Disc Method: Revolving Around Other Axes
Help Questions
AP Calculus AB › Disc Method: Revolving Around Other Axes
What is the correct setup for the volume when the region under $y=\ln(x+1)$ from $x=0$ to $x=2$ is revolved about $y=-1$?
$V=\pi\int_{0}^{2}(\ln(x+1))^2,dx$
$V=\pi\int_{0}^{2}(\ln(x+1)+1)^2,dx$
$V=\pi\int_{0}^{\ln 3}(y+1)^2,dy$
$V=\pi\int_{0}^{2}(-1-\ln(x+1))^2,dx$
$V=\pi\int_{0}^{2}(1-\ln(x+1))^2,dx$
Explanation
This problem involves the disc method for finding volumes of solids of revolution when revolving around a shifted axis, specifically y=-1. The standard disc method uses the radius as the function value when revolving around y=0, but here the axis is shifted down to y=-1. Since the function y = ln(x+1) is above y=-1, the radius becomes ln(x+1) + 1 to reflect the distance. This addition accounts for the negative shift of the axis. A tempting distractor is choice B, which uses (-1 - $ln(x+1))^2$, but while equivalent when squared, it may confuse the distance concept. To generalize for revolutions around y = k, always compute the radius as |f(x) - k| and square it in the integral setup.
What is the correct setup for the volume when the region under $y=x+2$ from $x=1$ to $x=4$ is revolved about $y=1$?
$V=\pi\int_{1}^{4}((x+2)-1)^2,dx$
$V=\pi\int_{1}^{4}(x+2)^2,dx$
$V=\pi\int_{1}^{4}(x+1)^2,dx$
$V=\pi\int_{1}^{4}(1-(x+2))^2,dx$
$V=\pi\int_{1}^{4}(1+x+2)^2,dx$
Explanation
This problem involves using the disc method to find the volume of a solid of revolution when revolving around a horizontal axis shifted from the x-axis, specifically y=1. The radius of each disc is the vertical distance from the curve y=x+2 to the axis y=1, given by |(x+2) - 1| = |x+1|. Since the curve is above y=1 throughout [1,4], x+1 is positive, matching (x+2) - 1. Therefore, the volume is correctly set up as π ∫ from 1 to 4 of ((x+2) - 1)² dx. The distractor choice A fails because it uses (x+2)², which would be correct for rotation around y=0, but ignores the shift to y=1. A general strategy for shifting the axis to y=k is to use the radius |k - f(x)| in the disc method integral, or equivalently (k - f(x))² to handle the absolute value through squaring.
Which integral represents the volume when the region under $y=\sin x$ from $x=0$ to $x=\pi$ is revolved about $y=2$?
$V=\pi\int_{0}^{2}(\sin x)^2,dy$
$V=\pi\int_{0}^{\pi}(2\sin x)^2,dx$
$V=\pi\int_{0}^{\pi}(\sin x-2)^2,dx$
$V=\pi\int_{0}^{\pi}(2-\sin x)^2,dx$
$V=\pi\int_{0}^{\pi}(\sin x)^2,dx$
Explanation
This problem involves the disc method with rotation about a horizontal line above the curve. When revolving the region under y = sin x from x = 0 to x = π about y = 2, we need the distance from the curve to the line y = 2. Since sin x ≤ 1 < 2 on [0,π], the radius is 2 - sin x, giving us π∫₀^π(2 - sin x)² dx. Option D incorrectly uses (sin x - 2)², which would give negative values inside the square. For rotation about a line above the curve, use (axis value - function value) to ensure positive radii.
Select the correct disc-method setup for the volume when the region under $y=1-x^2$ from $x=-1$ to $x=1$ is revolved about $y=-3$.
$V=\pi\int_{-1}^{1}(-3-(1-x^2))^2,dx$
$V=\pi\int_{-1}^{1}(3-x^2)^2,dx$
$V=\pi\int_{-3}^{1}(1-x^2)^2,dy$
$V=\pi\int_{-1}^{1}(1-x^2)^2,dx$
$V=\pi\int_{-1}^{1}((1-x^2)+3)^2,dx$
Explanation
This problem requires the disc method with rotation about a horizontal line below the parabola. When revolving the region under y = 1 - x² from x = -1 to x = 1 about y = -3, the radius is the distance from the curve to y = -3. Since 1 - x² ≥ 0 on [-1,1] and we're rotating about y = -3, the radius is (1 - x²) - (-3) = (1 - x²) + 3, giving us π∫₋₁¹((1 - x²) + 3)² dx. Option B incorrectly computes -3 - (1 - x²), which misunderstands the direction of measurement. When rotating about y = -c, add c to the function value: radius = f(x) + c.
What is the correct setup for the volume when the region under $y=\cos x+2$ from $x=0$ to $x=\pi$ is revolved about $y=1$?
$V=\pi\int_{1}^{3}(y-1)^2,dy$
$V=\pi\int_{0}^{\pi}((\cos x+2)-1)^2,dx$
$V=\pi\int_{0}^{\pi}(\cos x+1)^2,dx$
$V=\pi\int_{0}^{\pi}(\cos x+2)^2,dx$
$V=\pi\int_{0}^{\pi}(1-(\cos x+2))^2,dx$
Explanation
This problem involves the disc method for finding volumes of solids of revolution when revolving around a shifted axis, specifically y=1. The standard disc method uses the radius as the function value when revolving around y=0, but here the axis is shifted to y=1. Since the function y = cos x + 2 is above y=1, the radius becomes (cos x + 2) - 1 to reflect the distance. This subtraction ensures the correct perpendicular distance for each disc. A tempting distractor is choice A, which uses (cos x + $2)^2$, but this ignores the shift, as if revolving around y=0. To generalize for revolutions around y = k, always compute the radius as |f(x) - k| and square it in the integral setup.
What is the correct volume setup when the region under $y=1-x$ from $x=0$ to $x=1$ is revolved about $y=2$?
$V=\pi\int_{0}^{1}(2-(1-x))^2,dx$
$V=\pi\int_{0}^{2}(2-y)^2,dy$
$V=\pi\int_{0}^{1}(2+1-x)^2,dx$
$V=\pi\int_{0}^{1}(1-x)^2,dx$
$V=\pi\int_{0}^{1}((1-x)-2)^2,dx$
Explanation
This problem involves using the disc method to find the volume of a solid of revolution when revolving around a horizontal axis shifted from the x-axis, specifically y=2. The radius of each disc is the vertical distance from the curve y=1-x to the axis y=2, given by 2 - (1-x) since 1-x ≤ 1 < 2 throughout [0,1]. This expression is always positive in the interval, directly providing the radius. Therefore, the volume is correctly set up as π ∫ from 0 to 1 of (2 - (1-x))² dx. The distractor choice A fails because it uses (1-x)², which would be correct for rotation around y=0, but ignores the shift to y=2. A general strategy for shifting the axis to y=k is to use the radius |k - f(x)| in the disc method integral, or equivalently (k - f(x))² to handle the absolute value through squaring.
What is the correct volume setup when the region under $y=e^x$ from $x=0$ to $x=1$ is revolved about $y=0$?
$V=\pi\int_{0}^{1}(e^x+1)^2,dx$
$V=\pi\int_{0}^{1}(e^x-0)^2,dx$
$V=\pi\int_{0}^{1}(1-e^x)^2,dx$
$V=\pi\int_{0}^{1}(e^x)^2,dx$
$V=\pi\int_{0}^{1}(0-e^x)^2,dx$
Explanation
This problem involves using the disc method to find the volume of a solid of revolution when revolving around a horizontal axis shifted from the x-axis, specifically y=0. The radius of each disc is the vertical distance from the curve $y=e^x$ to the axis y=0, given by $e^x$ - 0 = $e^x$. Since the axis is at y=0 and the curve is above it throughout [0,1], this expression is positive and directly gives the radius. Therefore, the volume is correctly set up as π ∫ from 0 to 1 of $(e^x$)² dx. The distractor choice D fails because it uses (1 - $e^x$)², which would be appropriate if revolving around y=1 above the curve, but not for y=0. A general strategy for shifting the axis to y=k is to use the radius |k - f(x)| in the disc method integral, or equivalently (k - f(x))² to handle the absolute value through squaring.
What is the correct setup for the volume when the region under $y=x^2+1$ from $x=0$ to $x=2$ is revolved about $y=0$?
$V=\pi\int_{0}^{2}(x^2+1)^2,dx$
$V=\pi\int_{0}^{2}(x^2+1-2)^2,dx$
$V=\pi\int_{0}^{2}(0-(x^2+1))^2,dx$
$V=\pi\int_{0}^{2}(1-x^2)^2,dx$
$V=\pi\int_{0}^{2}(x^2-1)^2,dx$
Explanation
This problem involves the disc method for finding volumes of solids of revolution when revolving around a shifted axis, here the x-axis at y=0. The standard disc method uses the radius as the function value y = $x^2$ + 1 when revolving around y=0, with no shift adjustment needed. Since the axis is at y=0 and the function is above it, the radius remains $x^2$ + 1 without modification. No subtraction or addition is required because the shift is zero. A tempting distractor is choice D, which uses (1 - $x^2$$)^2$, but this incorrectly subtracts as if revolving around y=1, leading to a wrong radius. To generalize for revolutions around y = k, always compute the radius as |f(x) - k| and square it in the integral setup.
What is the correct volume setup when the region under $y=5-2x$ from $x=0$ to $x=2$ is revolved about $y=0$?
$V=\pi\int_{0}^{2}(5-2x-0)^2,dx$
$V=\pi\int_{0}^{2}(5-2x+1)^2,dx$
$V=\pi\int_{0}^{2}(5-2x)^2,dx$
$V=\pi\int_{0}^{2}(2x-5)^2,dx$
$V=\pi\int_{0}^{2}(0-(5-2x))^2,dx$
Explanation
This problem involves using the disc method to find the volume of a solid of revolution when revolving around a horizontal axis shifted from the x-axis, specifically y=0. The radius of each disc is the vertical distance from the curve y=5-2x to the axis y=0, given by 5-2x - 0 = 5-2x. Since the axis is at y=0 and the curve is above it throughout [0,2], this expression is positive and directly gives the radius. Therefore, the volume is correctly set up as π ∫ from 0 to 2 of (5-2x)² dx. The distractor choice D fails because it uses (2x-5)², which is mathematically equivalent but may confuse the sign without context. A general strategy for shifting the axis to y=k is to use the radius |k - f(x)| in the disc method integral, or equivalently (k - f(x))² to handle the absolute value through squaring.
What is the correct setup for the volume when the region under $y=4-x^2$ from $x=-1$ to $x=1$ is revolved about $y=6$?
$V=\pi\int_{-1}^{1}(6+4-x^2)^2,dx$
$V=\pi\int_{-1}^{1}((4-x^2)-6)^2,dx$
$V=\pi\int_{-1}^{1}(6-(4-x^2))^2,dx$
$V=\pi\int_{0}^{6}(6-y)^2,dy$
$V=\pi\int_{-1}^{1}(4-x^2)^2,dx$
Explanation
This problem involves using the disc method to find the volume of a solid of revolution when revolving around a horizontal axis shifted from the x-axis, specifically y=6. The radius of each disc is the vertical distance from the curve y=4-x² to the axis y=6, given by 6 - (4-x²) since 4-x² ≤ 3 < 6 throughout [-1,1]. This expression is always positive in the interval, directly providing the radius. Therefore, the volume is correctly set up as π ∫ from -1 to 1 of (6 - (4-x²))² dx. The distractor choice A fails because it uses (4-x²)², which would be correct for rotation around y=0, but ignores the shift to y=6. A general strategy for shifting the axis to y=k is to use the radius |k - f(x)| in the disc method integral, or equivalently (k - f(x))² to handle the absolute value through squaring.