - y)^2 dy$
$\frac{\sqrt{3}}{4} \int_{0}^{1} (y - \sqrt[3]{y})^2 dy$

Explanation: Since cross-sections are perpendicular to the y-axis, we must express the boundaries as functions of

y

. The curves are

x = y

and

x = \sqrt[3]{y}

. The intersection points are (0,0) and (1,1). For

y \in [0, 1]

\sqrt[3]{y} \ge y

, so the right boundary is

x = \sqrt[3]{y}

and the left is

x = y

. The side length of the equilateral triangle is

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

. The area is

A(y) = \frac{\sqrt{3}}{4}s(y)^2 = \frac{\sqrt{3}}{4}(\sqrt[3]{y} - y)^2

. The volume is the integral of this area from

y=0

y=1

Question 9

The base of a solid is the region bounded by $y = \cos(x)$ and $y = -\cos(x)$ for $x \in [-\pi/2, \pi/2]$ . If cross-sections perpendicular to the x-axis are semicircles, what is the volume of the solid?

$\frac{\pi^2}{8}$
$\frac{\pi^2}{4}$ (correct answer)
$\frac{\pi^2}{2}$
$\pi^2$

Explanation: The diameter of the semicircle at a given

x

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

. The area of a semicircle with diameter

s

A = \frac{\pi}{8}s^2

. Thus,

A(x) = \frac{\pi}{8}(2\cos(x))^2 = \frac{\pi}{8}(4\cos^2(x)) = \frac{\pi}{2}\cos^2(x)

. The volume is

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

. Using the identity

\cos^2(x) = \frac{1+\cos(2x)}{2}

, we have

V = \frac{\pi}{2} \int_{-\pi/2}^{\pi/2} \frac{1+\cos(2x)}{2} dx = \frac{\pi}{4} \left[ x + \frac{1}{2}\sin(2x) \right]_{-\pi/2}^{\pi/2}

. Evaluating at the limits gives

\frac{\pi}{4} \left( (\frac{\pi}{2} + 0) - (-\frac{\pi}{2} + 0) \right) = \frac{\pi}{4}(\pi) = \frac{\pi^2}{4}

Question 10

The base of a solid is the region enclosed by $y=2$ and $y=\sec(x)$ for $x$ in the interval $[-\pi/3, \pi/3]$ . Cross-sections perpendicular to the x-axis are equilateral triangles. Which integral represents the volume of the solid?

$\frac{\sqrt{3}}{4} \int_{-\pi/3}^{\pi/3} (2 - \sec(x))^2 dx$ (correct answer)
$\frac{\sqrt{3}}{4} \int_{-\pi/3}^{\pi/3} (4 - \sec^2(x)) dx$
$\frac{1}{2} \int_{-\pi/3}^{\pi/3} (2 - \sec(x))^2 dx$
$\frac{\sqrt{3}}{4} \int_{-\pi/3}^{\pi/3} (\sec(x) - 2)^2 dx$

Explanation: In the interval

[-\pi/3, \pi/3]

\sec(x)

ranges from

\sec(0)=1

\sec(\pi/3)=2

. Thus,

2 \ge \sec(x)

in this region. The side length of the equilateral triangle cross-section is

s(x) = \text{top} - \text{bottom} = 2 - \sec(x)

. The area of an equilateral triangle is

A(s) = \frac{\sqrt{3}}{4}s^2

. Therefore, the cross-sectional area is

A(x) = \frac{\sqrt{3}}{4}(2 - \sec(x))^2

. The volume is the integral of this area over the given interval:

V = \int_{-\pi/3}^{\pi/3} \frac{\sqrt{3}}{4}(2 - \sec(x))^2 dx

Question 11

The base of a solid is the triangular region bounded by the line $x+2y=2$ and the coordinate axes. The cross-sections perpendicular to the x-axis are isosceles right triangles with their hypotenuse on the base. What is the volume of the solid?

$\frac{1}{12}$
$\frac{2}{3}$
$\frac{1}{3}$
$\frac{1}{6}$ (correct answer)

Explanation: This problem tests your ability to find volumes of solids with known cross-sections, a key application of integration in Calculus 2. When you see "cross-sections perpendicular to an axis," think about setting up an integral where each slice contributes an area that depends on position. First, you need to understand the base region. The triangle is bounded by

x + 2y = 2

, the x-axis (

y = 0

), and the y-axis (

x = 0

). This gives vertices at

(0,0)

(2,0)

, and

(0,1)

. For any x-value from 0 to 2, the height of the triangular base at that position is

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

. Each cross-section is an isosceles right triangle with hypotenuse lying on the base. Since the hypotenuse has length

\frac{2-x}{2}

, and in an isosceles right triangle the legs have length

\frac{\text{hypotenuse}}{\sqrt{2}}

, each leg has length

\frac{2-x}{2\sqrt{2}}

. The area of each triangular cross-section is

\frac{1}{2} \times \text{leg}^2 = \frac{1}{2} \times \left(\frac{2-x}{2\sqrt{2}}\right)^2 = \frac{(2-x)^2}{16}

. The volume is

\int_0^2 \frac{(2-x)^2}{16} dx = \frac{1}{16} \int_0^2 (2-x)^2 dx = \frac{1}{16} \left[-\frac{(2-x)^3}{3}\right]_0^2 = \frac{1}{16} \times \frac{8}{3} = \frac{1}{6}

. Choice A (

\frac{1}{12}

) likely comes from incorrectly using the full triangle area instead of the isosceles right triangle. Choice B (

\frac{2}{3}

) and C (

\frac{1}{3}

) represent computational errors in the integration. Remember: always visualize the cross-section geometry carefully and double-check your area formula before integrating.

Question 12

A solid with volume $V$ is generated from a base region R with semicircular cross-sections perpendicular to the x-axis. A new solid is created using the same base region R, but with cross-sections that are equilateral triangles. What is the volume of the new solid in terms of $V$ ?

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

Explanation: Let

s(x)

be the width of the base R at

x

. The original volume is

V = V_{semi} = \int \frac{\pi}{8}(s(x))^2 dx

. The new volume is

V_{new} = V_{tri} = \int \frac{\sqrt{3}}{4}(s(x))^2 dx

. We can write

V_{new} = \frac{\sqrt{3}}{4} \int (s(x))^2 dx

. From the first equation,

\int (s(x))^2 dx = \frac{8}{\pi}V

. Substituting this into the second equation gives

V_{new} = \frac{\sqrt{3}}{4} \left( \frac{8}{\pi}V \right) = \frac{2\sqrt{3}}{\pi}V

Question 13

The base of a solid is the region enclosed by the parabola $y = 1 - x^2$ and the x-axis. The cross-sections perpendicular to the x-axis are semicircles with their diameters on the base. Which of the following integrals represents the volume of the solid?

$\frac{\pi}{8} \int_{-1}^{1} (1 - x^2)^2 dx$ (correct answer)
$\frac{\pi}{2} \int_{-1}^{1} (1 - x^2)^2 dx$
$\frac{\pi}{8} \int_{-1}^{1} (1 - x^2) dx$
$\frac{\pi}{4} \int_{0}^{1} (1 - y)^2 dy$

Explanation: The base is bounded by

y = 1 - x^2

and

y = 0

. The intersection points are at

x = -1

and

x = 1

, which are the limits of integration. For cross-sections perpendicular to the x-axis, the diameter of a semicircle at a given

x

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

. The area of a semicircle is

A = \frac{1}{2}\pi r^2

. Since the diameter is

s

, the radius is

r = s/2

, so

A(s) = \frac{1}{2}\pi (s/2)^2 = \frac{\pi}{8}s^2

. Therefore, the area of a cross-section is

A(x) = \frac{\pi}{8}(1 - x^2)^2

. The volume is the integral of this area function from -1 to 1:

V = \int_{-1}^{1} \frac{\pi}{8}(1 - x^2)^2 dx

Question 14

The base of a solid is the region in the first quadrant bounded by $y = e^x$ , $y = 1$ , and $x = 2$ . Cross-sections perpendicular to the x-axis are semicircles. Which integral represents the volume of the solid?

$\frac{\pi}{8} \int_{0}^{2} (e^{2x} - 1) dx$
$\frac{\pi}{8} \int_{0}^{2} e^{2x} dx$
$\frac{\pi}{2} \int_{0}^{2} (e^x - 1)^2 dx$
$\frac{\pi}{8} \int_{0}^{2} (e^x - 1)^2 dx$ (correct answer)

Explanation: The region is bounded above by

y=e^x

and below by

y=1

. The limits of integration are from

x=0

(where

e^x=1

) to

x=2

. The diameter of a semicircular cross-section is

s(x) = e^x - 1

. The area of a semicircle with diameter

s

A = \frac{\pi}{8}s^2

. Thus, the area of a cross-section is

A(x) = \frac{\pi}{8}(e^x - 1)^2

. The volume is given by the integral

V = \int_{0}^{2} \frac{\pi}{8}(e^x - 1)^2 dx

Question 15

Let R be a planar region. Let $V_S$ be the volume of a solid with base R and semicircular cross-sections. Let $V_T$ be the volume of a solid with the same base R and isosceles right triangle cross-sections with the hypotenuse on the base. What is the value of the ratio $\frac{V_S}{V_T}$ ?

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

Explanation: This problem tests your understanding of volumes with cross-sections, where you integrate the area of each cross-section along the base region. The key insight is comparing how the areas of semicircular and triangular cross-sections relate when they share the same base width. Let's say the width of the base at any point is

w

. For the semicircular cross-sections, the diameter equals

w

, so the radius is

w/2

. The area of a semicircle is

\frac{1}{2}\pi r^2 = \frac{1}{2}\pi \left(\frac{w}{2}\right)^2 = \frac{\pi w^2}{8}

. For the isosceles right triangle with hypotenuse

w

on the base, you need to find the legs. In an isosceles right triangle, if the hypotenuse is

w

, each leg has length

\frac{w}{\sqrt{2}}

. The area is

\frac{1}{2} \times \frac{w}{\sqrt{2}} \times \frac{w}{\sqrt{2}} = \frac{w^2}{4}

. The ratio of areas at each cross-section is:

\frac{\frac{\pi w^2}{8}}{\frac{w^2}{4}} = \frac{\pi w^2}{8} \times \frac{4}{w^2} = \frac{\pi}{2}

Since this ratio is constant for every cross-section, the volume ratio

\frac{V_S}{V_T} = \frac{\pi}{2}

, which is choice D. Choice A (

\frac{\pi}{4}

) likely comes from incorrectly using the full circle area instead of semicircle. Choice B (

\frac{4}{\pi}

) is the reciprocal of A. Choice C (

\frac{2}{\pi}

) is the reciprocal of the correct answer. Remember: when comparing cross-sectional volumes, focus on the area ratio at each slice—the volume ratio will be identical.

Question 16

The base of a solid is the region bounded by the parabola $x = y^2$ and the line $x = 4$ . The cross-sections perpendicular to the y-axis are isosceles right triangles with their hypotenuse on the base. Which integral gives the volume of this solid?

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

Explanation: The cross-sections are perpendicular to the y-axis, so we integrate with respect to

y

. The region is bounded by

x = y^2

(right boundary for

y<0

, left for

y>0

) and

x=4

(right boundary). The intersections are at

y^2=4

, so

y = \pm 2

. The length of a horizontal cross-section of the base is

s(y) = \text{right} - \text{left} = 4 - y^2

. For an isosceles right triangle with hypotenuse

s

, the legs are

s/\sqrt{2}

, and the area is

A = \frac{1}{2}(\frac{s}{\sqrt{2}})^2 = \frac{1}{4}s^2

. So,

A(y) = \frac{1}{4}(4 - y^2)^2

. The volume is

V = \int_{-2}^{2} \frac{1}{4}(4 - y^2)^2 dy

Question 17

The base of a solid is the region bounded by $y=k$ and $y=x^2$ for some constant $k>0$ . Cross-sections perpendicular to the y-axis are equilateral triangles. If the volume of the solid is $8\sqrt{3}$ , what is the value of $k$ ?

2
$2\sqrt{2}$
4 (correct answer)
8

Explanation: We integrate with respect to

y

from

y=0

y=k

. At a given height

y

x = \pm\sqrt{y}

. The horizontal side length of the triangular cross-section is

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

. The area of an equilateral triangle is

A = \frac{\sqrt{3}}{4}s^2

. So,

A(y) = \frac{\sqrt{3}}{4}(2\sqrt{y})^2 = \frac{\sqrt{3}}{4}(4y) = \sqrt{3}y

. The volume is

V = \int_{0}^{k} \sqrt{3}y \,dy = \sqrt{3} \left[ \frac{y^2}{2} \right]_{0}^{k} = \frac{\sqrt{3}}{2}k^2

. We are given

V = 8\sqrt{3}

. So,

\frac{\sqrt{3}}{2}k^2 = 8\sqrt{3} \implies \frac{k^2}{2} = 8 \implies k^2 = 16 \implies k=4

Question 18

The volume of a solid is described by the integral $V = \frac{1}{4} \int_{1}^{5} (g(y))^2 dy$ . If the cross-sections are perpendicular to the y-axis, which of the following accurately describes the solid's cross-sections?

Semicircles with diameter $g(y)$ on the base.
Equilateral triangles with side $g(y)$ on the base.
Isosceles right triangles with a leg $g(y)$ on the base.
Isosceles right triangles with hypotenuse $g(y)$ on the base. (correct answer)

Explanation: The volume of a solid with cross-sections perpendicular to the y-axis is given by

V = \int_{c}^{d} A(y) dy

, where

A(y)

is the area of the cross-section at

y

. From the given integral, we can identify the area function as

A(y) = \frac{1}{4}(g(y))^2

. We must match this to the area formula for the given shapes, assuming

s = g(y)

is the side length on the base. Semicircle:

A = \frac{\pi}{8}s^2

. Equilateral triangle:

A = \frac{\sqrt{3}}{4}s^2

. Isosceles right triangle with leg on base:

A = \frac{1}{2}s^2

. Isosceles right triangle with hypotenuse on base:

A = \frac{1}{4}s^2

. The formula

A(y) = \frac{1}{4}(g(y))^2

matches that of an isosceles right triangle with its hypotenuse on the base.

Question 19

The base of a solid is the region bounded by $y = \sqrt{x-1}$ , the line $x=5$ , and the x-axis. Cross-sections perpendicular to the y-axis are semicircles. Which integral represents the volume of the solid?

$\frac{\pi}{8} \int_{1}^{5} (x-1) dx$
$\frac{\pi}{2} \int_{0}^{2} (4-y^2)^2 dy$
$\frac{\pi}{8} \int_{0}^{2} (5 - (y^2+1)) dy$
$\frac{\pi}{8} \int_{0}^{2} (4-y^2)^2 dy$ (correct answer)

Explanation: When finding volumes with varying cross-sections, you need to set up an integral where each slice has area equal to the cross-sectional area at that point. Since the cross-sections are perpendicular to the y-axis, you'll integrate with respect to y. First, understand the base region. The curve

y = \sqrt{x-1}

can be rewritten as

x = y^2 + 1

. The region is bounded by this curve,

x = 5

, and the x-axis. Since

y = \sqrt{x-1}

starts at

(1,0)

and reaches

(5,2)

, your y-values range from 0 to 2. At any height y, the cross-section is a semicircle whose diameter extends from

x = y^2 + 1

(the curve) to

x = 5

(the vertical line). The diameter length is

5 - (y^2 + 1) = 4 - y^2

. Since the area of a semicircle is

\frac{\pi r^2}{2} = \frac{\pi d^2}{8}

, each cross-sectional area is

\frac{\pi}{8}(4-y^2)^2

. The volume integral becomes

\frac{\pi}{8} \int_{0}^{2} (4-y^2)^2 dy

, which is choice D. Choice A uses the wrong variable of integration and incorrect limits. Choice B has the correct integrand form but uses

\frac{\pi}{2}

instead of

\frac{\pi}{8}

—this comes from confusing the semicircle area formula with the full circle formula. Choice C fails to square the diameter, giving linear rather than area units. Study tip: Always check your cross-sectional area formula carefully—semicircles use

\frac{\pi d^2}{8}

, not

\frac{\pi d}{4}

. Sketch the region and identify which variable makes the cross-sections simpler to describe.

Question 20

The base of a solid is a region R in the xy-plane. For any such base R, which of the following cross-sectional shapes, taken perpendicular to the x-axis, will always produce the solid with the greatest volume?

Semicircles with their diameter on the base.
Equilateral triangles with a side on the base.
Isosceles right triangles with a leg on the base. (correct answer)
Isosceles right triangles with the hypotenuse on the base.

Explanation: The volume is given by

V = \int_a^b A(x) dx

, where

A(x)

is the cross-sectional area. Let

s(x)

be the width of the base. The area for each shape is a constant multiple of

s(x)^2

. To maximize the volume, we must maximize this constant. (A) Semicircle:

A = \frac{\pi}{8}s^2 \approx 0.393 s^2

. (B) Equilateral triangle:

A = \frac{\sqrt{3}}{4}s^2 \approx 0.433 s^2

. (C) Isosceles right triangle with leg on base:

A = \frac{1}{2}s^2 = 0.5 s^2

. (D) Isosceles right triangle with hypotenuse on base:

A = \frac{1}{4}s^2 = 0.25 s^2

. The largest constant is

1/2

, so isosceles right triangles with a leg on the base will always produce the greatest volume.