Disc Method: Revolving Around Other Axes
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AP Calculus BC › Disc Method: Revolving Around Other Axes
Which integral represents the volume when the region bounded by $y=3x-x^2$ and $y=0$ is revolved about $y=4$?
$V=\pi\int_{0}^{3}(4-(3x-x^2))^2,dx$
$V=\pi\int_{0}^{3}\big((4-(3x-x^2))^2-4^2\big),dx$
$V=\pi\int_{0}^{4}(4-y)^2,dy$
$V=\pi\int_{0}^{3}(3x-x^2)^2,dx$
$V=\pi\int_{0}^{3}\big(4^2-(4-(3x-x^2))^2\big),dx$
Explanation
This problem involves the disc method for volumes of revolution around a shifted axis, specifically around y=4. With the axis shifted to y=4, the radii become distances from y=4 to the boundaries. The outer radius is 4 - 0 =4, to y=0, and the inner radius is 4 - (3x - $x^2$), to the upper curve. The volume is π times the integral of $(4^2$ - (4 - (3x - $x^2$$))^2$) from 0 to 3. A tempting distractor is choice A, which uses only the inner radius without subtracting from the outer. A transferable axis-shift strategy is to subtract the axis value from the curve functions to find the effective radii.
Let $R$ be bounded by $y=\cos x$ and $y=0$ on $0\le x\le\tfrac{\pi}{2}$. Revolve $R$ about $y=2$. Which setup is correct?
$V=\pi\int_{0}^{\pi/2}\Big(\big(\cos x-2\big)^2-\big(0-2\big)^2\Big),dx$
$V=\pi\int_{0}^{2}\Big(\big(2-\arccos y\big)^2-\big(2-0\big)^2\Big),dy$
$V=\pi\int_{0}^{\pi/2}\Big(\big(2-0\big)^2-\big(2-\cos x\big)^2\Big),dx$
$V=\pi\int_{0}^{\pi/2}\big(\cos x\big)^2,dx$
$V=\pi\int_{0}^{\pi/2}\big(2-\cos x\big)^2,dx$
Explanation
This problem involves the disc method with revolution around y = 2, above the region. The region R is bounded by y = cos x and y = 0 on [0, π/2]. When revolving around y = 2, the outer radius extends from y = 2 down to y = 0: R_outer = 2 - 0 = 2, and the inner radius extends from y = 2 down to y = cos x: R_inner = 2 - cos x. The volume formula is V = π∫₀^(π/2)[2² - (2-cos x)²]dx. Choice A incorrectly uses only the inner radius, missing the washer structure. For revolution around a horizontal line above the region, subtract each boundary's y-value from the axis to find radii, with the lower boundary giving the larger radius.
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(2^2-(2-\sqrt{x})^2\big),dx$
$V=\pi\int_{0}^{2}(4-y^2)^2,dy$
$V=\pi\int_{0}^{4}(2-\sqrt{x})^2,dx$
$V=\pi\int_{0}^{4}(\sqrt{x})^2,dx$
$V=\pi\int_{0}^{4}(2+\sqrt{x})^2,dx$
Explanation
This problem involves the disc method for volumes of revolution around a shifted axis, specifically around y=2. When the axis is shifted to y=2, the radii change to reflect distances from this line to the region boundaries. The outer radius is 2 - 0 =2, the distance to y=0, and the inner radius is 2 - √x, the distance to y=√x. Thus, the volume is π times the integral of $(2^2$ - (2 - $√x)^2$) from 0 to 4. A tempting distractor is choice B, which uses only the inner radius, ignoring the washer hole structure. A transferable axis-shift strategy is to subtract the axis value from the curve functions to find the effective radii.
Select the correct disc/washer setup for revolving the region bounded by $y=\sin x$, $y=0$, $x=0$, $x=\pi$ about $y=-1$.
$V=\pi\int_{0}^{\pi}(1-\sin x)^2,dx$
$V=\pi\int_{0}^{1}(\arcsin y)^2,dy$
$V=\pi\int_{0}^{\pi}(\sin x)^2,dx$
$V=\pi\int_{0}^{\pi}(1+\sin x)^2,dx$
$V=\pi\int_{0}^{\pi}\big((1+\sin x)^2-1^2\big),dx$
Explanation
This problem involves the disc method for volumes of revolution around a shifted axis, specifically around y=-1. When the axis is shifted to y=-1, the radii are adjusted to distances from this line. The outer radius is sin x +1, the distance to y=sin x, and the inner radius is 1, the distance to y=0. The volume integral is π times the integral of ((sin x $+1)^2$ - $1^2$) from 0 to π. A tempting distractor is choice B, which subtracts instead of adding the shift, leading to incorrect radii. A transferable axis-shift strategy is to add or subtract the shift value to the curve functions depending on the direction.
What is the correct washer-method setup when the region bounded by $y=x^2+1$ and $y=5$ is revolved about $y=0$?
$V=\pi\int_{-2}^{2}\big((x^2+1)^2-5^2\big),dx$
$V=\pi\int_{-2}^{2}\big(5^2-(x^2+1)^2\big),dx$
$V=\pi\int_{-2}^{2}(x^2+1)^2,dx$
$V=\pi\int_{1}^{5}y^2,dy$
$V=\pi\int_{-2}^{2}(5-(x^2+1))^2,dx$
Explanation
This problem involves the disc method for volumes of revolution around a shifted axis, specifically around y=0. Although y=0 is the standard x-axis, the setup treats it as a base case for shifts. The outer radius is 5, distance to y=5, and inner is $x^2$ +1, to $y=x^2$ +1. The volume integral is π times $(5^2$ - $(x^2$ $+1)^2$) from -2 to 2. A tempting distractor is choice C, which uses an incorrect outer radius by shifting wrongly. A transferable axis-shift strategy is to subtract the axis value from the curve functions to find the effective radii.
Region $R$ is bounded by $y=\sqrt{x}$, $y=0$, and $x=9$. Revolve $R$ about $y=2$. Which volume setup is correct?
$V=\pi\int_{0}^{9}\big(\sqrt{x}\big)^2,dx$
$V=\pi\int_{0}^{9}\Big(\big(\sqrt{x}-2\big)^2-\big(0-2\big)^2\Big),dx$
$V=\pi\int_{0}^{9}\big(2-\sqrt{x}\big)^2,dx$
$V=\pi\int_{0}^{3}\Big(\big(2-y^2\big)^2-\big(2-0\big)^2\Big),dy$
$V=\pi\int_{0}^{9}\Big(\big(2-0\big)^2-\big(2-\sqrt{x}\big)^2\Big),dx$
Explanation
This problem involves the disc method with revolution around y = 2, which is above the region. The region R is bounded by y = √x (top curve), y = 0 (bottom curve), and x = 9. When revolving around y = 2, the outer radius is from y = 2 down to y = 0: R_outer = 2 - 0 = 2, and the inner radius is from y = 2 down to y = √x: R_inner = 2 - √x. The volume formula becomes V = π∫₀⁹[(2)² - (2-√x)²]dx. Choice B incorrectly uses only the inner radius, forgetting about the washer shape. When the axis of revolution is outside the region, identify which curve is closer to create the inner hole and which is farther to create the outer disc.
Let $R$ be bounded by $y=\sin x$ and $y=0$ on $0\le x\le\pi$. Revolve $R$ about $y=-1$. Which setup gives the volume?
$V=\pi\int_{0}^{\pi}\big(\sin x\big)^2,dx$
$V=\pi\int_{-1}^{0}\big(\pi\big)^2,dy$
$V=\pi\int_{0}^{\pi}\Big(1^2-\big(\sin x+1\big)^2\Big),dx$
$V=\pi\int_{0}^{\pi}\big(\sin x+1\big)^2,dx$
$V=\pi\int_{0}^{\pi}\Big(\big(\sin x+1\big)^2-1^2\Big),dx$
Explanation
This problem uses the disc method with revolution around y = -1, below the region. The region R is bounded by y = sin x and y = 0 on [0, π]. When revolving around y = -1, the outer radius extends from y = -1 up to y = sin x: R_outer = sin x - (-1) = sin x + 1, and the inner radius extends from y = -1 up to y = 0: R_inner = 0 - (-1) = 1. The volume is V = π∫₀^π[(sin x + 1)² - 1²]dx. Choice E incorrectly omits the inner radius term, treating it as a solid disc rather than a washer. For revolution around a line below the region, use (curve - axis) for distances and remember that both boundaries create radii.
Which integral represents the volume when the region under $y=\ln(x+1)$ from $x=0$ to $x=e-1$ is revolved about $y=-1$?
$\pi\displaystyle\int_{0}^{e-1}(\ln(x+1)+1)^2,dx$
$\pi\displaystyle\int_{0}^{e-1}(\ln(x+1)+1),dx$
$\pi\displaystyle\int_{0}^{e-1}(\ln(x+1)-1)^2,dx$
$\pi\displaystyle\int_{0}^{e-1}(\ln(x+1))^2,dx$
$\pi\displaystyle\int_{0}^{e-1}(1-\ln(x+1))^2,dx$
Explanation
This problem involves the disc method for finding volumes of revolution around a shifted axis, specifically around y = -1 instead of the x-axis. When the axis is shifted down to y = -1, the radius of each disc is the distance from the curve y = ln(x+1) to the axis, which is ln(x+1) - (-1) = ln(x+1) + 1. This addition reflects the downward shift, making the radius larger by 1. The integrand is π(ln(x+1) + 1)², integrated from 0 to e-1. A tempting distractor is choice B, π ∫ (ln(x+1) - 1)² dx, which would apply to an upward shift to y = 1 but not here. A transferable strategy for axis shifts in the disc method is to calculate the radius as the absolute difference between the function value and the axis position, ordering it to ensure a positive value.
Choose the correct volume setup for revolving the region bounded by $y=x^3$, $y=0$, $x=0$, $x=1$ about $y=2$.
$V=\pi\int_{0}^{1}(2-x^3)^2,dx$
$V=\pi\int_{0}^{1}\big(2^2-(2-x^3)^2\big),dx$
$V=\pi\int_{0}^{2}(1-y^{1/3})^2,dy$
$V=\pi\int_{0}^{1}(2+x^3)^2,dx$
$V=\pi\int_{0}^{1}(x^3)^2,dx$
Explanation
This problem involves the disc method for volumes of revolution around a shifted axis, specifically around y=2. With the axis at y=2, radii are distances from this line to the boundaries. The outer radius is 2 - 0 =2, to y=0, and inner is 2 - $x^3$, to $y=x^3$. The volume is π times $(2^2$ - (2 - $x^3$$)^2$) from 0 to 1. A tempting distractor is choice A, which uses only the inner radius without the washer setup. A transferable axis-shift strategy is to subtract the axis value from the curve functions to find the effective radii.
Choose the disc-method setup for revolving the region under $y=e^x$ from $x=0$ to $x=1$ about $y=3$.
$\pi\displaystyle\int_{0}^{1}(3+e^x)^2,dx$
$\pi\displaystyle\int_{0}^{1}(e^x)^2,dx$
$\pi\displaystyle\int_{0}^{1}(e^x-3)^2,dx$
$\pi\displaystyle\int_{0}^{1}(3-e^x)^2,dx$
$\pi\displaystyle\int_{0}^{1}(3-e^x),dx$
Explanation
This problem involves the disc method for finding volumes of revolution around a shifted axis, specifically around y = 3 instead of the x-axis. When the axis is shifted up to y = 3, the radius of each disc is the distance from the curve y = $e^x$ to the axis, which is 3 - $e^x$. This subtraction is used because the axis is above the curve, ensuring a positive radius since $e^x$ < 3 in [0,1]. The integrand becomes π(3 - $e^x$)², integrated from 0 to 1. A tempting distractor is choice D, π ∫ (3 + $e^x$)² dx, which might arise from incorrectly adding for an upward shift instead of subtracting. A transferable strategy for axis shifts in the disc method is to calculate the radius as the absolute difference between the function value and the axis position, ordering it to ensure a positive value.