AP Calculus BC Quiz: Disc Method Revolving Around Other Axes

What this quiz covers

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

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

Explanation: This problem involves the disc method for volumes of revolution around a shifted axis, specifically around y=4. With the axis shifted to y=4, the radii become distances from y=4 to the boundaries. The outer radius is 4 - 0 =4, to y=0, and the inner radius is 4 - (3x - x^2), to the upper curve. The volume is π times the integral of (4^2 - (4 - (3x - x^2))^2) from 0 to 3. A tempting distractor is choice A, which uses only the inner radius without subtracting from the outer. A transferable axis-shift strategy is to subtract the axis value from the curve functions to find the effective radii.

Question 2

1. $V=\pi\int_{0}^{\pi/2}\big(2-\cos x\big)^2\,dx$
2. $V=\pi\int_{0}^{\pi/2}\Big(\big(2-0\big)^2-\big(2-\cos x\big)^2\Big)\,dx$ (correct answer)
3. $V=\pi\int_{0}^{\pi/2}\big(\cos x\big)^2\,dx$
4. $V=\pi\int_{0}^{2}\Big(\big(2-\arccos y\big)^2-\big(2-0\big)^2\Big)\,dy$
5. $V=\pi\int_{0}^{\pi/2}\Big(\big(\cos x-2\big)^2-\big(0-2\big)^2\Big)\,dx$

Explanation: This problem involves the disc method with revolution around y = 2, above the region. The region R is bounded by y = cos x and y = 0 on [0, π/2]. When revolving around y = 2, the outer radius extends from y = 2 down to y = 0: R_outer = 2 - 0 = 2, and the inner radius extends from y = 2 down to y = cos x: R_inner = 2 - cos x. The volume formula is V = π∫₀^(π/2)[2² - (2-cos x)²]dx. Choice A incorrectly uses only the inner radius, missing the washer structure. For revolution around a horizontal line above the region, subtract each boundary's y-value from the axis to find radii, with the lower boundary giving the larger radius.

Question 3

1. $V=\pi\int_{-2}^{2}\big(5^2-(x^2+1)^2\big)\,dx$ (correct answer)
2. $V=\pi\int_{-2}^{2}(x^2+1)^2\,dx$
3. $V=\pi\int_{-2}^{2}(5-(x^2+1))^2\,dx$
4. $V=\pi\int_{-2}^{2}\big((x^2+1)^2-5^2\big)\,dx$
5. $V=\pi\int_{1}^{5}y^2\,dy$

Explanation: This problem involves the disc method for volumes of revolution around a shifted axis, specifically around y=0. Although y=0 is the standard x-axis, the setup treats it as a base case for shifts. The outer radius is 5, distance to y=5, and inner is x^2 +1, to y=x^2 +1. The volume integral is π times (5^2 - (x^2 +1)^2) from -2 to 2. A tempting distractor is choice C, which uses an incorrect outer radius by shifting wrongly. A transferable axis-shift strategy is to subtract the axis value from the curve functions to find the effective radii.

Question 4

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

Explanation: This problem involves the disc method with revolution around y = -2, below the region. The region R is bounded by y = 1 - x, y = 0, and x = 0, forming a triangle. When revolving around y = -2, the outer radius extends from y = -2 up to y = 1 - x: R_outer = (1-x) - (-2) = (1-x) + 2 = 3 - x, and the inner radius extends from y = -2 up to y = 0: R_inner = 0 - (-2) = 2. The volume is V = π∫₀¹[(3-x)² - 2²]dx = π∫₀¹[((1-x)+2)² - 2²]dx. Choice B incorrectly reverses the order of subtraction in the washer formula. When the axis is below the region, use (curve - axis) for distances and apply R_outer² - R_inner² in the integral.

Question 5

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

Explanation: This problem requires the disc method with a horizontal axis of revolution at y = 5, which is above the region. The region R is bounded by the parabola y = x² and the horizontal line y = 4, with intersection points at x = -2 and x = 2. When revolving around y = 5, the outer radius from the axis to the bottom boundary is (5 - x²), while the inner radius from the axis to the top boundary is (5 - 4) = 1. The washer method gives V = π∫[-2 to 2][(5 - x²)² - 1²]dx. Choice A incorrectly uses only the outer radius, treating this as a solid disc rather than a washer with a hollow center. The key strategy when revolving around a shifted axis is to measure all distances from that new axis, then apply outer² - inner² for washers.

Question 6

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

Explanation: This problem uses the disc method with revolution around y = -3, below the region. The region R is bounded by y = x³, y = 0, and x = 2. When revolving around y = -3, the outer radius extends from y = -3 up to y = x³: R_outer = x³ - (-3) = x³ + 3, and the inner radius extends from y = -3 up to y = 0: R_inner = 0 - (-3) = 3. The volume is V = π∫₀²[(x³ + 3)² - 3²]dx. Choice E incorrectly omits the inner radius term 3², treating the solid as having no hole. When revolving around a line below the region, use (curve - axis) for distances and remember that the x-axis boundary creates an inner hole.

Question 7

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

Explanation: This problem uses the disc method with revolution around the x-axis (y = 0), which is below the trapezoidal region. The region R is bounded by the line y = 3 - x and y = 1 for 0 ≤ x ≤ 2. When revolving around y = 0, the outer radius from the axis up to y = 3 - x is (3 - x), while the inner radius from the axis up to y = 1 is 1. Using the washer method, V = π∫[0 to 2][(3 - x)² - 1²]dx. Choice C incorrectly computes 1² - (3 - x)², reversing the subtraction order, which would give negative values since 1 < 3 - x for x in [0, 2]. The fundamental principle is that outer radius squared minus inner radius squared gives the area of the washer cross-section.

Question 8

1. $V=\pi\displaystyle\int_{-2}^{2}(4-x^2)^2\,dx$
2. $V=\pi\displaystyle\int_{-2}^{2}(x^2-1)^2\,dx$
3. $V=\pi\displaystyle\int_{-2}^{2}\big[(4-1)^2-(x^2-1)^2\big]dx$ (correct answer)
4. $V=\pi\displaystyle\int_{1}^{4}(y-1)^2\,dy$
5. $V=\pi\displaystyle\int_{-2}^{2}\big[(x^2-1)^2-(4-1)^2\big]dx$

Explanation: This problem requires the disc method with a horizontal axis of revolution at y = 1, which is below the given region. When revolving about y = 1, the radius at each x-value extends from the axis y = 1 to the boundary curves, giving outer radius (4 - 1) = 3 from the axis to y = 4, and inner radius (x² - 1) from the axis to y = x². Since we have a washer with outer radius 3 and inner radius (x² - 1), the volume integral is π∫[(3)² - (x² - 1)²]dx = π∫[(4 - 1)² - (x² - 1)²]dx from x = -2 to x = 2. Choice E incorrectly reverses the order of subtraction, which would give negative volume since the outer radius must be larger. When revolving about a horizontal line y = k, always measure radii as |curve - k| and ensure the outer radius is subtracted by the inner radius.

Question 9

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

Explanation: This problem involves the disc method for volumes of revolution around a shifted axis, specifically around y=2. When the axis is shifted to y=2, the radii change to reflect distances from this line to the region boundaries. The outer radius is 2 - 0 =2, the distance to y=0, and the inner radius is 2 - √x, the distance to y=√x. Thus, the volume is π times the integral of (2^2 - (2 - √x)^2) from 0 to 4. A tempting distractor is choice B, which uses only the inner radius, ignoring the washer hole structure. A transferable axis-shift strategy is to subtract the axis value from the curve functions to find the effective radii.

Question 10

1. $V=\pi\int_{0}^{\pi}(\sin x)^2\,dx$
2. $V=\pi\int_{0}^{\pi}(1-\sin x)^2\,dx$
3. $V=\pi\int_{0}^{\pi}(1+\sin x)^2\,dx$
4. $V=\pi\int_{0}^{\pi}\big((1+\sin x)^2-1^2\big)\,dx$ (correct answer)
5. $V=\pi\int_{0}^{1}(\arcsin y)^2\,dy$

Explanation: This problem involves the disc method for volumes of revolution around a shifted axis, specifically around y=-1. When the axis is shifted to y=-1, the radii are adjusted to distances from this line. The outer radius is sin x +1, the distance to y=sin x, and the inner radius is 1, the distance to y=0. The volume integral is π times the integral of ((sin x +1)^2 - 1^2) from 0 to π. A tempting distractor is choice B, which subtracts instead of adding the shift, leading to incorrect radii. A transferable axis-shift strategy is to add or subtract the shift value to the curve functions depending on the direction.

Question 11

1. $V=\pi\int_{0}^{9}\big(\sqrt{x}\big)^2\,dx$
2. $V=\pi\int_{0}^{9}\big(2-\sqrt{x}\big)^2\,dx$
3. $V=\pi\int_{0}^{9}\Big(\big(2-0\big)^2-\big(2-\sqrt{x}\big)^2\Big)\,dx$ (correct answer)
4. $V=\pi\int_{0}^{3}\Big(\big(2-y^2\big)^2-\big(2-0\big)^2\Big)\,dy$
5. $V=\pi\int_{0}^{9}\Big(\big(\sqrt{x}-2\big)^2-\big(0-2\big)^2\Big)\,dx$

Explanation: This problem involves the disc method with revolution around y = 2, which is above the region. The region R is bounded by y = √x (top curve), y = 0 (bottom curve), and x = 9. When revolving around y = 2, the outer radius is from y = 2 down to y = 0: R_outer = 2 - 0 = 2, and the inner radius is from y = 2 down to y = √x: R_inner = 2 - √x. The volume formula becomes V = π∫₀⁹[(2)² - (2-√x)²]dx. Choice B incorrectly uses only the inner radius, forgetting about the washer shape. When the axis of revolution is outside the region, identify which curve is closer to create the inner hole and which is farther to create the outer disc.

Question 12

1. $V=\pi\int_{0}^{\pi}\big(\sin x\big)^2\,dx$
2. $V=\pi\int_{0}^{\pi}\Big(\big(\sin x+1\big)^2-1^2\Big)\,dx$ (correct answer)
3. $V=\pi\int_{-1}^{0}\big(\pi\big)^2\,dy$
4. $V=\pi\int_{0}^{\pi}\Big(1^2-\big(\sin x+1\big)^2\Big)\,dx$
5. $V=\pi\int_{0}^{\pi}\big(\sin x+1\big)^2\,dx$

Explanation: This problem uses the disc method with revolution around y = -1, below the region. The region R is bounded by y = sin x and y = 0 on [0, π]. When revolving around y = -1, the outer radius extends from y = -1 up to y = sin x: R_outer = sin x - (-1) = sin x + 1, and the inner radius extends from y = -1 up to y = 0: R_inner = 0 - (-1) = 1. The volume is V = π∫₀^π[(sin x + 1)² - 1²]dx. Choice E incorrectly omits the inner radius term, treating it as a solid disc rather than a washer. For revolution around a line below the region, use (curve - axis) for distances and remember that both boundaries create radii.

Question 13

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

Explanation: This problem involves the disc method for finding volumes of revolution around a shifted axis, specifically around y = -1 instead of the x-axis. When the axis is shifted down to y = -1, the radius of each disc is the distance from the curve y = ln(x+1) to the axis, which is ln(x+1) - (-1) = ln(x+1) + 1. This addition reflects the downward shift, making the radius larger by 1. The integrand is π(ln(x+1) + 1)², integrated from 0 to e-1. A tempting distractor is choice B, π ∫ (ln(x+1) - 1)² dx, which would apply to an upward shift to y = 1 but not here. A transferable strategy for axis shifts in the disc method is to calculate the radius as the absolute difference between the function value and the axis position, ordering it to ensure a positive value.

Question 14

1. $V=\pi\int_{0}^{1}(2-x^3)^2\,dx$
2. $V=\pi\int_{0}^{1}\big(2^2-(2-x^3)^2\big)\,dx$ (correct answer)
3. $V=\pi\int_{0}^{1}(x^3)^2\,dx$
4. $V=\pi\int_{0}^{1}(2+x^3)^2\,dx$
5. $V=\pi\int_{0}^{2}(1-y^{1/3})^2\,dy$

Explanation: This problem involves the disc method for volumes of revolution around a shifted axis, specifically around y=2. With the axis at y=2, radii are distances from this line to the boundaries. The outer radius is 2 - 0 =2, to y=0, and inner is 2 - x^3, to y=x^3. The volume is π times (2^2 - (2 - x^3)^2) from 0 to 1. A tempting distractor is choice A, which uses only the inner radius without the washer setup. A transferable axis-shift strategy is to subtract the axis value from the curve functions to find the effective radii.

Question 15

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

Explanation: This problem involves the disc method for finding volumes of revolution around a shifted axis, specifically around y = 3 instead of the x-axis. When the axis is shifted up to y = 3, the radius of each disc is the distance from the curve y = e^x to the axis, which is 3 - e^x. This subtraction is used because the axis is above the curve, ensuring a positive radius since e^x < 3 in [0,1]. The integrand becomes π(3 - e^x)², integrated from 0 to 1. A tempting distractor is choice D, π ∫ (3 + e^x)² dx, which might arise from incorrectly adding for an upward shift instead of subtracting. A transferable strategy for axis shifts in the disc method is to calculate the radius as the absolute difference between the function value and the axis position, ordering it to ensure a positive value.

Question 16

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

Explanation: This problem involves the disc method with revolution around y = -2, below the region. The region R is bounded by y = 1 - x, y = 0, and x = 0, forming a triangle. When revolving around y = -2, the outer radius extends from y = -2 up to y = 1 - x: R_outer = (1-x) - (-2) = (1-x) + 2 = 3 - x, and the inner radius extends from y = -2 up to y = 0: R_inner = 0 - (-2) = 2. The volume is V = π∫₀¹[(3-x)² - 2²]dx = π∫₀¹[((1-x)+2)² - 2²]dx. Choice B incorrectly reverses the order of subtraction in the washer formula. When the axis is below the region, use (curve - axis) for distances and apply R_outer² - R_inner² in the integral.

Question 17

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

Explanation: This problem involves the disc method for finding volumes of revolution around a shifted axis, specifically around y = -1 instead of the x-axis. When the axis is shifted down to y = -1, the radius of each disc is the distance from the curve y = x² to the axis, which is x² - (-1) = x² + 1. This adjustment adds the shift amount because the axis is below the region, increasing the effective radius. Consequently, the integrand becomes π(x² + 1)², integrated from 0 to 2. A tempting distractor is choice C, π ∫ (x² - 1)² dx, which might result from incorrectly subtracting instead of adding for a downward shift, but that would apply to an upward shift to y = 1. A transferable strategy for axis shifts in the disc method is to calculate the radius as the absolute difference between the function value and the axis position, ordering it to ensure a positive value.

Question 18

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

Explanation: This problem involves the disc method for finding volumes of revolution around a shifted axis, specifically around $y = 1$ instead of the x-axis. When the axis is shifted up to $y = 1$, the radius of each disc is the distance from the curve $y = \cos x$ to the axis, which is $1 - \cos x$. This subtraction is appropriate because the axis is above or at the curve, yielding a non-negative radius. The integrand becomes $\pi (1 - \cos x)^2$, integrated from $0$ to $\pi/2$. A tempting distractor is choice A, $\pi \int (\cos x - 1)^2 \, dx$, which is the same mathematically but may suggest a negative radius before squaring, potentially confusing the shift direction. A transferable strategy for axis shifts in the disc method is to calculate the radius as the absolute difference between the function value and the axis position, ordering it to ensure a positive value.

Question 19

1. $V=\pi\int_{0}^{\pi}\big(\sin x\big)^2\,dx$
2. $V=\pi\int_{0}^{\pi}\Big(\big(\sin x+1\big)^2-1^2\Big)\,dx$ (correct answer)
3. $V=\pi\int_{-1}^{0}\big(\pi\big)^2\,dy$
4. $V=\pi\int_{0}^{\pi}\Big(1^2-\big(\sin x+1\big)^2\Big)\,dx$
5. $V=\pi\int_{0}^{\pi}\big(\sin x+1\big)^2\,dx$

Explanation: This problem uses the disc method with revolution around y = -1, below the region. The region R is bounded by y = sin x and y = 0 on [0, π]. When revolving around y = -1, the outer radius extends from y = -1 up to y = sin x: R_outer = sin x - (-1) = sin x + 1, and the inner radius extends from y = -1 up to y = 0: R_inner = 0 - (-1) = 1. The volume is V = π∫₀^π[(sin x + 1)² - 1²]dx. Choice E incorrectly omits the inner radius term, treating it as a solid disc rather than a washer. For revolution around a line below the region, use (curve - axis) for distances and remember that both boundaries create radii.

Question 20

1. $V=\pi\int_{0}^{9}\big(\sqrt{x}\big)^2\,dx$
2. $V=\pi\int_{0}^{9}\big(2-\sqrt{x}\big)^2\,dx$
3. $V=\pi\int_{0}^{9}\Big(\big(2-0\big)^2-\big(2-\sqrt{x}\big)^2\Big)\,dx$ (correct answer)
4. $V=\pi\int_{0}^{3}\Big(\big(2-y^2\big)^2-\big(2-0\big)^2\Big)\,dy$
5. $V=\pi\int_{0}^{9}\Big(\big(\sqrt{x}-2\big)^2-\big(0-2\big)^2\Big)\,dx$

Explanation: This problem involves the disc method with revolution around y = 2, which is above the region. The region R is bounded by y = √x (top curve), y = 0 (bottom curve), and x = 9. When revolving around y = 2, the outer radius is from y = 2 down to y = 0: R_outer = 2 - 0 = 2, and the inner radius is from y = 2 down to y = √x: R_inner = 2 - √x. The volume formula becomes V = π∫₀⁹[(2)² - (2-√x)²]dx. Choice B incorrectly uses only the inner radius, forgetting about the washer shape. When the axis of revolution is outside the region, identify which curve is closer to create the inner hole and which is farther to create the outer disc.