Washer Method: Revolving Around Other Axes
Help Questions
AP Calculus BC › Washer Method: Revolving Around Other Axes
What integral gives the volume when the region between $y=\sqrt{x}$ and $y=0$, $0\le x\le4$, is revolved about $y=-2$?
$\pi\int_{0}^{4}(\sqrt{x})^2,dx$
$\pi\int_{0}^{4}\big[(2)^2-(\sqrt{x}+2)^2\big],dx$
$\pi\int_{0}^{4}\big[(\sqrt{x}+2)-2\big]^2,dx$
$\pi\int_{0}^{4}\big[(\sqrt{x}+2)^2-(2)^2\big],dx$
$\pi\int_{0}^{4}\big[(\sqrt{x}-2)^2-(0-2)^2\big],dx$
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 = -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.
Which integral gives the volume when the region between $y=1-x^2$ and $y=0$ on $-1,1$ rotates about $y=-3$?
$V=\pi\displaystyle\int_{-1}^{1}\Big[(0+3)^2-(1-x^2+3)^2\Big]dx$
$V=\pi\displaystyle\int_{-1}^{1}\Big[(1-x^2-3)^2-(0-3)^2\Big]dx$
$V=\pi\displaystyle\int_{-1}^{1}\Big[(1-x^2+3)^2-(0+3)^2\Big]dx$
$V=\pi\displaystyle\int_{-1}^{1}\Big[(1-x^2)^2-0^2\Big]dx$
$V=\pi\displaystyle\int_{0}^{1}\Big[(3)^2-(3-y)^2\Big]dy$
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.
Which integral represents the volume when the region between $y=\sqrt{x}$ and $y=0$ on $0,4$ rotates about $y=-2$?
$V=\pi\displaystyle\int_{0}^{4}\Big[(0+2)^2-(\sqrt{x}+2)^2\Big]dx$
$V=\pi\displaystyle\int_{0}^{4}\Big[(\sqrt{x})^2-0^2\Big]dx$
$V=\pi\displaystyle\int_{0}^{4}\Big[(\sqrt{x}+2)^2-(0+2)^2\Big]dx$
$V=\pi\displaystyle\int_{0}^{4}\Big[(\sqrt{x}-2)^2-(0-2)^2\Big]dx$
$V=\pi\displaystyle\int_{0}^{2}\Big[(4-(-2))^2-(y^2-(-2))^2\Big]dy$
Explanation
This problem uses the washer method with rotation about the line $y = -2$. When rotating about $y = -2$, we calculate distances from this axis: the outer radius extends from $y = -2$ to $y = \sqrt{x}$, giving $R = \sqrt{x} - (-2) = \sqrt{x} + 2$, and the inner radius extends from $y = -2$ to $y = 0$, giving $r = 0 - (-2) = 2$. The volume integral is $\pi\int_{0}^{4}[(\sqrt{x} + 2)^2 - (0 + 2)^2], dx = \pi\int_{0}^{4}[(\sqrt{x} + 2)^2 - 4], dx$. 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.
What integral gives the volume when the region between $y=x^2$ and $y=2x$ on $0,2$ is revolved about $y=-1$?
$\pi\displaystyle\int_{0}^{2}\big[(2x-1)^2-(x^2-1)^2\big],dx$
$\pi\displaystyle\int_{0}^{2}\big[(2x)^2-(x^2)^2\big],dx$
$\pi\displaystyle\int_{0}^{2}\big[(x^2+1)^2-(2x+1)^2\big],dx$
$\pi\displaystyle\int_{0}^{2}\big[(2x+1)^2-(x^2+1)^2\big],dx$
$\pi\displaystyle\int_{0}^{2}\big[(2x+1)-(x^2+1)\big]^2,dx$
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 $2x + 1$ from the upper curve, and the inner radius is $x^2 + 1$ from the lower curve. Therefore, the integral is $\pi \int_{0}^{2} [(2x + 1)^2 - (x^2 + 1)^2] , dx$. 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 $y = k$, compute radii as $|y - k|$ for each curve and identify the outer as the larger distance and inner as the smaller.
Choose the correct volume integral for revolving the region between $y=\sin x$ and $y=0$ on $0,\pi$ about $y=2$.
$V=\pi\displaystyle\int_{0}^{\pi}\Big[(\sin x)^2-0^2\Big]dx$
$V=\pi\displaystyle\int_{0}^{\pi}\Big[(2-\sin x)^2-(2-0)^2\Big]dx$
$V=\pi\displaystyle\int_{0}^{2}\Big[(\arcsin y)^2-(0)^2\Big]dy$
$V=\pi\displaystyle\int_{0}^{\pi}\Big[(2+\sin x)^2-(2+0)^2\Big]dx$
$V=\pi\displaystyle\int_{0}^{\pi}\Big[(2-0)^2-(2-\sin x)^2\Big]dx$
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.
Choose the correct setup for revolving the region between $y=x^3$ and $y=x$ on $0\le x\le1$ about $y=2$.
$V=\pi\displaystyle\int_{0}^{1}\big[(x-2)^2-(x^3-2)^2\big]dx$
$V=\pi\displaystyle\int_{0}^{1}\big[(2-x^3)^2-(2-x)^2\big]dx$
$V=\pi\displaystyle\int_{0}^{1}\big[(2-x)^2-(2)^2\big]dx$
$V=\pi\displaystyle\int_{0}^{1}\big[(x-x^3)^2\big]dx$
$V=\pi\displaystyle\int_{0}^{1}\big[(2-x)^2-(2-x^3)^2\big]dx$
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.
Which integral gives the volume when the region between $y=\sqrt{x}$ and $y=0$ on $0,4$ is revolved about $y=-2$?
$V=\pi\int_{0}^{4}\big[(\sqrt{x}+2)^2-(0+2)^2\big]dx$
$V=\pi\int_{0}^{4}\big[(\sqrt{x}-2)^2-(0-2)^2\big]dx$
$V=\pi\int_{0}^{4}\big[(\sqrt{x}+2)^2-(2)^2\big]dx$
$V=\pi\int_{0}^{4}\big[(2)^2-(\sqrt{x}+2)^2\big]dx$
$V=\pi\int_{0}^{4}\big[(\sqrt{x})^2-(0)^2\big]dx$
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.
What integral gives the volume when the region between $y=\ln(x+1)$ and $y=0$, $0\le x\le e-1$, is revolved about $y=-1$?
$\pi\int_{0}^{e-1}\big[(\ln(x+1))^2-(0)^2\big],dx$
$\pi\int_{0}^{e-1}\big[(1)^2-(\ln(x+1)+1)^2\big],dx$
$\pi\int_{0}^{e-1}\big[(\ln(x+1)+1)-1\big]^2,dx$
$\pi\int_{0}^{e-1}\big[(\ln(x+1)+1)^2-(1)^2\big],dx$
$\pi\int_{0}^{e-1}\big[(\ln(x+1)-1)^2-(0-1)^2\big],dx$
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 = -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.
What integral gives the volume when the region between $y=x$ and $y=x^2$, $0\le x\le1$, is revolved about $y=-1$?
$\pi\int_{0}^{1}\big[(x^2+1)-(x+1)\big]^2,dx$
$\pi\int_{0}^{1}\big[(x^2+1)^2-(x+1)^2\big],dx$
$\pi\int_{0}^{1}\big[(x+1)^2-(x^2+1)^2\big],dx$
$\pi\int_{0}^{1}\big[(x^2-1)^2-(x-1)^2\big],dx$
$\pi\int_{0}^{1}\big[x^2-(x^2)^2\big],dx$
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 = -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.
What integral gives the volume when the region between $y=x^3$ and $y=x$, $0\le x\le1$, is revolved about $y=2$?
$\pi\int_{0}^{1}\big[(x)^2-(x^3)^2\big],dx$
$\pi\int_{0}^{1}\big[(2-x)^2-(2-x^3)^2\big],dx$
$\pi\int_{0}^{1}\big[(2-x)-(2-x^3)\big]^2,dx$
$\pi\int_{0}^{1}\big[(x-2)^2-(x^3-2)^2\big],dx$
$\pi\int_{0}^{1}\big[(2-x^3)^2-(2-x)^2\big],dx$
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 = 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.