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AP Calculus BC Quiz

AP Calculus BC Quiz: Washer Method Revolving Around Other Axes

Practice Washer Method Revolving Around Other Axes in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

What integral gives the volume when the region between y=xy=\sqrt{x}y=x​ and y=0y=0y=0, 0≤x≤40\le x\le40≤x≤4, is revolved about y=−2y=-2y=−2?

Select an answer to continue

What this quiz covers

This quiz focuses on Washer 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.

All questions

Question 1

What integral gives the volume when the region between y=xy=\sqrt{x}y=x​ and y=0y=0y=0, 0≤x≤40\le x\le40≤x≤4, is revolved about y=−2y=-2y=−2?

  1. π∫04[(x+2)2−(2)2] dx\pi\int_{0}^{4}\big[(\sqrt{x}+2)^2-(2)^2\big]\,dxπ∫04​[(x​+2)2−(2)2]dx (correct answer)
  2. π∫04[(2)2−(x+2)2] dx\pi\int_{0}^{4}\big[(2)^2-(\sqrt{x}+2)^2\big]\,dxπ∫04​[(2)2−(x​+2)2]dx
  3. π∫04(x)2 dx\pi\int_{0}^{4}(\sqrt{x})^2\,dxπ∫04​(x​)2dx
  4. π∫04[(x−2)2−(0−2)2] dx\pi\int_{0}^{4}\big[(\sqrt{x}-2)^2-(0-2)^2\big]\,dxπ∫04​[(x​−2)2−(0−2)2]dx
  5. π∫04[(x+2)−2]2 dx\pi\int_{0}^{4}\big[(\sqrt{x}+2)-2\big]^2\,dxπ∫04​[(x​+2)−2]2dx

Explanation: This problem involves using the washer method to find the volume of the solid formed by revolving the region around the shifted axis y = -2. To adjust for this axis, the radii are the distances from the curves to y = -2, which means adding 2 to each y-value. The outer radius is from the upper curve y = √x to the axis, giving √x + 2. The inner radius is from the lower curve y = 0 to the axis, giving 2. A tempting distractor is choice B, which reverses the outer and inner radii, resulting in a negative integrand and incorrect volume. In general, when revolving around a horizontal axis y = k, the radius for a curve y = f(x) is |f(x) - k|, and ensure the outer radius is the larger one in the integral.

Question 2

Which integral gives the volume when the region between y=1−x2y=1-x^2y=1−x2 and y=0y=0y=0 on [−1,1][-1,1][−1,1] rotates about y=−3y=-3y=−3?

  1. V=π∫−11[(1−x2+3)2−(0+3)2]dxV=\pi\displaystyle\int_{-1}^{1}\Big[(1-x^2+3)^2-(0+3)^2\Big]dxV=π∫−11​[(1−x2+3)2−(0+3)2]dx (correct answer)
  2. V=π∫−11[(0+3)2−(1−x2+3)2]dxV=\pi\displaystyle\int_{-1}^{1}\Big[(0+3)^2-(1-x^2+3)^2\Big]dxV=π∫−11​[(0+3)2−(1−x2+3)2]dx
  3. V=π∫−11[(1−x2)2−02]dxV=\pi\displaystyle\int_{-1}^{1}\Big[(1-x^2)^2-0^2\Big]dxV=π∫−11​[(1−x2)2−02]dx
  4. V=π∫−11[(1−x2−3)2−(0−3)2]dxV=\pi\displaystyle\int_{-1}^{1}\Big[(1-x^2-3)^2-(0-3)^2\Big]dxV=π∫−11​[(1−x2−3)2−(0−3)2]dx
  5. V=π∫01[(3)2−(3−y)2]dyV=\pi\displaystyle\int_{0}^{1}\Big[(3)^2-(3-y)^2\Big]dyV=π∫01​[(3)2−(3−y)2]dy

Explanation: This problem uses the washer method with rotation about y = -3. When rotating about y = -3 (below the region), we measure distances upward: the outer radius is from y = -3 to y = 1-x², giving R = (1-x²) - (-3) = 1-x² + 3 = 4-x², and the inner radius is from y = -3 to y = 0, giving r = 0 - (-3) = 3. The volume integral is π∫[-1,1][(1-x²+3)² - (0+3)²]dx = π∫[-1,1][(4-x²)² - 9]dx. Choice D incorrectly subtracts 3 instead of adding it when adjusting for the negative axis value. When the axis of rotation has a negative y-value, add its absolute value to all y-coordinates.

Question 3

Which integral represents the volume when the region between y=xy=\sqrt{x}y=x​ and y=0y=0y=0 on [0,4][0,4][0,4] rotates about y=−2y=-2y=−2?

  1. V=π∫04[(x)2−02]dxV=\pi\displaystyle\int_{0}^{4}\Big[(\sqrt{x})^2-0^2\Big]dxV=π∫04​[(x​)2−02]dx
  2. V=π∫04[(x−2)2−(0−2)2]dxV=\pi\displaystyle\int_{0}^{4}\Big[(\sqrt{x}-2)^2-(0-2)^2\Big]dxV=π∫04​[(x​−2)2−(0−2)2]dx
  3. V=π∫04[(x+2)2−(0+2)2]dxV=\pi\displaystyle\int_{0}^{4}\Big[(\sqrt{x}+2)^2-(0+2)^2\Big]dxV=π∫04​[(x​+2)2−(0+2)2]dx (correct answer)
  4. V=π∫04[(0+2)2−(x+2)2]dxV=\pi\displaystyle\int_{0}^{4}\Big[(0+2)^2-(\sqrt{x}+2)^2\Big]dxV=π∫04​[(0+2)2−(x​+2)2]dx
  5. V=π∫02[(4−(−2))2−(y2−(−2))2]dyV=\pi\displaystyle\int_{0}^{2}\Big[(4-(-2))^2-(y^2-(-2))^2\Big]dyV=π∫02​[(4−(−2))2−(y2−(−2))2]dy

Explanation: This problem uses the washer method with rotation about the line y=−2y = -2y=−2. When rotating about y=−2y = -2y=−2, we calculate distances from this axis: the outer radius extends from y=−2y = -2y=−2 to y=xy = \sqrt{x}y=x​, giving R=x−(−2)=x+2R = \sqrt{x} - (-2) = \sqrt{x} + 2R=x​−(−2)=x​+2, and the inner radius extends from y=−2y = -2y=−2 to y=0y = 0y=0, giving r=0−(−2)=2r = 0 - (-2) = 2r=0−(−2)=2. The volume integral is π∫04[(x+2)2−(0+2)2] dx=π∫04[(x+2)2−4] dx\pi\int_{0}^{4}[(\sqrt{x} + 2)^2 - (0 + 2)^2]\, dx = \pi\int_{0}^{4}[(\sqrt{x} + 2)^2 - 4]\, dxπ∫04​[(x​+2)2−(0+2)2]dx=π∫04​[(x​+2)2−4]dx. Choice B incorrectly subtracts 2 instead of adding it when adjusting for the axis below the region. Remember that when the axis is below the region, you add the absolute value of the axis position to find radii.

Question 4

What integral gives the volume when the region between y=x2y=x^2y=x2 and y=2xy=2xy=2x on [0,2][0,2][0,2] is revolved about y=−1y=-1y=−1?

  1. π∫02[(2x+1)2−(x2+1)2] dx\pi\displaystyle\int_{0}^{2}\big[(2x+1)^2-(x^2+1)^2\big]\,dxπ∫02​[(2x+1)2−(x2+1)2]dx (correct answer)
  2. π∫02[(2x)2−(x2)2] dx\pi\displaystyle\int_{0}^{2}\big[(2x)^2-(x^2)^2\big]\,dxπ∫02​[(2x)2−(x2)2]dx
  3. π∫02[(x2+1)2−(2x+1)2] dx\pi\displaystyle\int_{0}^{2}\big[(x^2+1)^2-(2x+1)^2\big]\,dxπ∫02​[(x2+1)2−(2x+1)2]dx
  4. π∫02[(2x−1)2−(x2−1)2] dx\pi\displaystyle\int_{0}^{2}\big[(2x-1)^2-(x^2-1)^2\big]\,dxπ∫02​[(2x−1)2−(x2−1)2]dx
  5. π∫02[(2x+1)−(x2+1)]2 dx\pi\displaystyle\int_{0}^{2}\big[(2x+1)-(x^2+1)\big]^2\,dxπ∫02​[(2x+1)−(x2+1)]2dx

Explanation: This problem utilizes the washer method to compute the volume of a solid formed by revolving a region around the shifted axis y=−1y = -1y=−1. To adjust for this axis, add 1 to each y-value to get the distances, since the curves are above the axis. The outer radius is 2x+12x + 12x+1 from the upper curve, and the inner radius is x2+1x^2 + 1x2+1 from the lower curve. Therefore, the integral is π∫02[(2x+1)2−(x2+1)2] dx\pi \int_{0}^{2} [(2x + 1)^2 - (x^2 + 1)^2] \, dxπ∫02​[(2x+1)2−(x2+1)2]dx. A tempting distractor is choice C, which swaps the inner and outer radii, resulting in a negative integrand and incorrect volume. In general, when revolving around y=ky = ky=k, compute radii as ∣y−k∣|y - k|∣y−k∣ for each curve and identify the outer as the larger distance and inner as the smaller.

Question 5

Choose the correct volume integral for revolving the region between y=sin⁡xy=\sin xy=sinx and y=0y=0y=0 on [0,π][0,\pi][0,π] about y=2y=2y=2.

  1. V=π∫0π[(2−sin⁡x)2−(2−0)2]dxV=\pi\displaystyle\int_{0}^{\pi}\Big[(2-\sin x)^2-(2-0)^2\Big]dxV=π∫0π​[(2−sinx)2−(2−0)2]dx
  2. V=π∫0π[(2+sin⁡x)2−(2+0)2]dxV=\pi\displaystyle\int_{0}^{\pi}\Big[(2+\sin x)^2-(2+0)^2\Big]dxV=π∫0π​[(2+sinx)2−(2+0)2]dx
  3. V=π∫0π[(2−0)2−(2−sin⁡x)2]dxV=\pi\displaystyle\int_{0}^{\pi}\Big[(2-0)^2-(2-\sin x)^2\Big]dxV=π∫0π​[(2−0)2−(2−sinx)2]dx (correct answer)
  4. V=π∫0π[(sin⁡x)2−02]dxV=\pi\displaystyle\int_{0}^{\pi}\Big[(\sin x)^2-0^2\Big]dxV=π∫0π​[(sinx)2−02]dx
  5. V=π∫02[(arcsin⁡y)2−(0)2]dyV=\pi\displaystyle\int_{0}^{2}\Big[(\arcsin y)^2-(0)^2\Big]dyV=π∫02​[(arcsiny)2−(0)2]dy

Explanation: This problem requires the washer method with rotation about y = 2. Since y = 2 is above the region bounded by y = sin x and y = 0 on [0,π], we measure distances downward: the outer radius is from y = 2 to y = 0, giving R = 2 - 0 = 2, and the inner radius is from y = 2 to y = sin x, giving r = 2 - sin x. The volume integral is π∫[0,π][(2-0)² - (2-sin x)²]dx = π∫[0,π][4 - (2-sin x)²]dx. Choice A incorrectly orders the radii, placing the smaller radius as the outer one. Always ensure the outer radius is larger than the inner radius in the washer method.

Question 6

Choose the correct setup for revolving the region between y=x3y=x^3y=x3 and y=xy=xy=x on 0≤x≤10\le x\le10≤x≤1 about y=2y=2y=2.​​

  1. V=π∫01[(2−x)2−(2−x3)2]dxV=\pi\displaystyle\int_{0}^{1}\big[(2-x)^2-(2-x^3)^2\big]dxV=π∫01​[(2−x)2−(2−x3)2]dx (correct answer)
  2. V=π∫01[(2−x3)2−(2−x)2]dxV=\pi\displaystyle\int_{0}^{1}\big[(2-x^3)^2-(2-x)^2\big]dxV=π∫01​[(2−x3)2−(2−x)2]dx
  3. V=π∫01[(x−x3)2]dxV=\pi\displaystyle\int_{0}^{1}\big[(x-x^3)^2\big]dxV=π∫01​[(x−x3)2]dx
  4. V=π∫01[(x−2)2−(x3−2)2]dxV=\pi\displaystyle\int_{0}^{1}\big[(x-2)^2-(x^3-2)^2\big]dxV=π∫01​[(x−2)2−(x3−2)2]dx
  5. V=π∫01[(2−x)2−(2)2]dxV=\pi\displaystyle\int_{0}^{1}\big[(2-x)^2-(2)^2\big]dxV=π∫01​[(2−x)2−(2)2]dx

Explanation: This problem requires the washer method with revolution about y = 2, where the region is between y = x³ and y = x on [0,1]. Since x ≥ x³ on [0,1] and y = 2 is above both curves, the outer radius is R = 2 - x³ (from axis to the curve farther below) and the inner radius is r = 2 - x (from axis to the curve closer to axis). The volume setup is V = π∫[(2-x³)² - (2-x)²]dx from 0 to 1. Choice B incorrectly reverses these radii, but since x ≥ x³ on [0,1], we need 2-x³ ≥ 2-x, making (2-x³)² the outer radius term. To verify the radius order, check which original function has smaller values, as this determines which curve is farther from an axis above the region.

Question 7

Which integral gives the volume when the region between y=xy=\sqrt{x}y=x​ and y=0y=0y=0 on [0,4][0,4][0,4] is revolved about y=−2y=-2y=−2?

  1. V=π∫04[(x)2−(0)2]dxV=\pi\int_{0}^{4}\big[(\sqrt{x})^2-(0)^2\big]dxV=π∫04​[(x​)2−(0)2]dx
  2. V=π∫04[(x+2)2−(2)2]dxV=\pi\int_{0}^{4}\big[(\sqrt{x}+2)^2-(2)^2\big]dxV=π∫04​[(x​+2)2−(2)2]dx (correct answer)
  3. V=π∫04[(2)2−(x+2)2]dxV=\pi\int_{0}^{4}\big[(2)^2-(\sqrt{x}+2)^2\big]dxV=π∫04​[(2)2−(x​+2)2]dx
  4. V=π∫04[(x−2)2−(0−2)2]dxV=\pi\int_{0}^{4}\big[(\sqrt{x}-2)^2-(0-2)^2\big]dxV=π∫04​[(x​−2)2−(0−2)2]dx
  5. V=π∫04[(x+2)2−(0+2)2]dxV=\pi\int_{0}^{4}\big[(\sqrt{x}+2)^2-(0+2)^2\big]dxV=π∫04​[(x​+2)2−(0+2)2]dx

Explanation: This problem involves the washer method with revolution about y = -2, a horizontal line below the x-axis. For revolution about y = -2, the outer radius extends from y = -2 up to y = √x, giving R = √x - (-2) = √x + 2, and the inner radius extends from y = -2 up to y = 0, giving r = 0 - (-2) = 2. The washer formula becomes π∫[(√x + 2)² - (2)²]dx from 0 to 4. Choice D incorrectly subtracts 2 from the functions instead of adding 2, which would give negative radii. When the axis of revolution is below the region, add the absolute value of the axis to all y-values to find positive radii.

Question 8

What integral gives the volume when the region between y=ln⁡(x+1)y=\ln(x+1)y=ln(x+1) and y=0y=0y=0, 0≤x≤e−10\le x\le e-10≤x≤e−1, is revolved about y=−1y=-1y=−1?

  1. π∫0e−1[(ln⁡(x+1)+1)2−(1)2] dx\pi\int_{0}^{e-1}\big[(\ln(x+1)+1)^2-(1)^2\big]\,dxπ∫0e−1​[(ln(x+1)+1)2−(1)2]dx (correct answer)
  2. π∫0e−1[(1)2−(ln⁡(x+1)+1)2] dx\pi\int_{0}^{e-1}\big[(1)^2-(\ln(x+1)+1)^2\big]\,dxπ∫0e−1​[(1)2−(ln(x+1)+1)2]dx
  3. π∫0e−1[(ln⁡(x+1))2−(0)2] dx\pi\int_{0}^{e-1}\big[(\ln(x+1))^2-(0)^2\big]\,dxπ∫0e−1​[(ln(x+1))2−(0)2]dx
  4. π∫0e−1[(ln⁡(x+1)−1)2−(0−1)2] dx\pi\int_{0}^{e-1}\big[(\ln(x+1)-1)^2-(0-1)^2\big]\,dxπ∫0e−1​[(ln(x+1)−1)2−(0−1)2]dx
  5. π∫0e−1[(ln⁡(x+1)+1)−1]2 dx\pi\int_{0}^{e-1}\big[(\ln(x+1)+1)-1\big]^2\,dxπ∫0e−1​[(ln(x+1)+1)−1]2dx

Explanation: This problem involves using the washer method to find the volume of the solid formed by revolving the region around the shifted axis y = -1. To adjust for this axis, the radii are the distances from the curves to y = -1, which means adding 1 to each y-value. The outer radius is from the upper curve y = ln(x + 1) to the axis, giving ln(x + 1) + 1. The inner radius is from the lower curve y = 0 to the axis, giving 1. A tempting distractor is choice B, which reverses the outer and inner radii, resulting in a negative integrand and incorrect volume. In general, when revolving around a horizontal axis y = k, the radius for a curve y = f(x) is |f(x) - k|, and ensure the outer radius is the larger one in the integral.

Question 9

What integral gives the volume when the region between y=xy=xy=x and y=x2y=x^2y=x2, 0≤x≤10\le x\le10≤x≤1, is revolved about y=−1y=-1y=−1?

  1. π∫01[(x+1)2−(x2+1)2] dx\pi\int_{0}^{1}\big[(x+1)^2-(x^2+1)^2\big]\,dxπ∫01​[(x+1)2−(x2+1)2]dx (correct answer)
  2. π∫01[(x2+1)2−(x+1)2] dx\pi\int_{0}^{1}\big[(x^2+1)^2-(x+1)^2\big]\,dxπ∫01​[(x2+1)2−(x+1)2]dx
  3. π∫01[x2−(x2)2] dx\pi\int_{0}^{1}\big[x^2-(x^2)^2\big]\,dxπ∫01​[x2−(x2)2]dx
  4. π∫01[(x2−1)2−(x−1)2] dx\pi\int_{0}^{1}\big[(x^2-1)^2-(x-1)^2\big]\,dxπ∫01​[(x2−1)2−(x−1)2]dx
  5. π∫01[(x2+1)−(x+1)]2 dx\pi\int_{0}^{1}\big[(x^2+1)-(x+1)\big]^2\,dxπ∫01​[(x2+1)−(x+1)]2dx

Explanation: This problem involves using the washer method to find the volume of the solid formed by revolving the region around the shifted axis y = -1. To adjust for this axis, the radii are the distances from the curves to y = -1, which means adding 1 to each y-value. The outer radius is from the upper curve y = x to the axis, giving x + 1. The inner radius is from the lower curve y = x² to the axis, giving x² + 1. A tempting distractor is choice B, which reverses the outer and inner radii, resulting in a negative integrand and incorrect volume. In general, when revolving around a horizontal axis y = k, the radius for a curve y = f(x) is |f(x) - k|, and ensure the outer radius is the larger one in the integral.

Question 10

What integral gives the volume when the region between y=x3y=x^3y=x3 and y=xy=xy=x, 0≤x≤10\le x\le10≤x≤1, is revolved about y=2y=2y=2?

  1. π∫01[(2−x)2−(2−x3)2] dx\pi\int_{0}^{1}\big[(2-x)^2-(2-x^3)^2\big]\,dxπ∫01​[(2−x)2−(2−x3)2]dx
  2. π∫01[(2−x3)2−(2−x)2] dx\pi\int_{0}^{1}\big[(2-x^3)^2-(2-x)^2\big]\,dxπ∫01​[(2−x3)2−(2−x)2]dx (correct answer)
  3. π∫01[(x)2−(x3)2] dx\pi\int_{0}^{1}\big[(x)^2-(x^3)^2\big]\,dxπ∫01​[(x)2−(x3)2]dx
  4. π∫01[(x−2)2−(x3−2)2] dx\pi\int_{0}^{1}\big[(x-2)^2-(x^3-2)^2\big]\,dxπ∫01​[(x−2)2−(x3−2)2]dx
  5. π∫01[(2−x)−(2−x3)]2 dx\pi\int_{0}^{1}\big[(2-x)-(2-x^3)\big]^2\,dxπ∫01​[(2−x)−(2−x3)]2dx

Explanation: This problem involves using the washer method to find the volume of the solid formed by revolving the region around the shifted axis y = 2. To adjust for this axis, the radii are the distances from the curves to y = 2, which means subtracting each y-value from 2 since the region is below. The outer radius is from the lower curve y = x³ to the axis, giving 2 - x³. The inner radius is from the upper curve y = x to the axis, giving 2 - x. A tempting distractor is choice A, which reverses the outer and inner radii, resulting in a negative integrand and incorrect volume. In general, when revolving around a horizontal axis y = k, the radius for a curve y = f(x) is |f(x) - k|, and ensure the outer radius is the larger one in the integral.

Question 11

What integral gives the volume when the region between y=x+1y=x+1y=x+1 and y=1y=1y=1, 0≤x≤30 \le x \le 30≤x≤3, is revolved about y=−1y=-1y=−1?

  1. π∫03[(x+2)2−(2)2] dx\pi\int_{0}^{3}\big[(x+2)^2-(2)^2\big]\,dxπ∫03​[(x+2)2−(2)2]dx (correct answer)
  2. π∫03[(2)2−(x+2)2] dx\pi\int_{0}^{3}\big[(2)^2-(x+2)^2\big]\,dxπ∫03​[(2)2−(x+2)2]dx
  3. π∫03[(x+1)2−(1)2] dx\pi\int_{0}^{3}\big[(x+1)^2-(1)^2\big]\,dxπ∫03​[(x+1)2−(1)2]dx
  4. π∫03[(x+1−(−1))−(1−(−1))]2 dx\pi\int_{0}^{3}\big[(x+1-(-1))-(1-(-1))\big]^2\,dxπ∫03​[(x+1−(−1))−(1−(−1))]2dx
  5. π∫03[(x)2−(0)2] dx\pi\int_{0}^{3}\big[(x)^2-(0)^2\big]\,dxπ∫03​[(x)2−(0)2]dx

Explanation: This problem involves using the washer method to find the volume of the solid formed by revolving the region around the shifted axis y=−1y = -1y=−1. To adjust for this axis, the radii are the distances from the curves to y=−1y = -1y=−1, which means adding 1 to each y-value. The outer radius is from the upper curve y=x+1y = x + 1y=x+1 to the axis, giving x+2x + 2x+2. The inner radius is from the lower curve y=1y = 1y=1 to the axis, giving 222. A tempting distractor is choice B, which reverses the outer and inner radii, resulting in a negative integrand and incorrect volume. In general, when revolving around a horizontal axis y=ky = ky=k, the radius for a curve y=f(x)y = f(x)y=f(x) is ∣f(x)−k∣|f(x) - k|∣f(x)−k∣, and ensure the outer radius is the larger one in the integral.

Question 12

Select the correct washer setup for revolving the region between y=ln⁡xy=\ln xy=lnx and y=0y=0y=0 on [1,e][1,e][1,e] about y=1y=1y=1.

  1. V=π∫1e[(1−ln⁡x)2−(1−0)2]dxV=\pi\displaystyle\int_{1}^{e}\Big[(1-\ln x)^2-(1-0)^2\Big]dxV=π∫1e​[(1−lnx)2−(1−0)2]dx
  2. V=π∫1e[(1−0)2−(1−ln⁡x)2]dxV=\pi\displaystyle\int_{1}^{e}\Big[(1-0)^2-(1-\ln x)^2\Big]dxV=π∫1e​[(1−0)2−(1−lnx)2]dx (correct answer)
  3. V=π∫1e[(ln⁡x)2−02]dxV=\pi\displaystyle\int_{1}^{e}\Big[(\ln x)^2-0^2\Big]dxV=π∫1e​[(lnx)2−02]dx
  4. V=π∫1e[(ln⁡x+1)2−(0+1)2]dxV=\pi\displaystyle\int_{1}^{e}\Big[(\ln x+1)^2-(0+1)^2\Big]dxV=π∫1e​[(lnx+1)2−(0+1)2]dx
  5. V=π∫01[(ey−1)2−(1−1)2]dyV=\pi\displaystyle\int_{0}^{1}\Big[(e^y-1)^2-(1-1)^2\Big]dyV=π∫01​[(ey−1)2−(1−1)2]dy

Explanation: This problem requires the washer method with rotation about y = 1. Since y = 1 is above the region between y = ln x and y = 0 on [1,e], we measure distances downward: the outer radius is from y = 1 to y = 0, giving R = 1 - 0 = 1, and the inner radius is from y = 1 to y = ln x, giving r = 1 - ln x. The volume integral is π∫[1,e][(1-0)² - (1-ln x)²]dx = π∫[1,e][1 - (1-ln x)²]dx. Choice A reverses the radii, placing the smaller radius as outer. Remember that ln x < 1 on [1,e], so 1 - ln x > 0, making it the valid inner radius.

Question 13

What integral setup gives the volume when the region between y=2y=2y=2 and y=x3y=x^3y=x3 on [0,23][0,\sqrt[3]{2}][0,32​] rotates about y=−1y=-1y=−1?

  1. V=π∫023[(2+1)2−(x3+1)2]dxV=\pi\displaystyle\int_{0}^{\sqrt[3]{2}}\Big[(2+1)^2-(x^3+1)^2\Big]dxV=π∫032​​[(2+1)2−(x3+1)2]dx
  2. V=π∫023[(x3+1)2−(2+1)2]dxV=\pi\displaystyle\int_{0}^{\sqrt[3]{2}}\Big[(x^3+1)^2-(2+1)^2\Big]dxV=π∫032​​[(x3+1)2−(2+1)2]dx
  3. V=π∫023[(2−(−1))2−(x3−(−1))2]dxV=\pi\displaystyle\int_{0}^{\sqrt[3]{2}}\Big[(2-(-1))^2-(x^3-(-1))^2\Big]dxV=π∫032​​[(2−(−1))2−(x3−(−1))2]dx (correct answer)
  4. V=π∫023[22−(x3)2]dxV=\pi\displaystyle\int_{0}^{\sqrt[3]{2}}\Big[2^2-(x^3)^2\Big]dxV=π∫032​​[22−(x3)2]dx
  5. V=π∫−12[(23−0)2−(y3−0)2]dyV=\pi\displaystyle\int_{-1}^{2}\Big[(\sqrt[3]{2}-0)^2-(\sqrt[3]{y}-0)^2\Big]dyV=π∫−12​[(32​−0)2−(3y​−0)2]dy

Explanation: This problem involves the washer method with rotation about y = -1. When rotating about y = -1, we measure distances upward from this axis: the outer radius is from y = -1 to y = 2, giving R = 2 - (-1) = 3, and the inner radius is from y = -1 to y = x³, giving r = x³ - (-1) = x³ + 1. The volume integral is π∫[0,∛2][(2-(-1))² - (x³-(-1))²]dx = π∫[0,∛2][9 - (x³+1)²]dx. Choice A incorrectly adds 1 to both functions instead of subtracting the axis value -1. When the axis is below the region, subtract the negative axis value (which adds its absolute value) to find radii.

Question 14

Which integral represents the volume when the region between y=sin⁡xy=\sin xy=sinx and y=0y=0y=0 on [0,π][0,\pi][0,π] is revolved about y=−2y=-2y=−2?

  1. π∫0π[(sin⁡x+2)2−(2)2]dx\pi\displaystyle\int_{0}^{\pi}\big[(\sin x+2)^2-(2)^2\big]dxπ∫0π​[(sinx+2)2−(2)2]dx (correct answer)
  2. π∫0π[(sin⁡x)2−(0)2]dx\pi\displaystyle\int_{0}^{\pi}\big[(\sin x)^2-(0)^2\big]dxπ∫0π​[(sinx)2−(0)2]dx
  3. π∫0π[(2)2−(sin⁡x+2)2]dx\pi\displaystyle\int_{0}^{\pi}\big[(2)^2-(\sin x+2)^2\big]dxπ∫0π​[(2)2−(sinx+2)2]dx
  4. π∫0π[(sin⁡x−2)2−(0−2)2]dx\pi\displaystyle\int_{0}^{\pi}\big[(\sin x-2)^2-(0-2)^2\big]dxπ∫0π​[(sinx−2)2−(0−2)2]dx
  5. π∫0π[(sin⁡x+2)−(2)]2dx\pi\displaystyle\int_{0}^{\pi}\big[(\sin x+2)-(2)\big]^2dxπ∫0π​[(sinx+2)−(2)]2dx

Explanation: This problem utilizes the washer method to compute the volume of a solid formed by revolving a region around the shifted axis y=−2y = -2y=−2. To adjust for this axis, add 2 to each y-value to get the distances, since the curves are above the axis. The outer radius is sin⁡x+2\sin x + 2sinx+2 from the upper curve, and the inner radius is 0+2=20 + 2 = 20+2=2 from the lower curve. Therefore, the integral is π∫0π[(sin⁡x+2)2−(2)2] dx\pi \int_0^\pi [(\sin x + 2)^2 - (2)^2] \, dxπ∫0π​[(sinx+2)2−(2)2]dx. A tempting distractor is choice C, which swaps the inner and outer radii, resulting in a negative integrand. In general, when revolving around y=ky = ky=k, compute radii as ∣y−k∣|y - k|∣y−k∣ for each curve and identify the outer as the larger distance and inner as the smaller.

Question 15

What integral represents the volume when the region between y=cos⁡xy=\cos xy=cosx and y=0y=0y=0 on [0,π2][0,\tfrac{\pi}{2}][0,2π​] is revolved about y=1y=1y=1?

  1. π∫0π/2[(1−cos⁡x)2−(1−0)2]dx\pi\displaystyle\int_{0}^{\pi/2}\big[(1-\cos x)^2-(1-0)^2\big]dxπ∫0π/2​[(1−cosx)2−(1−0)2]dx
  2. π∫0π/2[(cos⁡x)2−(0)2]dx\pi\displaystyle\int_{0}^{\pi/2}\big[(\cos x)^2-(0)^2\big]dxπ∫0π/2​[(cosx)2−(0)2]dx
  3. π∫0π/2[(1−0)2−(1−cos⁡x)2]dx\pi\displaystyle\int_{0}^{\pi/2}\big[(1-0)^2-(1-\cos x)^2\big]dxπ∫0π/2​[(1−0)2−(1−cosx)2]dx (correct answer)
  4. π∫0π/2[(1+cos⁡x)2−(1+0)2]dx\pi\displaystyle\int_{0}^{\pi/2}\big[(1+\cos x)^2-(1+0)^2\big]dxπ∫0π/2​[(1+cosx)2−(1+0)2]dx
  5. π∫0π/2[(1−0)−(1−cos⁡x)]2dx\pi\displaystyle\int_{0}^{\pi/2}\big[(1-0)-(1-\cos x)\big]^2dxπ∫0π/2​[(1−0)−(1−cosx)]2dx

Explanation: This problem utilizes the washer method to compute the volume of a solid formed by revolving a region around the shifted axis y = 1. To adjust for this axis, subtract each y-value from 1 to get the distances, since the curves are below or at the axis. The outer radius is 1 - 0 = 1 from the lower curve, and the inner radius is 1 - cos x from the upper curve. Therefore, the integral is π ∫ from 0 to π/2 of [(1 - 0)² - (1 - cos x)²] dx. A tempting distractor is choice A, which swaps the inner and outer radii, resulting in a negative integrand in some parts. In general, when revolving around y = k, compute radii as |y - k| for each curve and identify the outer as the larger distance and inner as the smaller.

Question 16

Which integral sets up the volume when the region between y=ln⁡xy=\ln xy=lnx and y=0y=0y=0 on [1,e][1,e][1,e] is revolved about y=−1y=-1y=−1?

  1. π∫1e[(ln⁡x+1)2−(1)2]dx\pi\displaystyle\int_{1}^{e}\big[(\ln x+1)^2-(1)^2\big]dxπ∫1e​[(lnx+1)2−(1)2]dx (correct answer)
  2. π∫1e[(ln⁡x)2−(0)2]dx\pi\displaystyle\int_{1}^{e}\big[(\ln x)^2-(0)^2\big]dxπ∫1e​[(lnx)2−(0)2]dx
  3. π∫1e[(1)2−(ln⁡x+1)2]dx\pi\displaystyle\int_{1}^{e}\big[(1)^2-(\ln x+1)^2\big]dxπ∫1e​[(1)2−(lnx+1)2]dx
  4. π∫1e[(ln⁡x−1)2−(0−1)2]dx\pi\displaystyle\int_{1}^{e}\big[(\ln x-1)^2-(0-1)^2\big]dxπ∫1e​[(lnx−1)2−(0−1)2]dx
  5. π∫1e[(ln⁡x+1)−(1)]2dx\pi\displaystyle\int_{1}^{e}\big[(\ln x+1)-(1)\big]^2dxπ∫1e​[(lnx+1)−(1)]2dx

Explanation: This problem utilizes the washer method to compute the volume of a solid formed by revolving a region around the shifted axis y = -1. To adjust for this axis, add 1 to each y-value to get the distances, since the curves are above the axis. The outer radius is ln x + 1 from the upper curve, and the inner radius is 0 + 1 = 1 from the lower curve. Therefore, the integral is π ∫ from 1 to e of [(ln x + 1)² - (1)²] dx. A tempting distractor is choice C, which swaps the inner and outer radii, resulting in a negative integrand. In general, when revolving around y = k, compute radii as |y - k| for each curve and identify the outer as the larger distance and inner as the smaller.

Question 17

What integral gives the volume when the region between y=exy=e^xy=ex and y=1y=1y=1, 0≤x≤10\le x\le10≤x≤1, is revolved about y=0y=0y=0?

  1. π∫01[(ex)2−(1)2] dx\pi\int_{0}^{1}\big[(e^x)^2-(1)^2\big]\,dxπ∫01​[(ex)2−(1)2]dx (correct answer)
  2. π∫01[(1)2−(ex)2] dx\pi\int_{0}^{1}\big[(1)^2-(e^x)^2\big]\,dxπ∫01​[(1)2−(ex)2]dx
  3. π∫01[(ex−0)2−(1−0)2] dx\pi\int_{0}^{1}\big[(e^x-0)^2-(1-0)^2\big]\,dxπ∫01​[(ex−0)2−(1−0)2]dx
  4. π∫01[(ex+0)2−(1+0)2] dx\pi\int_{0}^{1}\big[(e^x+0)^2-(1+0)^2\big]\,dxπ∫01​[(ex+0)2−(1+0)2]dx
  5. π∫01[(ex−1)]2 dx\pi\int_{0}^{1}\big[(e^x-1)\big]^2\,dxπ∫01​[(ex−1)]2dx

Explanation: This problem involves using the washer method to find the volume of the solid formed by revolving the region around the shifted axis y = 0. To adjust for this axis, the radii are the distances from the curves to y = 0, which is simply the y-value since the region is above. The outer radius is from the upper curve y = e^x to the axis, giving e^x. The inner radius is from the lower curve y = 1 to the axis, giving 1. A tempting distractor is choice B, which reverses the outer and inner radii, resulting in a negative integrand and incorrect volume. In general, when revolving around a horizontal axis y = k, the radius for a curve y = f(x) is |f(x) - k|, and ensure the outer radius is the larger one in the integral.

Question 18

Find the volume setup when the region between y=x2y=x^2y=x2 and y=4y=4y=4 for 0≤x≤20\le x\le20≤x≤2 is revolved about y=5y=5y=5.

  1. V=π∫02[(4−5)2−(x2−5)2]dxV=\pi\displaystyle\int_{0}^{2}\big[(4-5)^2-(x^2-5)^2\big]dxV=π∫02​[(4−5)2−(x2−5)2]dx
  2. V=π∫02[(5−x2)2−(5−4)2]dxV=\pi\displaystyle\int_{0}^{2}\big[(5-x^2)^2-(5-4)^2\big]dxV=π∫02​[(5−x2)2−(5−4)2]dx (correct answer)
  3. V=π∫02[(4−x2)2]dxV=\pi\displaystyle\int_{0}^{2}\big[(4-x^2)^2\big]dxV=π∫02​[(4−x2)2]dx
  4. V=π∫02[(5−4)2−(5−x2)2]dxV=\pi\displaystyle\int_{0}^{2}\big[(5-4)^2-(5-x^2)^2\big]dxV=π∫02​[(5−4)2−(5−x2)2]dx
  5. V=π∫02[(4−5)2−(4−x2)2]dxV=\pi\displaystyle\int_{0}^{2}\big[(4-5)^2-(4-x^2)^2\big]dxV=π∫02​[(4−5)2−(4−x2)2]dx

Explanation: This problem requires the washer method with revolution about the horizontal line y = 5, which is above the bounded region. When revolving about y = 5, we measure distances from this axis: the outer radius is from y = 5 down to the parabola y = x², giving R = 5 - x², and the inner radius is from y = 5 down to the line y = 4, giving r = 5 - 4 = 1. The washer method formula becomes V = π∫[R² - r²]dx = π∫[(5-x²)² - 1²]dx from 0 to 2. Choice A incorrectly uses (4-5)² = 1 as the outer radius and (x²-5)² as the inner radius, reversing the distance calculations and producing negative radii. The key strategy when revolving about y = k is to always compute radii as |k - y(x)|, ensuring positive distances from the axis to each curve.

Question 19

Choose the washer-method setup for revolving the region between y=sin⁡xy=\sin xy=sinx and y=cos⁡xy=\cos xy=cosx on 0≤x≤π40\le x\le\frac{\pi}{4}0≤x≤4π​ about y=−1y=-1y=−1.​​

  1. V=π∫0π/4[(cos⁡x+1)2−(sin⁡x+1)2]dxV=\pi\displaystyle\int_{0}^{\pi/4}\big[(\cos x+1)^2-(\sin x+1)^2\big]dxV=π∫0π/4​[(cosx+1)2−(sinx+1)2]dx (correct answer)
  2. V=π∫0π/4[(sin⁡x+1)2−(cos⁡x+1)2]dxV=\pi\displaystyle\int_{0}^{\pi/4}\big[(\sin x+1)^2-(\cos x+1)^2\big]dxV=π∫0π/4​[(sinx+1)2−(cosx+1)2]dx
  3. V=π∫0π/4[(cos⁡x)2−(sin⁡x)2]dxV=\pi\displaystyle\int_{0}^{\pi/4}\big[(\cos x)^2-(\sin x)^2\big]dxV=π∫0π/4​[(cosx)2−(sinx)2]dx
  4. V=π∫0π/4[(1−sin⁡x)2−(1−cos⁡x)2]dxV=\pi\displaystyle\int_{0}^{\pi/4}\big[(1-\sin x)^2-(1-\cos x)^2\big]dxV=π∫0π/4​[(1−sinx)2−(1−cosx)2]dx
  5. V=π∫0π/4[(cos⁡x+1)2−(1)2]dxV=\pi\displaystyle\int_{0}^{\pi/4}\big[(\cos x+1)^2-(1)^2\big]dxV=π∫0π/4​[(cosx+1)2−(1)2]dx

Explanation: This problem requires the washer method for revolution about y = -1, with the region bounded by y = sin x and y = cos x on [0, π/4]. Since cos x ≥ sin x on this interval and both curves are above y = -1, the outer radius is R = cos x - (-1) = cos x + 1 and the inner radius is r = sin x - (-1) = sin x + 1. The volume setup is V = π∫[(cos x + 1)² - (sin x + 1)²]dx from 0 to π/4. Choice B reverses the radii, incorrectly assuming sin x is farther from the axis, which would give a negative integrand since cos x > sin x on this interval. To determine which function gives the outer radius, always check which curve is farther from the axis of revolution throughout the integration interval.

Question 20

What integral gives the volume when the region between y=sin⁡xy=\sin xy=sinx and y=0y=0y=0, 0≤x≤π0\le x\le\pi0≤x≤π, is revolved about y=1y=1y=1?

  1. π∫0π[(1−0)2−(1−sin⁡x)2] dx\pi\int_{0}^{\pi}\big[(1-0)^2-(1-\sin x)^2\big]\,dxπ∫0π​[(1−0)2−(1−sinx)2]dx (correct answer)
  2. π∫0π[(1−sin⁡x)2−(1−0)2] dx\pi\int_{0}^{\pi}\big[(1-\sin x)^2-(1-0)^2\big]\,dxπ∫0π​[(1−sinx)2−(1−0)2]dx
  3. π∫0π[(sin⁡x)2−(0)2] dx\pi\int_{0}^{\pi}\big[(\sin x)^2-(0)^2\big]\,dxπ∫0π​[(sinx)2−(0)2]dx
  4. π∫0π[(sin⁡x−1)2−(0−1)2] dx\pi\int_{0}^{\pi}\big[(\sin x-1)^2-(0-1)^2\big]\,dxπ∫0π​[(sinx−1)2−(0−1)2]dx
  5. π∫0π[(1−sin⁡x)−1]2 dx\pi\int_{0}^{\pi}\big[(1-\sin x)-1\big]^2\,dxπ∫0π​[(1−sinx)−1]2dx

Explanation: This problem involves using the washer method to find the volume of the solid formed by revolving the region around the shifted axis y = 1. To adjust for this axis, the radii are the distances from the curves to y = 1, which means subtracting each y-value from 1 since the region is below or touching. The outer radius is from the lower curve y = 0 to the axis, giving 1. The inner radius is from the upper curve y = sin x to the axis, giving 1 - sin x. A tempting distractor is choice B, which reverses the outer and inner radii, resulting in a negative integrand and incorrect volume. In general, when revolving around a horizontal axis y = k, the radius for a curve y = f(x) is |f(x) - k|, and ensure the outer radius is the larger one in the integral.