Washer Method: Revolving Around Other Axes
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AP Calculus AB › Washer Method: Revolving Around Other Axes
The region between $y=4$ and $y=\sqrt{x}$ for $0\le x\le16$ is revolved about $y=6$. Which integral gives the volume?
$V=\pi\int_{0}^{16}\big[(6-4)-(6-\sqrt{x})\big]^2dx$
$V=\pi\int_{0}^{16}\big[(6-\sqrt{x})^2-(6-4)^2\big]dx$
$V=\pi\int_{0}^{16}\big[(4-6)^2-(\sqrt{x}-6)^2\big]dx$
$V=\pi\int_{0}^{16}\big[(6-4)^2-(6-\sqrt{x})^2\big]dx$
$V=\pi\int_{0}^{16}\big[4^2-(\sqrt{x})^2\big]dx$
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.
Region bounded by $y=1-x$ and $y=1-x^2$ for $0\le x\le1$ is revolved about $y=3$. Choose the correct volume setup.
$V=\pi\int_{0}^{1}\big[(3-(1-x^2))^2-(3-(1-x))^2\big]dx$
$V=\pi\int_{0}^{1}\big[((1-x)-3)^2-((1-x^2)-3)^2\big]dx$
$V=\pi\int_{0}^{1}\big[(1-x)^2-(1-x^2)^2\big]dx$
$V=\pi\int_{0}^{1}\big[(3-(1-x))-(3-(1-x^2))\big]^2dx$
$V=\pi\int_{0}^{1}\big[(3-(1-x))^2-(3-(1-x^2))^2\big]dx$
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.
For $-1\le x\le1$, the region between $y=x^2+1$ and $y=1$ is revolved about $y=4$. Which integral represents the volume?
$V=\pi\int_{-1}^{1}\big[(4-(x^2+1))-(4-1)\big]^2dx$
$V=\pi\int_{-1}^{1}\big[(x^2+1-4)^2-(1-4)^2\big]dx$
$V=\pi\int_{-1}^{1}\big[(4-(x^2+1))^2-(4-1)^2\big]dx$
$V=\pi\int_{-1}^{1}\big[(4-1)^2-(4-(x^2+1))^2\big]dx$
$V=\pi\int_{-1}^{1}\big[(x^2+1)^2-1^2\big]dx$
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=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.
Let $R$ be bounded by $y=4-x$ and $y=x$ on $0 \leq x \leq 2$ and revolved about $y=5$. Which setup gives the volume?
$V=\pi\int_{0}^{2}\big[(5-(4-x))^2-(5-x)^2\big]dx$
$V=\pi\int_{0}^{2}\big[(5-(4-x))-(5-x)\big]^2dx$
$V=\pi\int_{0}^{2}\big[(4-x)^2-x^2\big]dx$
$V=\pi\int_{0}^{2}\big[(5-x)^2-(5-(4-x))^2\big]dx$
$V=\pi\int_{0}^{2}\big[(x-5)^2-((4-x)-5)^2\big]dx$
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 $y=5$, both radii are $5 - x$ and $5 - (4 - x)$ since below, giving $5 - x$ and $5 - (4 - x)$. The outer is $5 - x$ 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.
Region bounded by $y=\sqrt{4-x^2}$ and $y=0$ for $-2\le x\le2$ is revolved about $y=-3$. Which setup is correct?
$V=\pi\int_{-2}^{2}\big[(3)^2-(\sqrt{4-x^2}+3)^2\big]dx$
$V=\pi\int_{-2}^{2}\big[(\sqrt{4-x^2}+3)^2-(3)^2\big]dx$
$V=\pi\int_{-2}^{2}\big[(\sqrt{4-x^2})^2-0^2\big]dx$
$V=\pi\int_{-2}^{2}\big[(-3-\sqrt{4-x^2})^2-(-3-0)^2\big]dx$
$V=\pi\int_{-2}^{2}\big[(\sqrt{4-x^2}+3)-3\big]^2dx$
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.
The region between $y=\frac{x^2}{4}$ and $y=1$ for $0\le x\le2$ is revolved about $y=-1$. Which setup is correct?
$V=\pi\int_{0}^{2}\big[(-1-1)^2-(-1-x^2/4)^2\big]dx$
$V=\pi\int_{0}^{2}\big[(x^2/4+1)^2-(2)^2\big]dx$
$V=\pi\int_{0}^{2}\big[(1+1)^2-(x^2/4+1)^2\big]dx$
$V=\pi\int_{0}^{2}\big[(2)-(x^2/4+1)\big]^2dx$
$V=\pi\int_{0}^{2}\big[1^2-(x^2/4)^2\big]dx$
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 = -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.
For $0\le x\le\pi$, the region between $y=\sin x$ and $y=0$ is revolved about $y=1$. Which integral represents the volume?
$V=\pi\int_{0}^{\pi}\big[(\sin x-1)^2-(0-1)^2\big]dx$
$V=\pi\int_{0}^{\pi}\big[(1-\sin x)^2-(1-0)^2\big]dx$
$V=\pi\int_{0}^{\pi}\sin^2x,dx$
$V=\pi\int_{0}^{\pi}\big[(1-0)^2-(1-\sin x)^2\big]dx$
$V=\pi\int_{0}^{\pi}\big[(1-\sin x)-(1-0)\big]^2dx$
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.
Select the correct washer-method integral for revolving the region between $y=\frac{1}{x}$ and $y=\frac{1}{4}$ on $1,4$ about $y=0$.
$V=\pi\int_{1}^{4}\Big[\left(\frac{1}{4}\right)^2-\left(\frac{1}{x}\right)^2\Big]dx$
$V=\pi\int_{1}^{4}\Big[\left(\frac{1}{x}\right)^2-\left(\frac{1}{4}\right)^2\Big]dx$
$V=\pi\int_{1}^{4}\Big[\left(\frac{1}{x}-\frac{1}{4}\right)^2-(0)^2\Big]dx$
$V=\pi\int_{1}^{4}\Big[\left(\frac{1}{x}\right)^2-(0)^2\Big]dx$
$V=\pi\int_{1}^{4}\Big[\left(\frac{1}{4}-0\right)^2-\left(\frac{1}{x}-0\right)^2\Big]dx$
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.
What is the washer-method volume setup when the region between $y=\ln x$ and $y=1$ for $1\le x\le e$ is revolved about $y=-1$?
$V=\pi\int_{1}^{e}\Big[(\ln x+1)^2-(2)^2\Big]dx$
$V=\pi\int_{1}^{e}\Big[(1)^2-(\ln x)^2\Big]dx$
$V=\pi\int_{1}^{e}\Big[(1-(-1))^2-(1-\ln x)^2\Big]dx$
$V=\pi\int_{1}^{e}\Big[(1+1)^2-(\ln x+1)^2\Big]dx$
$V=\pi\int_{1}^{e}\Big[(1+1)^2-(\ln x)^2\Big]dx$
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.
Which integral gives the volume when the region between $y=\sqrt{x}$ and $y=0$ from $x=0$ to $x=4$ is revolved about $y=-2$?
$V=\pi\int_{0}^{4}\Big[(\sqrt{x})^2-(2)^2\Big]dx$
$V=\pi\int_{0}^{4}\Big[(\sqrt{x}-2)^2-(0-2)^2\Big]dx$
$V=\pi\int_{0}^{4}\Big[(\sqrt{x})^2-(0)^2\Big]dx$
$V=\pi\int_{0}^{4}\Big[(2)^2-(\sqrt{x}+2)^2\Big]dx$
$V=\pi\int_{0}^{4}\Big[(\sqrt{x}+2)^2-(2)^2\Big]dx$
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.