Washer Method: Revolving Around Other Axes
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AP Calculus BC › Washer Method: Revolving Around Other Axes
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 $y=-1$?
$\pi\int_{0}^{e-1}\big[(\ln(x+1)+1)-1\big]^2,dx$
$\pi\int_{0}^{e-1}\big[(1)^2-(\ln(x+1)+1)^2\big],dx$
$\pi\int_{0}^{e-1}\big[(\ln(x+1))^2-(0)^2\big],dx$
$\pi\int_{0}^{e-1}\big[(\ln(x+1)+1)^2-(1)^2\big],dx$
$\pi\int_{0}^{e-1}\big[(\ln(x+1)-1)^2-(0-1)^2\big],dx$
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 $y=-1$?
$\pi\int_{0}^{1}\big[(x+1)^2-(x^2+1)^2\big],dx$
$\pi\int_{0}^{1}\big[(x^2-1)^2-(x-1)^2\big],dx$
$\pi\int_{0}^{1}\big[(x^2+1)-(x+1)\big]^2,dx$
$\pi\int_{0}^{1}\big[(x^2+1)^2-(x+1)^2\big],dx$
$\pi\int_{0}^{1}\big[x^2-(x^2)^2\big],dx$
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 $y=\sqrt{x}$ and $y=0$, $0\le x\le4$, is revolved about $y=-2$?
$\pi\int_{0}^{4}(\sqrt{x})^2,dx$
$\pi\int_{0}^{4}\big[(\sqrt{x}+2)-2\big]^2,dx$
$\pi\int_{0}^{4}\big[(2)^2-(\sqrt{x}+2)^2\big],dx$
$\pi\int_{0}^{4}\big[(\sqrt{x}-2)^2-(0-2)^2\big],dx$
$\pi\int_{0}^{4}\big[(\sqrt{x}+2)^2-(2)^2\big],dx$
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.
What integral gives the volume when the region between $y=x^3$ and $y=x$, $0\le x\le1$, is revolved about $y=2$?
$\pi\int_{0}^{1}\big[(2-x)^2-(2-x^3)^2\big],dx$
$\pi\int_{0}^{1}\big[(2-x)-(2-x^3)\big]^2,dx$
$\pi\int_{0}^{1}\big[(x-2)^2-(x^3-2)^2\big],dx$
$\pi\int_{0}^{1}\big[(2-x^3)^2-(2-x)^2\big],dx$
$\pi\int_{0}^{1}\big[(x)^2-(x^3)^2\big],dx$
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.
What integral gives the volume when the region between $y=x+1$ and $y=1$, $0 \le x \le 3$, is revolved about $y=-1$?
$\pi\int_{0}^{3}\big[(x+1-(-1))-(1-(-1))\big]^2,dx$
$\pi\int_{0}^{3}\big[(x+1)^2-(1)^2\big],dx$
$\pi\int_{0}^{3}\big[(x)^2-(0)^2\big],dx$
$\pi\int_{0}^{3}\big[(x+2)^2-(2)^2\big],dx$
$\pi\int_{0}^{3}\big[(2)^2-(x+2)^2\big],dx$
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 $y = x + 1$ to the axis, giving $x + 2$. The inner radius is from the lower curve $y = 1$ 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.
Select the correct washer setup for revolving the region between $y=\ln x$ and $y=0$ on $1,e$ about $y=1$.
$V=\pi\displaystyle\int_{0}^{1}\Big[(e^y-1)^2-(1-1)^2\Big]dy$
$V=\pi\displaystyle\int_{1}^{e}\Big[(1-0)^2-(1-\ln x)^2\Big]dx$
$V=\pi\displaystyle\int_{1}^{e}\Big[(\ln x)^2-0^2\Big]dx$
$V=\pi\displaystyle\int_{1}^{e}\Big[(\ln x+1)^2-(0+1)^2\Big]dx$
$V=\pi\displaystyle\int_{1}^{e}\Big[(1-\ln x)^2-(1-0)^2\Big]dx$
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.
Which integral gives the volume when the region between $y=1-x^2$ and $y=0$ on $-1,1$ rotates about $y=-3$?
$V=\pi\displaystyle\int_{-1}^{1}\Big[(0+3)^2-(1-x^2+3)^2\Big]dx$
$V=\pi\displaystyle\int_{-1}^{1}\Big[(1-x^2)^2-0^2\Big]dx$
$V=\pi\displaystyle\int_{-1}^{1}\Big[(1-x^2+3)^2-(0+3)^2\Big]dx$
$V=\pi\displaystyle\int_{-1}^{1}\Big[(1-x^2-3)^2-(0-3)^2\Big]dx$
$V=\pi\displaystyle\int_{0}^{1}\Big[(3)^2-(3-y)^2\Big]dy$
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.
What integral setup gives the volume when the region between $y=2$ and $y=x^3$ on $0,\sqrt[3{2}]$ rotates about $y=-1$?
$V=\pi\displaystyle\int_{0}^{\sqrt[3]{2}}\Big[2^2-(x^3)^2\Big]dx$
$V=\pi\displaystyle\int_{0}^{\sqrt[3]{2}}\Big[(2-(-1))^2-(x^3-(-1))^2\Big]dx$
$V=\pi\displaystyle\int_{0}^{\sqrt[3]{2}}\Big[(2+1)^2-(x^3+1)^2\Big]dx$
$V=\pi\displaystyle\int_{0}^{\sqrt[3]{2}}\Big[(x^3+1)^2-(2+1)^2\Big]dx$
$V=\pi\displaystyle\int_{-1}^{2}\Big[(\sqrt[3]{2}-0)^2-(\sqrt[3]{y}-0)^2\Big]dy$
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.
Which integral represents the volume when the region between $y=\sqrt{x}$ and $y=0$ on $0,4$ rotates about $y=-2$?
$V=\pi\displaystyle\int_{0}^{4}\Big[(\sqrt{x}-2)^2-(0-2)^2\Big]dx$
$V=\pi\displaystyle\int_{0}^{2}\Big[(4-(-2))^2-(y^2-(-2))^2\Big]dy$
$V=\pi\displaystyle\int_{0}^{4}\Big[(0+2)^2-(\sqrt{x}+2)^2\Big]dx$
$V=\pi\displaystyle\int_{0}^{4}\Big[(\sqrt{x}+2)^2-(0+2)^2\Big]dx$
$V=\pi\displaystyle\int_{0}^{4}\Big[(\sqrt{x})^2-0^2\Big]dx$
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.