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 $y=4$?
$V=\pi\int_{0}^{3}(3x-x^2)^2,dx$
$V=\pi\int_{0}^{3}(4-(3x-x^2))^2,dx$
$V=\pi\int_{0}^{4}(4-y)^2,dy$
$V=\pi\int_{0}^{3}\big(4^2-(4-(3x-x^2))^2\big),dx$
$V=\pi\int_{0}^{3}\big((4-(3x-x^2))^2-4^2\big),dx$
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?
$V=\pi\int_{0}^{\pi/2}\big(\cos x\big)^2,dx$
$V=\pi\int_{0}^{\pi/2}\Big(\big(\cos x-2\big)^2-\big(0-2\big)^2\Big),dx$
$V=\pi\int_{0}^{\pi/2}\big(2-\cos x\big)^2,dx$
$V=\pi\int_{0}^{\pi/2}\Big(\big(2-0\big)^2-\big(2-\cos x\big)^2\Big),dx$
$V=\pi\int_{0}^{2}\Big(\big(2-\arccos y\big)^2-\big(2-0\big)^2\Big),dy$
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.
What is the correct washer-method setup when the region bounded by $y=x^2+1$ and $y=5$ is revolved about $y=0$?
$V=\pi\int_{-2}^{2}\big(5^2-(x^2+1)^2\big),dx$
$V=\pi\int_{-2}^{2}(5-(x^2+1))^2,dx$
$V=\pi\int_{-2}^{2}\big((x^2+1)^2-5^2\big),dx$
$V=\pi\int_{-2}^{2}(x^2+1)^2,dx$
$V=\pi\int_{1}^{5}y^2,dy$
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 $y=1-x$, $y=0$, and $x=0$. Revolve $R$ about $y=-2$. Which setup gives the volume?
$V=\pi\int_{0}^{1}\big(1-x\big)^2,dx$
$V=\pi\int_{0}^{1}\Big(\big(1-x\big)+2\Big)^2,dx$
$V=\pi\int_{-2}^{-1}\big(1\big)^2,dy$
$V=\pi\int_{0}^{1}\Big(2^2-\big((1-x)+2\big)^2\Big),dx$
$V=\pi\int_{0}^{1}\Big(\big((1-x)+2\big)^2-2^2\Big),dx$
Explanation
This problem involves the disc method with revolution around y = -2, below the region. The region R is bounded by y = 1 - x, y = 0, and x = 0, forming a triangle. When revolving around y = -2, the outer radius extends from y = -2 up to y = 1 - x: R_outer = (1-x) - (-2) = (1-x) + 2 = 3 - x, and the inner radius extends from y = -2 up to y = 0: R_inner = 0 - (-2) = 2. The volume is V = π∫₀¹[(3-x)² - 2²]dx = π∫₀¹[((1-x)+2)² - 2²]dx. Choice B incorrectly reverses the order of subtraction in the washer formula. When the axis is below the region, use (curve - axis) for distances and apply R_outer² - R_inner² in the integral.
Let $R$ be bounded by $y=x^2$ and $y=4$; revolving $R$ about $y=5$, which integral gives the volume?
$V=\pi\displaystyle\int_{-2}^{2}(5-x^2)^2,dx$
$V=\pi\displaystyle\int_{-2}^{2}(5-4)^2,dx$
$V=\pi\displaystyle\int_{0}^{4}(5-y)^2,dy$
$V=\pi\displaystyle\int_{-2}^{2}(x^2-4)^2,dx$
$V=\pi\displaystyle\int_{-2}^{2}\big((5-x^2)^2-(5-4)^2\big),dx$
Explanation
This problem requires the disc method with a horizontal axis of revolution at y = 5, which is above the region. The region R is bounded by the parabola y = x² and the horizontal line y = 4, with intersection points at x = -2 and x = 2. When revolving around y = 5, the outer radius from the axis to the bottom boundary is (5 - x²), while the inner radius from the axis to the top boundary is (5 - 4) = 1. The washer method gives V = π∫[-2 to 2][(5 - x²)² - 1²]dx. Choice A incorrectly uses only the outer radius, treating this as a solid disc rather than a washer with a hollow center. The key strategy when revolving around a shifted axis is to measure all distances from that new axis, then apply outer² - inner² for washers.
Region $R$ is bounded by $y=x^3$, $y=0$, and $x=2$. Revolve $R$ about $y=-3$. Which setup gives the volume?
$V=\pi\int_{-3}^{5}\big(2\big)^2,dy$
$V=\pi\int_{0}^{2}\big(x^3\big)^2,dx$
$V=\pi\int_{0}^{2}\big(x^3+3\big)^2,dx$
$V=\pi\int_{0}^{2}\Big(\big(x^3+3\big)^2-3^2\Big),dx$
$V=\pi\int_{0}^{2}\Big(3^2-\big(x^3+3\big)^2\Big),dx$
Explanation
This problem uses the disc method with revolution around y = -3, below the region. The region R is bounded by y = x³, y = 0, and x = 2. When revolving around y = -3, the outer radius extends from y = -3 up to y = x³: R_outer = x³ - (-3) = x³ + 3, and the inner radius extends from y = -3 up to y = 0: R_inner = 0 - (-3) = 3. The volume is V = π∫₀²[(x³ + 3)² - 3²]dx. Choice E incorrectly omits the inner radius term 3², treating the solid as having no hole. When revolving around a line below the region, use (curve - axis) for distances and remember that the x-axis boundary creates an inner hole.
Region $R$ is bounded by $y=3-x$ and $y=1$ for $0\le x\le2$; revolving about $y=0$, which volume integral is correct?
$V=\pi\displaystyle\int_{0}^{2}\big(1^2-(3-x)^2\big),dx$
$V=\pi\displaystyle\int_{1}^{3}(y)^2,dy$
$V=\pi\displaystyle\int_{0}^{2}(3-x)^2,dx$
$V=\pi\displaystyle\int_{0}^{2}\big((3-x)^2-1^2\big),dx$
$V=\pi\displaystyle\int_{0}^{2}(3-x-1)^2,dx$
Explanation
This problem uses the disc method with revolution around the x-axis (y = 0), which is below the trapezoidal region. The region R is bounded by the line y = 3 - x and y = 1 for 0 ≤ x ≤ 2. When revolving around y = 0, the outer radius from the axis up to y = 3 - x is (3 - x), while the inner radius from the axis up to y = 1 is 1. Using the washer method, V = π∫[0 to 2][(3 - x)² - 1²]dx. Choice C incorrectly computes 1² - (3 - x)², reversing the subtraction order, which would give negative values since 1 < 3 - x for x in [0, 2]. The fundamental principle is that outer radius squared minus inner radius squared gives the area of the washer cross-section.
Find the disc-method setup for the volume when the region between $y=x^2$ and $y=4$ is revolved about $y=1$.
$V=\pi\displaystyle\int_{-2}^{2}(4-x^2)^2,dx$
$V=\pi\displaystyle\int_{-2}^{2}(x^2-1)^2,dx$
$V=\pi\displaystyle\int_{1}^{4}(y-1)^2,dy$
$V=\pi\displaystyle\int_{-2}^{2}\big[(x^2-1)^2-(4-1)^2\big]dx$
$V=\pi\displaystyle\int_{-2}^{2}\big[(4-1)^2-(x^2-1)^2\big]dx$
Explanation
This problem requires the disc method with a horizontal axis of revolution at y = 1, which is below the given region. When revolving about y = 1, the radius at each x-value extends from the axis y = 1 to the boundary curves, giving outer radius (4 - 1) = 3 from the axis to y = 4, and inner radius (x² - 1) from the axis to y = x². Since we have a washer with outer radius 3 and inner radius (x² - 1), the volume integral is π∫[(3)² - (x² - 1)²]dx = π∫[(4 - 1)² - (x² - 1)²]dx from x = -2 to x = 2. Choice E incorrectly reverses the order of subtraction, which would give negative volume since the outer radius must be larger. When revolving about a horizontal line y = k, always measure radii as |curve - k| and ensure the outer radius is subtracted by the inner radius.
Which integral gives the volume when the region between $y=\sqrt{x}$ and $y=0$ on $0,4$ is revolved about $y=2$?
$V=\pi\int_{0}^{4}(2-\sqrt{x})^2,dx$
$V=\pi\int_{0}^{4}(2+\sqrt{x})^2,dx$
$V=\pi\int_{0}^{2}(4-y^2)^2,dy$
$V=\pi\int_{0}^{4}\big(2^2-(2-\sqrt{x})^2\big),dx$
$V=\pi\int_{0}^{4}(\sqrt{x})^2,dx$
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 $y=-1$.
$V=\pi\int_{0}^{\pi}(\sin x)^2,dx$
$V=\pi\int_{0}^{\pi}(1+\sin x)^2,dx$
$V=\pi\int_{0}^{\pi}(1-\sin x)^2,dx$
$V=\pi\int_{0}^{1}(\arcsin y)^2,dy$
$V=\pi\int_{0}^{\pi}\big((1+\sin x)^2-1^2\big),dx$
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.