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

Explanation: This problem involves the disc method for finding volumes of solids of revolution when revolving around a shifted axis, specifically y=-1. The standard disc method uses the radius as the function value when revolving around y=0, but here the axis is shifted down to y=-1. Since the function y = ln(x+1) is above y=-1, the radius becomes ln(x+1) + 1 to reflect the distance. This addition accounts for the negative shift of the axis. A tempting distractor is choice B, which uses (-1 - ln(x+1))^2, but while equivalent when squared, it may confuse the distance concept. To generalize for revolutions around y = k, always compute the radius as |f(x) - k| and square it in the integral setup.

Question 2

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

Explanation: This problem involves using the disc method to find the volume of a solid of revolution when revolving around a horizontal axis shifted from the x-axis, specifically y=1. The radius of each disc is the vertical distance from the curve y=x+2 to the axis y=1, given by |(x+2) - 1| = |x+1|. Since the curve is above y=1 throughout [1,4], x+1 is positive, matching (x+2) - 1. Therefore, the volume is correctly set up as π ∫ from 1 to 4 of ((x+2) - 1)² dx. The distractor choice A fails because it uses (x+2)², which would be correct for rotation around y=0, but ignores the shift to y=1. A general strategy for shifting the axis to y=k is to use the radius |k - f(x)| in the disc method integral, or equivalently (k - f(x))² to handle the absolute value through squaring.

Question 3

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

Explanation: This problem involves the disc method with rotation about a horizontal line above the curve. When revolving the region under y = sin x from x = 0 to x = π about y = 2, we need the distance from the curve to the line y = 2. Since sin x ≤ 1 < 2 on [0,π], the radius is 2 - sin x, giving us π∫₀^π(2 - sin x)² dx. Option D incorrectly uses (sin x - 2)², which would give negative values inside the square. For rotation about a line above the curve, use (axis value - function value) to ensure positive radii.

Question 4

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

Explanation: This problem requires the disc method with rotation about a horizontal line below the parabola. When revolving the region under y = 1 - x² from x = -1 to x = 1 about y = -3, the radius is the distance from the curve to y = -3. Since 1 - x² ≥ 0 on [-1,1] and we're rotating about y = -3, the radius is (1 - x²) - (-3) = (1 - x²) + 3, giving us π∫₋₁¹((1 - x²) + 3)² dx. Option B incorrectly computes -3 - (1 - x²), which misunderstands the direction of measurement. When rotating about y = -c, add c to the function value: radius = f(x) + c.

Question 5

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

Explanation: This problem involves the disc method for finding volumes of solids of revolution when revolving around a shifted axis, specifically y=1. The standard disc method uses the radius as the function value when revolving around y=0, but here the axis is shifted to y=1. Since the function y = cos x + 2 is above y=1, the radius becomes (cos x + 2) - 1 to reflect the distance. This subtraction ensures the correct perpendicular distance for each disc. A tempting distractor is choice A, which uses (cos x + 2)^2, but this ignores the shift, as if revolving around y=0. To generalize for revolutions around y = k, always compute the radius as |f(x) - k| and square it in the integral setup.

Question 6

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

Explanation: This problem involves using the disc method to find the volume of a solid of revolution when revolving around a horizontal axis shifted from the x-axis, specifically y=2. The radius of each disc is the vertical distance from the curve y=1-x to the axis y=2, given by 2 - (1-x) since 1-x ≤ 1 < 2 throughout [0,1]. This expression is always positive in the interval, directly providing the radius. Therefore, the volume is correctly set up as π ∫ from 0 to 1 of (2 - (1-x))² dx. The distractor choice A fails because it uses (1-x)², which would be correct for rotation around y=0, but ignores the shift to y=2. A general strategy for shifting the axis to y=k is to use the radius |k - f(x)| in the disc method integral, or equivalently (k - f(x))² to handle the absolute value through squaring.

Question 7

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

Explanation: This problem involves using the disc method to find the volume of a solid of revolution when revolving around a horizontal axis shifted from the x-axis, specifically y=0. The radius of each disc is the vertical distance from the curve y=e^x to the axis y=0, given by e^x - 0 = e^x. Since the axis is at y=0 and the curve is above it throughout [0,1], this expression is positive and directly gives the radius. Therefore, the volume is correctly set up as π ∫ from 0 to 1 of (e^x)² dx. The distractor choice D fails because it uses (1 - e^x)², which would be appropriate if revolving around y=1 above the curve, but not for y=0. A general strategy for shifting the axis to y=k is to use the radius |k - f(x)| in the disc method integral, or equivalently (k - f(x))² to handle the absolute value through squaring.

Question 8

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

Explanation: This problem involves the disc method for finding volumes of solids of revolution when revolving around a shifted axis, here the x-axis at y=0. The standard disc method uses the radius as the function value y = x^2 + 1 when revolving around y=0, with no shift adjustment needed. Since the axis is at y=0 and the function is above it, the radius remains x^2 + 1 without modification. No subtraction or addition is required because the shift is zero. A tempting distractor is choice D, which uses (1 - x^2)^2, but this incorrectly subtracts as if revolving around y=1, leading to a wrong radius. To generalize for revolutions around y = k, always compute the radius as |f(x) - k| and square it in the integral setup.

Question 9

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

Explanation: This problem involves using the disc method to find the volume of a solid of revolution when revolving around a horizontal axis shifted from the x-axis, specifically y=0. The radius of each disc is the vertical distance from the curve y=5-2x to the axis y=0, given by 5-2x - 0 = 5-2x. Since the axis is at y=0 and the curve is above it throughout [0,2], this expression is positive and directly gives the radius. Therefore, the volume is correctly set up as π ∫ from 0 to 2 of (5-2x)² dx. The distractor choice D fails because it uses (2x-5)², which is mathematically equivalent but may confuse the sign without context. A general strategy for shifting the axis to y=k is to use the radius |k - f(x)| in the disc method integral, or equivalently (k - f(x))² to handle the absolute value through squaring.

Question 10

1. $V=\pi\int_{-1}^{1}(4-x^2)^2\,dx$
2. $V=\pi\int_{-1}^{1}(6-(4-x^2))^2\,dx$ (correct answer)
3. $V=\pi\int_{-1}^{1}((4-x^2)-6)^2\,dx$
4. $V=\pi\int_{-1}^{1}(6+4-x^2)^2\,dx$
5. $V=\pi\int_{0}^{6}(6-y)^2\,dy$

Explanation: This problem involves using the disc method to find the volume of a solid of revolution when revolving around a horizontal axis shifted from the x-axis, specifically y=6. The radius of each disc is the vertical distance from the curve y=4-x² to the axis y=6, given by 6 - (4-x²) since 4-x² ≤ 3 < 6 throughout [-1,1]. This expression is always positive in the interval, directly providing the radius. Therefore, the volume is correctly set up as π ∫ from -1 to 1 of (6 - (4-x²))² dx. The distractor choice A fails because it uses (4-x²)², which would be correct for rotation around y=0, but ignores the shift to y=6. A general strategy for shifting the axis to y=k is to use the radius |k - f(x)| in the disc method integral, or equivalently (k - f(x))² to handle the absolute value through squaring.

Question 11

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

Explanation: This problem involves the disc method for finding volumes of solids of revolution when revolving around a shifted axis, specifically y=2. The standard disc method uses the radius as the function value when revolving around y=0, but here the axis is shifted up to y=2. Since the function y = √(x+1) is below y=2, the radius becomes 2 - √(x+1) to account for the distance. This reversal in subtraction reflects that the axis is above the curve. A tempting distractor is choice D, which uses (2 + √(x+1))^2, but this adds instead of subtracting, overstating the radius and volume. To generalize for revolutions around y = k, always compute the radius as |f(x) - k| and square it in the integral setup.

Question 12

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

Explanation: This problem involves the disc method for finding volumes of solids of revolution when revolving around a shifted axis, specifically y=4. The standard disc method uses the radius as the function value when revolving around y=0, but here the axis is shifted up to y=4. Since the function y = (3/2)x is below y=4, the radius becomes 4 - (3/2)x to account for the distance. This reversal in subtraction reflects that the axis is above the curve. A tempting distractor is choice D, which uses (4 + (3/2)x)^2, but this adds instead of subtracting, leading to an overstated volume. To generalize for revolutions around y = k, always compute the radius as |f(x) - k| and square it in the integral setup.

Question 13

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

Explanation: This problem involves the disc method for finding volumes of solids of revolution when revolving around a shifted axis, here the x-axis at y=0. The standard disc method uses the radius as the function value y = sin x + 1 when revolving around y=0, with no shift adjustment needed. Since the axis is at y=0 and the function is above it, the radius remains sin x + 1 without modification. No subtraction or addition is required because the shift is zero. A tempting distractor is choice D, which uses (sin x - 1)^2, but this subtracts incorrectly as if shifting the axis up, understating the radius. To generalize for revolutions around y = k, always compute the radius as |f(x) - k| and square it in the integral setup.

Question 14

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

Explanation: This problem involves using the disc method to find the volume of a solid of revolution when revolving around a horizontal axis shifted from the x-axis, specifically y=2. The radius of each disc is the vertical distance from the curve y=2/(x+1) to the axis y=2, given by |2 - 2/(x+1)|. Since the curve is below y=2 throughout [0,3], 2 - 2/(x+1) is positive and directly gives the radius. Therefore, the volume is correctly set up as π ∫ from 0 to 3 of (2 - 2/(x+1))² dx. The distractor choice A fails because it uses (2/(x+1))², which would be correct for rotation around y=0, but ignores the shift to y=2. A general strategy for shifting the axis to y=k is to use the radius |k - f(x)| in the disc method integral, or equivalently (k - f(x))² to handle the absolute value through squaring.

Question 15

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

Explanation: This problem involves the disc method for finding volumes of solids of revolution when revolving around a shifted axis, specifically y=-2. The standard disc method uses the radius as the function value when revolving around y=0, but here the axis is shifted down to y=-2. Since the function y = 1 - x/3 is above y=-2, the radius becomes (1 - x/3) + 2 to reflect the distance. This addition accounts for the negative shift of the axis. A tempting distractor is choice B, which uses (-2 - (1 - x/3))^2, but while mathematically equivalent when squared, it may confuse the conceptual distance. To generalize for revolutions around y = k, always compute the radius as |f(x) - k| and square it in the integral setup.

Question 16

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

Explanation: This problem involves using the disc method to find the volume of a solid of revolution when revolving around a horizontal axis shifted from the x-axis, specifically y=-1. The radius of each disc is the vertical distance from the curve y=cos x to the axis y=-1, given by cos x - (-1) = cos x + 1. Since the axis is below the curve throughout [0,π/2], this expression is always positive and directly gives the radius. Therefore, the volume is correctly set up as π ∫ from 0 to π/2 of (cos x + 1)² dx. The distractor choice A fails because it uses (cos x)², which would be correct for rotation around y=0, but ignores the shift to y=-1. A general strategy for shifting the axis to y=k is to use the radius |k - f(x)| in the disc method integral, or equivalently (k - f(x))² to handle the absolute value through squaring.

Question 17

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

Explanation: This problem involves the disc method for finding volumes of solids of revolution when revolving around a shifted axis, specifically y=-1. The standard disc method uses the radius as the function value when revolving around y=0, but here the axis is shifted down to y=-1. Since the function y = 2 ln x is above y=-1, the radius becomes 2 ln x + 1 to reflect the distance. This addition accounts for the negative shift of the axis. A tempting distractor is choice B, which uses (-1 - 2 ln x)^2, but while equivalent when squared, it may obscure the positive radius. To generalize for revolutions around y = k, always compute the radius as |f(x) - k| and square it in the integral setup.

Question 18

1. $V=\pi\int_{0}^{6}\left(\frac{x}{2}\right)^2\,dx$
2. $V=\pi\int_{0}^{6}\left(4-\frac{x}{2}\right)^2\,dx$ (correct answer)
3. $V=\pi\int_{0}^{6}\left(\frac{x}{2}-4\right)^2\,dx$
4. $V=\pi\int_{0}^{4}(4-y)^2\,dy$
5. $V=\pi\int_{0}^{6}\left(4+\frac{x}{2}\right)^2\,dx$

Explanation: This problem involves using the disc method to find the volume of a solid of revolution when revolving around a horizontal axis shifted from the x-axis, specifically y=4. The radius of each disc is the vertical distance from the curve y=x/2 to the axis y=4, given by 4 - x/2 since x/2 ≤ 3 < 4 throughout [0,6]. This expression is always positive in the interval, directly providing the radius. Therefore, the volume is correctly set up as π ∫ from 0 to 6 of (4 - x/2)² dx. The distractor choice A fails because it uses (x/2)², which would be correct for rotation around y=0, but ignores the shift to y=4. A general strategy for shifting the axis to y=k is to use the radius |k - f(x)| in the disc method integral, or equivalently (k - f(x))² to handle the absolute value through squaring.

Question 19

1. $V=\pi\int_{0}^{2}\left(\frac{x^3}{8}\right)^2\,dx$
2. $V=\pi\int_{0}^{2}\left(1-\frac{x^3}{8}\right)^2\,dx$ (correct answer)
3. $V=\pi\int_{0}^{2}\left(\frac{x^3}{8}-1\right)^2\,dx$
4. $V=\pi\int_{0}^{2}\left(1+\frac{x^3}{8}\right)^2\,dx$
5. $V=\pi\int_{0}^{1}(1-y)^2\,dy$

Explanation: This problem involves the disc method for finding volumes of solids of revolution when revolving around a shifted axis, specifically y=1. The standard disc method uses the radius as the function value when revolving around y=0, but here the axis is shifted to y=1. Since the function y = x^3 / 8 is below y=1, the radius becomes 1 - x^3 / 8 to account for the distance. This reversal in subtraction reflects that the axis is above the curve. A tempting distractor is choice D, which uses (1 + x^3 / 8)^2, but this adds instead of subtracting, overstating the radius. To generalize for revolutions around y = k, always compute the radius as |f(x) - k| and square it in the integral setup.

Question 20

1. $V=\pi\int_{0}^{2}(3x^2)^2\,dx$
2. $V=\pi\int_{0}^{2}(3x^2-2)^2\,dx$
3. $V=\pi\int_{0}^{2}(-2-3x^2)^2\,dx$
4. $V=\pi\int_{0}^{2}(3x^2+2)^2\,dx$ (correct answer)
5. $V=\pi\int_{-2}^{12}(y+2)^2\,dy$

Explanation: This problem involves using the disc method to find the volume of a solid of revolution when revolving around a horizontal axis shifted from the x-axis, specifically y=-2. The radius of each disc is the vertical distance from the curve y=3x² to the axis y=-2, given by 3x² - (-2) = 3x² + 2. Since the axis is below the curve throughout [0,2], this expression is always positive and directly gives the radius. Therefore, the volume is correctly set up as π ∫ from 0 to 2 of (3x² + 2)² dx. The distractor choice A fails because it uses (3x²)², which would be correct for rotation around y=0, but ignores the shift to y=-2. A general strategy for shifting the axis to y=k is to use the radius |k - f(x)| in the disc method integral, or equivalently (k - f(x))² to handle the absolute value through squaring.