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AP Calculus AB Quiz

AP Calculus AB Quiz: Washer Method Revolving Around Other Axes

Practice Washer Method Revolving Around Other Axes in AP Calculus AB 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

The region between y=4y=4y=4 and y=xy=\sqrt{x}y=x​ for 0≤x≤160\le x\le160≤x≤16 is revolved about y=6y=6y=6. Which integral gives the volume?

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 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.

All questions

Question 1

The region between y=4y=4y=4 and y=xy=\sqrt{x}y=x​ for 0≤x≤160\le x\le160≤x≤16 is revolved about y=6y=6y=6. Which integral gives the volume?

  1. V=π∫016[(6−4)2−(6−x)2]dxV=\pi\int_{0}^{16}\big[(6-4)^2-(6-\sqrt{x})^2\big]dxV=π∫016​[(6−4)2−(6−x​)2]dx
  2. V=π∫016[(6−x)2−(6−4)2]dxV=\pi\int_{0}^{16}\big[(6-\sqrt{x})^2-(6-4)^2\big]dxV=π∫016​[(6−x​)2−(6−4)2]dx (correct answer)
  3. V=π∫016[42−(x)2]dxV=\pi\int_{0}^{16}\big[4^2-(\sqrt{x})^2\big]dxV=π∫016​[42−(x​)2]dx
  4. V=π∫016[(6−4)−(6−x)]2dxV=\pi\int_{0}^{16}\big[(6-4)-(6-\sqrt{x})\big]^2dxV=π∫016​[(6−4)−(6−x​)]2dx
  5. V=π∫016[(4−6)2−(x−6)2]dxV=\pi\int_{0}^{16}\big[(4-6)^2-(\sqrt{x}-6)^2\big]dxV=π∫016​[(4−6)2−(x​−6)2]dx

Explanation: This problem involves using the washer method to find the volume of a solid formed by revolving a region around a shifted horizontal axis, specifically y = 6. To adjust for the shifted axis, subtract y from 6, making outer 6 - √x for the lower curve. Inner is 6 - 4 = 2 for the upper. These handle axis above. A tempting distractor is choice A, swapping order. A transferable strategy for washer method with shifted axes is to always compute distances by subtracting the axis value from each curve's y-value, determine outer as the larger distance and inner as the smaller, and integrate π times (outer² - inner²) dx.

Question 2

Region bounded by y=1−xy=1-xy=1−x and y=1−x2y=1-x^2y=1−x2 for 0≤x≤10\le x\le10≤x≤1 is revolved about y=3y=3y=3. Choose the correct volume setup.

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

Explanation: This problem involves using the washer method to find the volume of a solid formed by revolving a region around a shifted horizontal axis, specifically y = 3. To adjust for the shifted axis, subtract each y-value from 3, making the outer radius 3 - (1 - x) for the upper curve. The inner radius is 3 - (1 - x²) for the lower curve. These adjustments handle the axis above. A tempting distractor is choice A, which swaps radii. A transferable strategy for washer method with shifted axes is to always compute distances by subtracting the axis value from each curve's y-value, determine outer as the larger distance and inner as the smaller, and integrate π times (outer² - inner²) dx.

Question 3

For −1≤x≤1-1\le x\le1−1≤x≤1, the region between y=x2+1y=x^2+1y=x2+1 and y=1y=1y=1 is revolved about y=4y=4y=4. Which integral represents the volume?

  1. V=π∫−11[(4−(x2+1))2−(4−1)2]dxV=\pi\int_{-1}^{1}\big[(4-(x^2+1))^2-(4-1)^2\big]dxV=π∫−11​[(4−(x2+1))2−(4−1)2]dx
  2. V=π∫−11[(4−1)2−(4−(x2+1))2]dxV=\pi\int_{-1}^{1}\big[(4-1)^2-(4-(x^2+1))^2\big]dxV=π∫−11​[(4−1)2−(4−(x2+1))2]dx (correct answer)
  3. V=π∫−11[(x2+1)2−12]dxV=\pi\int_{-1}^{1}\big[(x^2+1)^2-1^2\big]dxV=π∫−11​[(x2+1)2−12]dx
  4. V=π∫−11[(x2+1−4)2−(1−4)2]dxV=\pi\int_{-1}^{1}\big[(x^2+1-4)^2-(1-4)^2\big]dxV=π∫−11​[(x2+1−4)2−(1−4)2]dx
  5. V=π∫−11[(4−(x2+1))−(4−1)]2dxV=\pi\int_{-1}^{1}\big[(4-(x^2+1))-(4-1)\big]^2dxV=π∫−11​[(4−(x2+1))−(4−1)]2dx

Explanation: The washer method for volumes of revolution around a shifted horizontal axis involves adjusting the radii to account for the distance from each curve to the axis. For revolution around y=4, both radii are 4 minus curves since below, giving 4 - 1 and 4 - (x² + 1). The outer is 3. This works well. A tempting distractor like option A fails by swapping. A transferable strategy for shifted washer problems is to use axis minus curve for below regions.

Question 4

Let RRR be bounded by y=4−xy=4-xy=4−x and y=xy=xy=x on 0≤x≤20 \leq x \leq 20≤x≤2 and revolved about y=5y=5y=5. Which setup gives the volume?

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

Explanation: The washer method for volumes of revolution around a shifted horizontal axis involves adjusting the radii to account for the distance from each curve to the axis. For revolution around y=5y=5y=5, both radii are 5−x5 - x5−x and 5−(4−x)5 - (4 - x)5−(4−x) since below, giving 5−x5 - x5−x and 5−(4−x)5 - (4 - x)5−(4−x). The outer is 5−x5 - x5−x as lower curve is farther. This adjustment maintains correct distances. A tempting distractor like option B fails by reversing order, yielding negative integrand. A transferable strategy for shifted washer problems is to subtract curves from axis when region is below and verify positive difference.

Question 5

Region bounded by y=4−x2y=\sqrt{4-x^2}y=4−x2​ and y=0y=0y=0 for −2≤x≤2-2\le x\le2−2≤x≤2 is revolved about y=−3y=-3y=−3. Which setup is correct?

  1. V=π∫−22[(4−x2+3)2−(3)2]dxV=\pi\int_{-2}^{2}\big[(\sqrt{4-x^2}+3)^2-(3)^2\big]dxV=π∫−22​[(4−x2​+3)2−(3)2]dx (correct answer)
  2. V=π∫−22[(3)2−(4−x2+3)2]dxV=\pi\int_{-2}^{2}\big[(3)^2-(\sqrt{4-x^2}+3)^2\big]dxV=π∫−22​[(3)2−(4−x2​+3)2]dx
  3. V=π∫−22[(4−x2)2−02]dxV=\pi\int_{-2}^{2}\big[(\sqrt{4-x^2})^2-0^2\big]dxV=π∫−22​[(4−x2​)2−02]dx
  4. V=π∫−22[(−3−4−x2)2−(−3−0)2]dxV=\pi\int_{-2}^{2}\big[(-3-\sqrt{4-x^2})^2-(-3-0)^2\big]dxV=π∫−22​[(−3−4−x2​)2−(−3−0)2]dx
  5. V=π∫−22[(4−x2+3)−3]2dxV=\pi\int_{-2}^{2}\big[(\sqrt{4-x^2}+3)-3\big]^2dxV=π∫−22​[(4−x2​+3)−3]2dx

Explanation: This problem involves using the washer method to find the volume of a solid formed by revolving a region around a shifted horizontal axis, specifically y = -3. To adjust, add 3 to y, outer √(4-x^2) +3, inner 0+3=3? But since lower is 0, distance 0 - (-3)=3, upper √ +3 >3. Yes. A tempting distractor is choice B, reversing. A transferable strategy for washer method with shifted axes is to always compute distances by subtracting the axis value from each curve's y-value, determine outer as the larger distance and inner as the smaller, and integrate π times (outer² - inner²) dx.

Question 6

The region between y=x24y=\frac{x^2}{4}y=4x2​ and y=1y=1y=1 for 0≤x≤20\le x\le20≤x≤2 is revolved about y=−1y=-1y=−1. Which setup is correct?

  1. V=π∫02[(1+1)2−(x2/4+1)2]dxV=\pi\int_{0}^{2}\big[(1+1)^2-(x^2/4+1)^2\big]dxV=π∫02​[(1+1)2−(x2/4+1)2]dx (correct answer)
  2. V=π∫02[(x2/4+1)2−(2)2]dxV=\pi\int_{0}^{2}\big[(x^2/4+1)^2-(2)^2\big]dxV=π∫02​[(x2/4+1)2−(2)2]dx
  3. V=π∫02[12−(x2/4)2]dxV=\pi\int_{0}^{2}\big[1^2-(x^2/4)^2\big]dxV=π∫02​[12−(x2/4)2]dx
  4. V=π∫02[(−1−1)2−(−1−x2/4)2]dxV=\pi\int_{0}^{2}\big[(-1-1)^2-(-1-x^2/4)^2\big]dxV=π∫02​[(−1−1)2−(−1−x2/4)2]dx
  5. V=π∫02[(2)−(x2/4+1)]2dxV=\pi\int_{0}^{2}\big[(2)-(x^2/4+1)\big]^2dxV=π∫02​[(2)−(x2/4+1)]2dx

Explanation: This problem involves using the washer method to find the volume of a solid formed by revolving a region around a shifted horizontal axis, specifically y = -1. To adjust for the shifted axis, add 1 to each y-value since the axis is below the region, making the outer radius 1 - (-1) = 2 for the upper curve y=1. The inner radius is x²/4 - (-1) = x²/4 + 1 for the lower curve. These adjustments ensure the radii represent distances from the axis. A tempting distractor is choice B, which reverses the outer and inner radii, resulting in a negative integrand and incorrect volume. A transferable strategy for washer method with shifted axes is to always compute distances by subtracting the axis value from each curve's y-value, determine outer as the larger distance and inner as the smaller, and integrate π times (outer² - inner²) dx.

Question 7

For 0≤x≤π0\le x\le\pi0≤x≤π, the region between y=sin⁡xy=\sin xy=sinx and y=0y=0y=0 is revolved about y=1y=1y=1. Which integral represents the volume?

  1. V=π∫0π[(1−sin⁡x)2−(1−0)2]dxV=\pi\int_{0}^{\pi}\big[(1-\sin x)^2-(1-0)^2\big]dxV=π∫0π​[(1−sinx)2−(1−0)2]dx
  2. V=π∫0π[(1−0)2−(1−sin⁡x)2]dxV=\pi\int_{0}^{\pi}\big[(1-0)^2-(1-\sin x)^2\big]dxV=π∫0π​[(1−0)2−(1−sinx)2]dx (correct answer)
  3. V=π∫0πsin⁡2x dxV=\pi\int_{0}^{\pi}\sin^2x\,dxV=π∫0π​sin2xdx
  4. V=π∫0π[(sin⁡x−1)2−(0−1)2]dxV=\pi\int_{0}^{\pi}\big[(\sin x-1)^2-(0-1)^2\big]dxV=π∫0π​[(sinx−1)2−(0−1)2]dx
  5. V=π∫0π[(1−sin⁡x)−(1−0)]2dxV=\pi\int_{0}^{\pi}\big[(1-\sin x)-(1-0)\big]^2dxV=π∫0π​[(1−sinx)−(1−0)]2dx

Explanation: The washer method for volumes of revolution around a shifted horizontal axis involves adjusting the radii to account for the distance from each curve to the axis. For revolution around y=1, both radii are 1 minus curves since below, giving 1 - 0 and 1 - sin x. The outer is 1. This works as region touches but doesn't cross. A tempting distractor like option A fails by reversing order. A transferable strategy for shifted washer problems is to use axis minus curve and split if crossing.

Question 8

Select the correct washer-method integral for revolving the region between y=1xy=\frac{1}{x}y=x1​ and y=14y=\frac{1}{4}y=41​ on [1,4][1,4][1,4] about y=0y=0y=0.

  1. V=π∫14[(1x)2−(14)2]dxV=\pi\int_{1}^{4}\Big[\left(\frac{1}{x}\right)^2-\left(\frac{1}{4}\right)^2\Big]dxV=π∫14​[(x1​)2−(41​)2]dx (correct answer)
  2. V=π∫14[(1x−14)2−(0)2]dxV=\pi\int_{1}^{4}\Big[\left(\frac{1}{x}-\frac{1}{4}\right)^2-(0)^2\Big]dxV=π∫14​[(x1​−41​)2−(0)2]dx
  3. V=π∫14[(14)2−(1x)2]dxV=\pi\int_{1}^{4}\Big[\left(\frac{1}{4}\right)^2-\left(\frac{1}{x}\right)^2\Big]dxV=π∫14​[(41​)2−(x1​)2]dx
  4. V=π∫14[(1x)2−(0)2]dxV=\pi\int_{1}^{4}\Big[\left(\frac{1}{x}\right)^2-(0)^2\Big]dxV=π∫14​[(x1​)2−(0)2]dx
  5. V=π∫14[(14−0)2−(1x−0)2]dxV=\pi\int_{1}^{4}\Big[\left(\frac{1}{4}-0\right)^2-\left(\frac{1}{x}-0\right)^2\Big]dxV=π∫14​[(41​−0)2−(x1​−0)2]dx

Explanation: This problem uses the washer method revolving about y = 0 (the x-axis), making it a standard case without axis shifting. The region lies between y = 1/x and y = 1/4, where 1/x ≥ 1/4 on [1,4], so the outer radius is R = 1/x and the inner radius is r = 1/4. The washer formula gives V = π∫[(1/x)² - (1/4)²]dx from 1 to 4. Choice E incorrectly writes (1/4 - 0)² - (1/x - 0)², unnecessarily subtracting 0 and reversing the order. When rotating about y = 0, no shift is needed—just use the y-values directly as radii, with the larger value as the outer radius.

Question 9

What is the washer-method volume setup when the region between y=ln⁡xy=\ln xy=lnx and y=1y=1y=1 for 1≤x≤e1\le x\le e1≤x≤e is revolved about y=−1y=-1y=−1?

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

Explanation: This problem uses the washer method with rotation about y = -1, which is below both curves. When revolving about y = -1, the outer radius extends from y = -1 up to the horizontal line y = 1, giving R = 1 - (-1) = 2, while the inner radius goes from y = -1 up to the curve y = ln x, giving r = ln x - (-1) = ln x + 1. The washer formula becomes V = π∫[(1+1)² - (ln x + 1)²]dx from 1 to e. Choice B incorrectly uses (1)² - (ln x)², failing to account for the shift by 1 unit. Remember that when rotating about y = k, every y-coordinate must be adjusted by subtracting k to find the distance from the axis.

Question 10

Which integral gives the volume when the region between y=xy=\sqrt{x}y=x​ and y=0y=0y=0 from x=0x=0x=0 to x=4x=4x=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[(x)2−(2)2]dxV=\pi\int_{0}^{4}\Big[(\sqrt{x})^2-(2)^2\Big]dxV=π∫04​[(x​)2−(2)2]dx
  4. 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
  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 rotation about the line y = -2, below the x-axis. When revolving about y = -2, we need distances from this axis: the outer radius extends from y = -2 up to y = √x, giving R = √x - (-2) = √x + 2, while the inner radius is from y = -2 up to y = 0, giving r = 0 - (-2) = 2. The washer formula yields V = π∫[(√x + 2)² - (2)²]dx from 0 to 4. Choice A incorrectly uses (√x)² - (0)², completely ignoring the shift to y = -2. For rotation about y = k, always compute radii as |y - k|, where y represents the function values and k is the axis of rotation.

Question 11

Choose the washer-method setup for revolving the region between y=exy=e^xy=ex and y=2y=2y=2 on [0,ln⁡2][0,\ln 2][0,ln2] about y=1y=1y=1.

  1. V=π∫0ln⁡2[(2−1)2−(ex−1)2]dxV=\pi\int_{0}^{\ln 2}\Big[(2-1)^2-(e^x-1)^2\Big]dxV=π∫0ln2​[(2−1)2−(ex−1)2]dx (correct answer)
  2. V=π∫0ln⁡2[(2)2−(ex)2]dxV=\pi\int_{0}^{\ln 2}\Big[(2)^2-(e^x)^2\Big]dxV=π∫0ln2​[(2)2−(ex)2]dx
  3. V=π∫0ln⁡2[(2−ex)2−(1)2]dxV=\pi\int_{0}^{\ln 2}\Big[(2-e^x)^2-(1)^2\Big]dxV=π∫0ln2​[(2−ex)2−(1)2]dx
  4. V=π∫0ln⁡2[(ex−1)2−(2−1)2]dxV=\pi\int_{0}^{\ln 2}\Big[(e^x-1)^2-(2-1)^2\Big]dxV=π∫0ln2​[(ex−1)2−(2−1)2]dx
  5. V=π∫0ln⁡2[(2−1)2−(1−ex)2]dxV=\pi\int_{0}^{\ln 2}\Big[(2-1)^2-(1-e^x)^2\Big]dxV=π∫0ln2​[(2−1)2−(1−ex)2]dx

Explanation: This problem involves the washer method revolving about y = 1, which lies between the exponential curve and the horizontal line. Since y = 2 is above y = 1 and y = eˣ starts at e⁰ = 1 and increases to e^(ln 2) = 2, the outer radius is R = 2 - 1 = 1 throughout, while the inner radius is r = eˣ - 1. The washer formula gives V = π∫[(2-1)² - (eˣ-1)²]dx from 0 to ln 2. Choice B incorrectly uses (2)² - (eˣ)² without shifting by the axis location. When rotating about y = k between two curves, always measure distances as |y - k| from the axis to each curve.

Question 12

For 0≤x≤10\le x\le10≤x≤1, region between y=2−xy=2-xy=2−x and y=1y=1y=1 is revolved about y=−2y=-2y=−2. Which integral represents the volume?

  1. V=π∫01[(2−x+2)2−(1+2)2]dxV=\pi\int_{0}^{1}\big[(2-x+2)^2-(1+2)^2\big]dxV=π∫01​[(2−x+2)2−(1+2)2]dx (correct answer)
  2. V=π∫01[(1+2)2−(2−x+2)2]dxV=\pi\int_{0}^{1}\big[(1+2)^2-(2-x+2)^2\big]dxV=π∫01​[(1+2)2−(2−x+2)2]dx
  3. V=π∫01[(2−x)2−12]dxV=\pi\int_{0}^{1}\big[(2-x)^2-1^2\big]dxV=π∫01​[(2−x)2−12]dx
  4. V=π∫01[(−2−(2−x))2−(−2−1)2]dxV=\pi\int_{0}^{1}\big[(-2-(2-x))^2-(-2-1)^2\big]dxV=π∫01​[(−2−(2−x))2−(−2−1)2]dx
  5. V=π∫01[(4−x)−3]2dxV=\pi\int_{0}^{1}\big[(4-x)-3\big]^2dxV=π∫01​[(4−x)−3]2dx

Explanation: This problem involves using the washer method to find the volume of a solid formed by revolving a region around a shifted horizontal axis, specifically y = -2. To adjust for the shifted axis, add 2 to each y-value, making the outer radius (2 - x) + 2 = 4 - x for the upper curve. The inner radius is 1 + 2 = 3 for the lower curve. These adjustments reflect distances from below. A tempting distractor is choice B, which reverses the order. A transferable strategy for washer method with shifted axes is to always compute distances by subtracting the axis value from each curve's y-value, determine outer as the larger distance and inner as the smaller, and integrate π times (outer² - inner²) dx.

Question 13

For 1≤x≤31\le x\le31≤x≤3, the region between y=ln⁡xy=\ln xy=lnx and y=0y=0y=0 is revolved about y=2y=2y=2. Which is the correct washer setup?

  1. V=π∫13[(2−ln⁡x)2−(2−0)2]dxV=\pi\int_{1}^{3}\big[(2-\ln x)^2-(2-0)^2\big]dxV=π∫13​[(2−lnx)2−(2−0)2]dx
  2. V=π∫13[(ln⁡x)2−02]dxV=\pi\int_{1}^{3}\big[(\ln x)^2-0^2\big]dxV=π∫13​[(lnx)2−02]dx
  3. V=π∫13[(2−0)2−(2−ln⁡x)2]dxV=\pi\int_{1}^{3}\big[(2-0)^2-(2-\ln x)^2\big]dxV=π∫13​[(2−0)2−(2−lnx)2]dx (correct answer)
  4. V=π∫13[(2+ln⁡x)2−(2+0)2]dxV=\pi\int_{1}^{3}\big[(2+\ln x)^2-(2+0)^2\big]dxV=π∫13​[(2+lnx)2−(2+0)2]dx
  5. V=π∫13[(2−ln⁡x)−(2−0)]2dxV=\pi\int_{1}^{3}\big[(2-\ln x)-(2-0)\big]^2dxV=π∫13​[(2−lnx)−(2−0)]2dx

Explanation: The washer method for volumes of revolution around a shifted horizontal axis involves adjusting the radii to account for the distance from each curve to the axis. For revolution around y=2, both radii are found by subtracting the curve from 2 since the region is below, giving 2 - 0 and 2 - ln x. This makes the constant outer and the varying inner. The order ensures positive volume. A tempting distractor like option A fails by reversing the radii, causing a negative integrand. A transferable strategy for shifted washer problems is to use axis minus curve for regions below and confirm outer is larger distance.

Question 14

For 0≤x≤10\le x\le10≤x≤1, region between y=2xy=2xy=2x and y=x3y=x^3y=x3 is revolved about y=−1y=-1y=−1. Which integral represents the volume?

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

Explanation: This problem involves using the washer method to find the volume of a solid formed by revolving a region around a shifted horizontal axis, specifically y = -1. To adjust for the shifted axis, add 1 to y, outer 2x +1 for upper. Inner x^3 +1 for lower. Adjustments correct. A tempting distractor is choice B, reversing. A transferable strategy for washer method with shifted axes is to always compute distances by subtracting the axis value from each curve's y-value, determine outer as the larger distance and inner as the smaller, and integrate π times (outer² - inner²) dx.

Question 15

The region between y=x2y=x^2y=x2 and y=2xy=2xy=2x for 0≤x≤20\le x\le20≤x≤2 is revolved about y=−3y=-3y=−3. Which integral represents the volume?

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

Explanation: The washer method for volumes of revolution around a shifted horizontal axis involves adjusting the radii to account for the distance from each curve to the axis. For revolution around y=-3, both radii are curve plus 3 since above, giving 2x + 3 and x² + 3. The outer is 2x + 3 as upper. This ensures proper radius lengths. A tempting distractor like option A fails by swapping, causing negative volume. A transferable strategy for shifted washer problems is to add the shift magnitude when axis is below the region.

Question 16

For 0≤x≤10\le x\le10≤x≤1, the region between y=1−x2y=\sqrt{1-x^2}y=1−x2​ and y=0y=0y=0 is revolved about y=−1y=-1y=−1. Choose the volume integral.

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

Explanation: The washer method for volumes of revolution around a shifted horizontal axis involves adjusting the radii to account for the distance from each curve to the axis. For revolution around y=-1, both radii are curve plus 1 since above, giving √(1-x²) + 1 and 1. The outer is the varying one. This adjustment is key for shifted axis. A tempting distractor like option B fails by reversing, negative volume. A transferable strategy for shifted washer problems is to adjust by the shift and determine order by distances.

Question 17

Region bounded by y=exy=e^xy=ex and y=1y=1y=1 for 0≤x≤ln⁡30\le x\le\ln 30≤x≤ln3 is revolved about y=4y=4y=4. Which is the correct setup?

  1. V=π∫0ln⁡3[(4−ex)2−(4−1)2]dxV=\pi\int_{0}^{\ln 3}\big[(4-e^x)^2-(4-1)^2\big]dxV=π∫0ln3​[(4−ex)2−(4−1)2]dx
  2. V=π∫0ln⁡3[(4−1)2−(4−ex)2]dxV=\pi\int_{0}^{\ln 3}\big[(4-1)^2-(4-e^x)^2\big]dxV=π∫0ln3​[(4−1)2−(4−ex)2]dx (correct answer)
  3. V=π∫0ln⁡3[(ex)2−12]dxV=\pi\int_{0}^{\ln 3}\big[(e^x)^2-1^2\big]dxV=π∫0ln3​[(ex)2−12]dx
  4. V=π∫0ln⁡3[(ex−4)2−(1−4)2]dxV=\pi\int_{0}^{\ln 3}\big[(e^x-4)^2-(1-4)^2\big]dxV=π∫0ln3​[(ex−4)2−(1−4)2]dx
  5. V=π∫0ln⁡3[(4−ex)−(4−1)]2dxV=\pi\int_{0}^{\ln 3}\big[(4-e^x)-(4-1)\big]^2dxV=π∫0ln3​[(4−ex)−(4−1)]2dx

Explanation: The washer method for volumes of revolution around a shifted horizontal axis involves adjusting the radii to account for the distance from each curve to the axis. For revolution around y=4, both radii are 4 minus curves since below, giving 4 - 1 and 4 - e^x. The outer is constant 3. This captures the washer correctly. A tempting distractor like option A fails by reversing, negative integrand. A transferable strategy for shifted washer problems is to use axis minus curve for below-axis regions and outer as farther.

Question 18

Region bounded by y=x2y=\frac{x}{2}y=2x​ and y=x4+1y=\frac{x}{4}+1y=4x​+1 for 0≤x≤40\le x\le40≤x≤4 is revolved about y=−1y=-1y=−1. Which setup is correct?

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

Explanation: This problem involves using the washer method to find the volume of a solid formed by revolving a region around a shifted horizontal axis, specifically y = -1. To adjust, add 1, outer x/4 +1 +1 = x/4 +2, inner x/2 +1. Yes B. A tempting distractor is choice A, reversing. A transferable strategy for washer method with shifted axes is to always compute distances by subtracting the axis value from each curve's y-value, determine outer as the larger distance and inner as the smaller, and integrate π times (outer² - inner²) dx.

Question 19

The region between y=x+2y=x+2y=x+2 and y=x2+2y=x^2+2y=x2+2 on 0≤x≤10\le x\le10≤x≤1 is revolved about y=1y=1y=1. Choose the volume integral.

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

Explanation: The washer method for volumes of revolution around a shifted horizontal axis involves adjusting the radii to account for the distance from each curve to the axis. For revolution around y=1, both radii are calculated as the curve minus 1 since the region is above the axis, yielding x + 1 and x² + 1. This adjustment positions the radii relative to the shifted axis. The outer is the larger, which is x + 1 here. A tempting distractor like option D fails because it assumes the region is below the axis, leading to incorrect order. A transferable strategy for shifted washer problems is to subtract the axis value from the curves when above and ensure the larger radius is outer.

Question 20

For 0≤x≤10\le x\le10≤x≤1, region between y=3−xy=3-xy=3−x and y=2−x2y=2-x^2y=2−x2 is revolved about y=6y=6y=6. Which integral gives the volume?

  1. V=π∫01[(6−(3−x))2−(6−(2−x2))2]dxV=\pi\int_{0}^{1}\big[(6-(3-x))^2-(6-(2-x^2))^2\big]dxV=π∫01​[(6−(3−x))2−(6−(2−x2))2]dx
  2. V=π∫01[(6−(2−x2))2−(6−(3−x))2]dxV=\pi\int_{0}^{1}\big[(6-(2-x^2))^2-(6-(3-x))^2\big]dxV=π∫01​[(6−(2−x2))2−(6−(3−x))2]dx (correct answer)
  3. V=π∫01[(3−x)2−(2−x2)2]dxV=\pi\int_{0}^{1}\big[(3-x)^2-(2-x^2)^2\big]dxV=π∫01​[(3−x)2−(2−x2)2]dx
  4. V=π∫01[(6−(3−x))−(6−(2−x2))]2dxV=\pi\int_{0}^{1}\big[(6-(3-x))-(6-(2-x^2))\big]^2dxV=π∫01​[(6−(3−x))−(6−(2−x2))]2dx
  5. V=π∫01[((3−x)−6)2−((2−x2)−6)2]dxV=\pi\int_{0}^{1}\big[((3-x)-6)^2-((2-x^2)-6)^2\big]dxV=π∫01​[((3−x)−6)2−((2−x2)−6)2]dx

Explanation: This problem involves using the washer method to find the volume of a solid formed by revolving a region around a shifted horizontal axis, specifically y = 6. To adjust, subtract y from 6, outer 6 - (2 - x^2), inner 6 - (3 - x). Yes B. A tempting distractor is choice A, swapping. A transferable strategy for washer method with shifted axes is to always compute distances by subtracting the axis value from each curve's y-value, determine outer as the larger distance and inner as the smaller, and integrate π times (outer² - inner²) dx.