AP Calculus AB Quiz: Disc Method Revolving Around Xy Axes

What this quiz covers

This quiz focuses on Disc Method Revolving Around Xy Axes, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

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.

Question 1

1. $\pi\int_{-2}^{2} (\sqrt{4-x^2})^2\,dx$ (correct answer)
2. $\pi\int_{-2}^{2} \sqrt{4-x^2}\,dx$
3. $\pi\int_{-2}^{2} (4-x^2)^2\,dx$
4. $\pi\int_{0}^{2} (\sqrt{4-x^2})^2\,dx$
5. $\pi\int_{-2}^{2} (\sqrt{4-x^2})^2\,dy$

Explanation: This problem applies the disc method for revolution about the x-axis. The region bounded by y = √(4-x²) and the x-axis creates a solid with no hole, so discs rather than washers are appropriate. The radius of each disc at position x equals the distance from the x-axis to the curve: r(x) = √(4-x²). Using the disc method formula: V = π∫[a to b] [r(x)]² dx = π∫[-2 to 2] (√(4-x²))² dx. Choice B is incorrect because it omits the required squaring of the radius in the disc method formula. Use discs when the solid extends from the axis of revolution to a single boundary curve.

Question 2

1. $\pi\int_{0}^{4} (3\sqrt{y})^2\,dy$ (correct answer)
2. $\pi\int_{0}^{4} 3\sqrt{y}\,dy$
3. $\pi\int_{0}^{4} (\sqrt{y})^2\,dy$
4. $\pi\int_{0}^{2} (3\sqrt{y})^2\,dy$
5. $\pi\int_{0}^{4} (3\sqrt{y})^2\,dx$

Explanation: This problem uses the disc method for revolution about the y-axis. Since the region is bounded by x = 3√y and the y-axis with no inner boundary, discs are the appropriate method. The radius of each disc at position y is the distance from the y-axis to the curve: r(y) = 3√y. The volume formula becomes V = π∫[a to b] [r(y)]² dy = π∫[0 to 4] (3√y)² dy. Choice B fails because it doesn't square the radius function, which is essential in the disc method. Apply the disc method when there's a single boundary curve extending from the axis of revolution.

Question 3

1. $\pi\int_{0}^{1} (2+y^3)\,dy$
2. $\pi\int_{0}^{1} (2+y^3)^2\,dy$ (correct answer)
3. $\pi\int_{0}^{1} (y^3)^2\,dy$
4. $\pi\int_{0}^{1} (2+y^3)^2\,dx$
5. $\pi\int_{1}^{0} (2+y^3)^2\,dy$

Explanation: This problem uses the disc method for revolution about the y-axis. Since the region is bounded by x = 2+y³ and the y-axis with no inner boundary, discs are the correct approach. The radius of each disc at position y is r(y) = 2+y³, the distance from the y-axis to the curve. The volume is V = π∫[a to b] [r(y)]² dy = π∫[0 to 1] (2+y³)² dy. Choice A fails because it doesn't square the radius function, which is essential in the disc method. Apply discs when there's exactly one boundary curve extending from the axis of revolution.

Question 4

1. $\pi\int_{-2}^{2} (4-x^2)^2\,dx$ (correct answer)
2. $\pi\int_{-2}^{2} (4-x^2)\,dx$
3. $\pi\int_{0}^{2} (4-x^2)^2\,dx$
4. $\pi\int_{-2}^{2} (4-x^2)^2\,dy$
5. $\pi\int_{-2}^{2} (x^2-4)^2\,dx$

Explanation: This problem requires the disc method for revolution about the x-axis. The region bounded by y = 4-x² and the x-axis forms a solid without a hole, making discs appropriate rather than washers. The radius of each disc at position x equals the distance from the x-axis to the curve: r(x) = 4-x². Applying the disc method: V = π∫[a to b] [r(x)]² dx = π∫[-2 to 2] (4-x²)² dx. Choice B is tempting but incorrect because it omits the essential squaring of the radius required in the disc method formula. Use discs when the region extends from the axis of revolution to a single bounding curve.

Question 5

1. $\pi\int_{0}^{2} (y^2)^2\,dy$ (correct answer)
2. $\pi\int_{0}^{2} y^2\,dy$
3. $\pi\int_{0}^{2} (y^2)^2\,dx$
4. $\pi\int_{0}^{4} (y^2)^2\,dy$
5. $2\pi\int_{0}^{2} (y^2)^2\,dy$

Explanation: This problem applies the disc method for revolution about the y-axis. The region bounded by $x = y^2$ and the y-axis creates a solid with no hole, so discs rather than washers are appropriate. The radius of each disc at position y equals the distance from the y-axis to the curve: $r(y) = y^2$. Using the disc method formula: $V = \pi \int_{a}^{b} [r(y)]^2 \, dy = \pi \int_{0}^{2} (y^2)^2 \, dy$. Choice B is incorrect because it omits the required squaring of the radius in the disc method formula. Use discs when the solid extends from the axis of revolution to a single boundary curve.

Question 6

1. $\pi\int_{0}^{1} (x-x^3)^2\,dx$ (correct answer)
2. $\pi\int_{0}^{1} (x-x^3)\,dx$
3. $\pi\int_{0}^{1} (x^3-x)^2\,dx$
4. $\pi\int_{0}^{1} (x-x^3)^2\,dy$
5. $\pi\int_{-1}^{1} (x-x^3)^2\,dx$

Explanation: This problem applies the disc method for revolution about the x-axis. The region bounded by y = x-x³ and the x-axis creates a solid with no hole, so discs rather than washers are appropriate. The radius of each disc at position x equals the distance from the x-axis to the curve: r(x) = x-x³. Using the disc method formula: V = π∫[a to b] [r(x)]² dx = π∫[0 to 1] (x-x³)² dx. Choice B is incorrect because it omits the required squaring of the radius in the disc method formula. Use discs when the solid extends from the axis of revolution to a single boundary curve.

Question 7

1. $\pi\int_{0}^{\pi} (\sin y)^2\,dy$ (correct answer)
2. $\pi\int_{0}^{\pi} \sin y\,dy$
3. $\pi\int_{0}^{\pi} (\sin y)^2\,dx$
4. $\pi\int_{0}^{2\pi} (\sin y)^2\,dy$
5. $2\pi\int_{0}^{\pi} (\sin y)^2\,dy$

Explanation: This problem applies the disc method for revolution about the y-axis. The region bounded by x = sin y and the y-axis creates a solid with no hole, so discs rather than washers are appropriate. The radius of each disc at position y equals the distance from the y-axis to the curve: r(y) = sin y. Using the disc method formula: V = π∫[a to b] [r(y)]² dy = π∫[0 to π] (sin y)² dy. Choice B is incorrect because it omits the required squaring of the radius in the disc method formula. Use the disc method when the solid extends from the axis of revolution to a single boundary curve.

Question 8

1. $\pi\int_{0}^{1} e^x\,dx$
2. $\pi\int_{0}^{1} (e^x)^2\,dx$ (correct answer)
3. $\pi\int_{0}^{1} (e^x)^2\,dy$
4. $\pi\int_{1}^{0} (e^x)^2\,dx$
5. $2\pi\int_{0}^{1} (e^x)^2\,dx$

Explanation: This problem applies the disc method for revolution about the x-axis. The region bounded by y = eˣ and the x-axis creates a solid with no hole, so discs are appropriate. The radius of each disc at position x equals the distance from the x-axis to the curve: r(x) = eˣ. Using the disc method formula: V = π∫[a to b] [r(x)]² dx = π∫[0 to 1] (eˣ)² dx. Choice A is incorrect because it omits the essential squaring of the radius function required in the disc method. Apply discs when there's a single boundary curve extending from the axis of revolution.

Question 9

1. $\pi\int_{0}^{\pi/2} (\cos x)^2\,dx$ (correct answer)
2. $\pi\int_{0}^{\pi/2} \cos x\,dx$
3. $\pi\int_{0}^{\pi} (\cos x)^2\,dx$
4. $\pi\int_{0}^{\pi/2} (\cos x)^2\,dy$
5. $\pi\int_{0}^{\pi/2} (\sin x)^2\,dx$

Explanation: This problem uses the disc method for revolution about the x-axis. Since the region is bounded by y = cos x and the x-axis with no inner boundary, we apply discs rather than washers. The radius of each disc at position x is the distance from the x-axis to the curve: r(x) = cos x. The volume formula becomes V = π∫[a to b] [r(x)]² dx = π∫[0 to π/2] (cos x)² dx. Choice B is tempting but fails because it doesn't square the radius function, which is required by the disc method. Use the disc method when the region extends from the axis of revolution to exactly one bounding curve.

Question 10

1. $\pi\int_{0}^{2} (2-x)\,dx$
2. $\pi\int_{0}^{2} (2-x)^2\,dx$ (correct answer)
3. $\pi\int_{0}^{2} (2-x)^2\,dy$
4. $\pi\int_{-2}^{0} (2-x)^2\,dx$
5. $\pi\int_{0}^{2} (x-2)^2\,dx$

Explanation: This problem uses the disc method for revolution about the x-axis. The region bounded by y = 2-x and the x-axis creates a solid with no hole, so discs are appropriate rather than washers. The radius of each disc at position x equals the distance from the x-axis to the curve: r(x) = 2-x. Using the disc method formula: V = π∫[a to b] [r(x)]² dx = π∫[0 to 2] (2-x)² dx. Choice A fails because it doesn't square the radius function, which is essential in the disc method. Use discs when the region has a single boundary extending from the axis of revolution to one curve.

Question 11

1. $\pi\int_{0}^{\pi/2} \cos y\,dy$
2. $\pi\int_{0}^{\pi/2} (\cos y)^2\,dy$ (correct answer)
3. $\pi\int_{0}^{\pi/2} (\sin y)^2\,dy$
4. $\pi\int_{0}^{\pi} (\cos y)^2\,dy$
5. $\pi\int_{0}^{\pi/2} (\cos y)^2\,dx$

Explanation: This problem uses the disc method for revolution about the y-axis. Since the region is bounded by x = cos y and the y-axis with no inner boundary, discs are the appropriate approach. The radius of each disc at position y is r(y) = cos y, the distance from the y-axis to the curve. The volume is V = π∫[a to b] [r(y)]² dy = π∫[0 to π/2] (cos y)² dy. Choice A fails because it doesn't square the radius function, which is essential in the disc method. Apply discs when there's exactly one boundary curve extending from the axis of revolution.

Question 12

1. $\pi\int_{0}^{1} \left(\frac{1}{1+x^2}\right)^2\,dx$ (correct answer)
2. $\pi\int_{0}^{1} \frac{1}{1+x^2}\,dx$
3. $\pi\int_{0}^{1} \left(1+x^2\right)^2\,dx$
4. $\pi\int_{-1}^{1} \left(\frac{1}{1+x^2}\right)^2\,dx$
5. $\pi\int_{0}^{1} \left(\frac{1}{1+x^2}\right)^2\,dy$

Explanation: This problem applies the disc method for revolution about the x-axis. The region bounded by y = 1/(1+x²) and the x-axis creates a solid with no hole, so discs rather than washers are appropriate. The radius of each disc at position x equals the distance from the x-axis to the curve: r(x) = 1/(1+x²). Using the disc method formula: V = π∫[a to b] [r(x)]² dx = π∫[0 to 1] (1/(1+x²))² dx. Choice B is incorrect because it omits the required squaring of the radius in the disc method formula. Use discs when the solid extends from the axis of revolution to a single boundary curve.

Question 13

1. $\pi\int_{0}^{\pi/4} (\tan x)^2\,dx$ (correct answer)
2. $\pi\int_{0}^{\pi/4} \tan x\,dx$
3. $\pi\int_{0}^{\pi/2} (\tan x)^2\,dx$
4. $\pi\int_{0}^{\pi/4} (\tan x)^2\,dy$
5. $2\pi\int_{0}^{\pi/4} (\tan x)^2\,dx$

Explanation: This problem applies the disc method for revolution about the x-axis. The region bounded by y = tan x and the x-axis creates a solid with no hole, so discs rather than washers are appropriate. The radius of each disc at position x equals the distance from the x-axis to the curve: r(x) = tan x. Using the disc method formula: V = π∫[a to b] [r(x)]² dx = π∫[0 to π/4] (tan x)² dx. Choice B is incorrect because it omits the required squaring of the radius in the disc method formula. Use discs when the solid extends from the axis of revolution to a single boundary curve.

Question 14

1. $\pi\int_{0}^{1} (1+x^3)^2\,dx$ (correct answer)
2. $\pi\int_{0}^{1} (1+x^3)\,dx$
3. $\pi\int_{0}^{1} (x^3)^2\,dx$
4. $\pi\int_{0}^{1} (1+x^3)^2\,dy$
5. $\pi\int_{1}^{0} (1+x^3)^2\,dx$

Explanation: This problem uses the disc method for revolution about the x-axis. Since the region is bounded by y = 1+x³ and the x-axis with no inner boundary, discs are appropriate rather than washers. The radius of each disc at position x is the distance from the x-axis to the curve: r(x) = 1+x³. The volume formula becomes V = π∫[a to b] [r(x)]² dx = π∫[0 to 1] (1+x³)² dx. Choice B fails because it doesn't square the radius function, which is essential in the disc method. Apply discs when there's exactly one boundary curve extending from the axis of revolution.

Question 15

1. $\pi\int_{0}^{\pi/4} (\tan y)^2\,dy$ (correct answer)
2. $\pi\int_{0}^{\pi/4} \tan y\,dy$
3. $\pi\int_{0}^{\pi/2} (\tan y)^2\,dy$
4. $\pi\int_{0}^{\pi/4} (\tan y)^2\,dx$
5. $2\pi\int_{0}^{\pi/4} (\tan y)^2\,dy$

Explanation: This problem uses the disc method for revolution about the y-axis. Since the region is bounded by x = tan y and the y-axis with no inner boundary, discs are the correct approach. The radius of each disc at position y is r(y) = tan y, the distance from the y-axis to the curve. The volume is V = π∫[a to b] [r(y)]² dy = π∫[0 to π/4] (tan y)² dy. Choice B fails because it doesn't square the radius function, which is essential in the disc method. Apply discs when there's exactly one boundary curve extending from the axis of revolution.

Question 16

1. $\pi\int_{0}^{\pi} (1-\cos x)^2\,dx$ (correct answer)
2. $\pi\int_{0}^{\pi} (1-\cos x)\,dx$
3. $\pi\int_{0}^{\pi} (\cos x-1)^2\,dx$
4. $\pi\int_{0}^{\pi} (1-\cos x)^2\,dy$
5. $\pi\int_{0}^{2\pi} (1-\cos x)^2\,dx$

Explanation: This problem requires the disc method for revolution about the x-axis. The region bounded by y = 1-cos x and the x-axis forms a solid without a hole, making discs appropriate rather than washers. The radius of each disc at position x equals the distance from the x-axis to the curve: r(x) = 1-cos x. Applying the disc method: V = π∫[a to b] [r(x)]² dx = π∫[0 to π] (1-cos x)² dx. Choice B is tempting but incorrect because it omits the essential squaring of the radius required in the disc method formula. Use discs when the region extends from the axis of revolution to a single bounding curve.

Question 17

1. $V=\pi\displaystyle\int_{0}^{4}x\,dx$
2. $V=\pi\displaystyle\int_{0}^{4}(\sqrt{x})^{2}\,dx$ (correct answer)
3. $V=\pi\displaystyle\int_{0}^{4}\sqrt{x}\,dx$
4. $V=\pi\displaystyle\int_{0}^{2}(\sqrt{x})^{2}\,dx$
5. $V=\pi\displaystyle\int_{0}^{4}(x)^{2}\,dx$

Explanation: This problem requires the disc method to find the volume when revolving around the x-axis. Since we're revolving the region under y = √x from x = 0 to x = 4 around the x-axis, we create circular discs perpendicular to the x-axis. The radius of each disc is the y-value, which is √x, and the disc method formula gives V = π∫[radius]² dx. Therefore, we need π∫₀⁴(√x)² dx = π∫₀⁴ x dx. Choice C incorrectly uses √x instead of (√x)², forgetting that the disc method requires squaring the radius. When revolving around the x-axis, always square the function value to get the disc's area before integrating.

Question 18

1. $\pi\displaystyle\int_{0}^{2}(2-x)\,dx$
2. $\pi\displaystyle\int_{0}^{2}(2-x)^{2}\,dx$ (correct answer)
3. $\pi\displaystyle\int_{0}^{2}(x-2)^{2}\,dy$
4. $\pi\displaystyle\int_{0}^{2}x^{2}\,dx$
5. $\pi\displaystyle\int_{0}^{2}(2-(2-x))^{2}\,dx$

Explanation: This problem requires the disc method for revolving around the x-axis. When rotating the region under y = 2 - x from x = 0 to x = 2 around the x-axis, we create circular discs perpendicular to the x-axis. Each disc has radius equal to the y-value at that x-position, which is 2 - x, so the disc's area is π(2 - x)². The volume is the integral of these disc areas: π∫₀²(2 - x)²dx. Choice A incorrectly uses π∫₀²(2 - x)dx, which forgets to square the radius function—this is a common error that gives an incorrect result. The disc method for x-axis rotation always requires squaring the function: use π∫[f(x)]²dx, never just π∫f(x)dx.

Question 19

1. $V=\pi\displaystyle\int_{0}^{3}(2x)^{2}\,dx$ (correct answer)
2. $V=2\pi\displaystyle\int_{0}^{3}(2x)\,dx$
3. $V=\pi\displaystyle\int_{0}^{3}(2x)\,dx$
4. $V=\pi\displaystyle\int_{0}^{3}x^{2}\,dx$
5. $V=\pi\displaystyle\int_{0}^{2}(2x)^{2}\,dx$

Explanation: This problem uses the disc method for revolving the region under y = 2x around the x-axis. When revolving around the x-axis, we form discs perpendicular to the x-axis with radius equal to the y-value at each x. For y = 2x, the radius is 2x, so the disc method gives V = π∫₀³(2x)² dx = π∫₀³ 4x² dx. The limits are from x = 0 to x = 3 as specified in the problem. Choice C incorrectly uses (2x) without squaring, which would give the circumference times dx rather than the area of the disc. Remember: disc method always requires squaring the radius function when setting up the integral.

Question 20

1. $\pi\displaystyle\int_{1}^{3}\left(\frac{1}{x}\right)^{2}\,dx$ (correct answer)
2. $\pi\displaystyle\int_{1}^{3}\frac{1}{x}\,dx$
3. $\pi\displaystyle\int_{1}^{3}x^{2}\,dx$
4. $\pi\displaystyle\int_{0}^{3}\left(\frac{1}{x}\right)^{2}\,dx$
5. $\pi\displaystyle\int_{1}^{3}\left(\frac{1}{y}\right)^{2}\,dy$

Explanation: This problem applies the disc method for rotation about the x-axis. Revolving the region under y = 1/x from x = 1 to x = 3 around the x-axis creates circular discs perpendicular to the x-axis. The radius of each disc equals the y-value, which is 1/x, making the disc's area π(1/x)². We integrate from x = 1 to x = 3 to find the volume: π∫₁³(1/x)²dx = π∫₁³(1/x²)dx. Choice B incorrectly uses π∫₁³(1/x)dx, missing the crucial squaring step—this would give ln|x| evaluated from 1 to 3, which doesn't represent a volume. Remember: disc method for x-axis rotation always requires π times the square of the function, which accounts for the circular area of each disc.