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AP Calculus BC Practice Quiz: Volume with Disc Method: Revolving Around the $x$- or $y$-Axis

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 7 questions to check your progress.

Question 1 of 7

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 of the following definite integrals gives the volume of the solid generated when R is revolved about the x-axis?

All Questions (7)

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 of the following definite integrals gives the volume of the solid generated when R is revolved about the x-axis?

A) $\pi \int_{0}^{4} x \, dx$

B) $\pi \int_{0}^{4} \sqrt{x} \, dx$

C) $\int_{0}^{4} x \, dx$

D) $\pi \int_{0}^{2} y^4 \, dy$

Correct Answer: A

When revolving a region about the x-axis using the disc method, the volume is given by the integral $V = \pi \int_{a}^{b} [R(x)]^2 \, dx$. In this case, the region is bounded by $x=0$ and $x=4$, so the limits of integration are from 0 to 4. The radius of each disc, $R(x)$, is the function value, which is $y = \sqrt{x}$. Therefore, the integral is $V = \pi \int_{0}^{4} (\sqrt{x})^2 \, dx = \pi \int_{0}^{4} x \, dx$.

The region in the first quadrant bounded by the graph of $y = 9 - x^2$ and the coordinate axes is revolved about the x-axis. What is the volume of the solid generated?

A) $18\pi$

B) $\frac{243\pi}{5}$

C) $\frac{648\pi}{5}$

D) $54\pi$

Correct Answer: C

First, find the bounds of integration by setting $y=0$: $9 - x^2 = 0 \implies x=3$ (in the first quadrant). The volume is found using the disc method for revolution about the x-axis: $V = \pi \int_{a}^{b} [R(x)]^2 \, dx$. Here, $R(x) = 9 - x^2$, $a=0$, and $b=3$. So, $V = \pi \int_{0}^{3} (9 - x^2)^2 \, dx = \pi \int_{0}^{3} (81 - 18x^2 + x^4) \, dx$. Evaluating the integral gives $\pi [81x - 6x^3 + \frac{x^5}{5}]_{0}^{3} = \pi [(81(3) - 6(3)^3 + \frac{3^5}{5}) - 0] = \pi [243 - 162 + \frac{243}{5}] = \pi [81 + \frac{243}{5}] = \pi [\frac{405+243}{5}] = \frac{648\pi}{5}$.

Let R be the region bounded by the graph of $y = x^3$, the y-axis, and the line $y=8$. Which of the following gives the volume of the solid formed by revolving R about the y-axis?

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

B) $\pi \int_{0}^{8} (y^{1/3})^2 \, dy$

C) $\pi \int_{0}^{8} y^3 \, dy$

D) $\pi \int_{0}^{2} (8 - x^3)^2 \, dx$

Correct Answer: B

For revolution about the y-axis, the disc method formula is $V = \pi \int_{c}^{d} [R(y)]^2 \, dy$. The radius, $R(y)$, is the horizontal distance from the y-axis to the curve, which is the x-value. We must express x in terms of y: if $y = x^3$, then $x = y^{1/3}$. The region is bounded by the y-axis ($x=0$) and extends vertically from $y=0$ to $y=8$. Therefore, the integral for the volume is $V = \pi \int_{0}^{8} (y^{1/3})^2 \, dy$.

The volume of a solid generated by revolving a region bounded by a continuous function $y=f(x)$, the x-axis, and the lines $x=a$ and $x=b$ about the x-axis is found using the disc method. The integral for this volume is $\pi \int_{a}^{b} [R(x)]^2 dx$. What does $R(x)$ represent in this context?

A) The area of the region being revolved.

B) The thickness of a representative disc.

C) The radius of a representative disc, given by the function $f(x)$.

D) The circumference of a representative disc.

Correct Answer: C

In the disc method for revolution around the x-axis, the solid is conceptualized as a stack of infinitesimally thin discs. The radius of each disc at a given x-value is the distance from the axis of revolution (the x-axis) to the outer edge of the region. This distance is given by the function value, $f(x)$. Therefore, $R(x) = f(x)$ represents the radius of a representative disc.

The region enclosed by the graph of $y = e^x$, the y-axis, and the line $y=e$ is revolved about the y-axis. Which integral represents the volume of the resulting solid?

A) $\pi \int_{1}^{e} (\ln y)^2 \, dy$

B) $\pi \int_{0}^{1} (e^x)^2 \, dx$

C) $\pi \int_{0}^{e} (\ln y)^2 \, dy$

D) $\pi \int_{1}^{e} e^{2y} \, dy$

Correct Answer: A

To find the volume of a solid revolved about the y-axis, we use the integral $V = \pi \int_{c}^{d} [R(y)]^2 \, dy$. The radius $R(y)$ is the x-coordinate. We must solve the function for x: $y = e^x \implies x = \ln y$. The bounds of integration are in terms of y. The region starts at the y-intercept of $y=e^x$, which is $y=e^0=1$, and goes up to the line $y=e$. Therefore, the integral is $V = \pi \int_{1}^{e} (\ln y)^2 \, dy$.

What is the volume of the solid generated when the region bounded by the line $y = 2x$, the y-axis, and the line $y=4$ is revolved about the y-axis?

A) $\frac{8\pi}{3}$

B) $8\pi$

C) $\frac{16\pi}{3}$

D) $16\pi$

Correct Answer: C

For revolution about the y-axis, we need to express the radius $R(y)$ in terms of y. The function is $y=2x$, so $x = \frac{y}{2}$. This is our radius, $R(y) = \frac{y}{2}$. The bounds of integration are from $y=0$ to $y=4$. The volume integral is $V = \pi \int_{0}^{4} (\frac{y}{2})^2 \, dy = \pi \int_{0}^{4} \frac{y^2}{4} \, dy$. Evaluating the integral: $V = \frac{\pi}{4} [\frac{y^3}{3}]_{0}^{4} = \frac{\pi}{4} [\frac{4^3}{3} - 0] = \frac{\pi}{4} (\frac{64}{3}) = \frac{16\pi}{3}$.

The base of a solid is the region in the first quadrant enclosed by the graph of $y = \sin(x)$ from $x=0$ to $x=\pi$. If the solid is generated by revolving this region about the x-axis, which of the following integrals represents the volume of the solid?

A) $\pi \int_{0}^{\pi} \sin(x) \, dx$

B) $\int_{0}^{\pi} \sin^2(x) \, dx$

C) $\pi \int_{0}^{\pi} \sin^2(x) \, dx$

D) $\pi \int_{0}^{1} (\arcsin y)^2 \, dy$

Correct Answer: C

The volume of a solid of revolution around the x-axis is given by the disc method formula $V = \pi \int_{a}^{b} [f(x)]^2 \, dx$. In this problem, the function is $f(x) = \sin(x)$ and the interval is from $a=0$ to $b=\pi$. The radius of a representative disc is $R(x) = \sin(x)$. Therefore, the integral that represents the volume is $V = \pi \int_{0}^{\pi} (\sin(x))^2 \, dx$, which is written as $\pi \int_{0}^{\pi} \sin^2(x) \, dx$.