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 $y=x^2$ from $x=0$ to $x=2$ is revolved about $y=3$?
$V=\pi\int_{0}^{2}(3-x^2)^2,dx$
$V=\pi\int_{0}^{2}(3+x^2)^2,dx$
$V=\pi\int_{0}^{2}(x^2-3)^2,dx$
$V=\pi\int_{0}^{2}(x^2)^2,dx$
$V=\pi\int_{0}^{3}(3-y)^2,dy$
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=3. The radius of each disc is the vertical distance from the curve y=x² to the axis y=3, given by |3 - x²|. Since the curve crosses y=3 within the interval [0,2], the expression 3 - x² changes sign, but squaring it provides the correct (distance)² for the integral. Therefore, the volume is correctly set up as π ∫ from 0 to 2 of (3 - x²)² dx. The distractor choice A fails because it uses the radius as x², which would be correct for rotation around y=0, but ignores the shift to y=3. 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 $y=0$?
$V=\pi\int_{0}^{1}(e^x+1)^2,dx$
$V=\pi\int_{0}^{1}(e^x)^2,dx$
$V=\pi\int_{0}^{1}(0-e^x)^2,dx$
$V=\pi\int_{0}^{1}(e^x-0)^2,dx$
$V=\pi\int_{0}^{1}(1-e^x)^2,dx$
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 $y=4-x^2$ from $x=-1$ to $x=1$ is revolved about $y=6$?
$V=\pi\int_{-1}^{1}(6+4-x^2)^2,dx$
$V=\pi\int_{-1}^{1}((4-x^2)-6)^2,dx$
$V=\pi\int_{-1}^{1}(6-(4-x^2))^2,dx$
$V=\pi\int_{-1}^{1}(4-x^2)^2,dx$
$V=\pi\int_{0}^{6}(6-y)^2,dy$
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.
What is the correct volume setup when the region under $y=5-2x$ from $x=0$ to $x=2$ is revolved about $y=0$?
$V=\pi\int_{0}^{2}(5-2x)^2,dx$
$V=\pi\int_{0}^{2}(5-2x+1)^2,dx$
$V=\pi\int_{0}^{2}(5-2x-0)^2,dx$
$V=\pi\int_{0}^{2}(2x-5)^2,dx$
$V=\pi\int_{0}^{2}(0-(5-2x))^2,dx$
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 $y=\sqrt{4-x^2}$ from $x=0$ to $x=2$ is revolved about $y=1$?
$V=\pi\int_{0}^{2}(\sqrt{4-x^2}-1)^2,dx$
$V=\pi\int_{0}^{2}(\sqrt{4-x^2})^2,dx$
$V=\pi\int_{0}^{1}(1-y)^2,dy$
$V=\pi\int_{0}^{2}(\sqrt{4-x^2}+1)^2,dx$
$V=\pi\int_{0}^{2}(1-\sqrt{4-x^2})^2,dx$
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=1. The radius of each disc is the vertical distance from the curve y=√(4-x²) to the axis y=1, given by |1 - √(4-x²)|. Since the curve crosses y=1 within [0,2], the expression 1 - √(4-x²) changes sign, but squaring it provides the correct (distance)². Therefore, the volume is correctly set up as π ∫ from 0 to 2 of (1 - √(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=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.
What is the correct volume setup when the region under $y=\sqrt{x}$ from $x=1$ to $x=5$ is revolved about $y=-2$?
$V=\pi\int_{1}^{5}(\sqrt{x})^2,dx$
$V=\pi\int_{1}^{5}(\sqrt{x}+2)^2,dx$
$V=\pi\int_{1}^{5}(2-\sqrt{x})^2,dx$
$V=\pi\int_{-2}^{\sqrt{5}}(y+2)^2,dy$
$V=\pi\int_{1}^{5}(-2-\sqrt{x})^2,dx$
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=-2. The radius of each disc is the vertical distance from the curve y=√x to the axis y=-2, given by √x - (-2) = √x + 2. Since the axis is below the curve throughout [1,5], this expression is always positive and directly gives the radius. Therefore, the volume is correctly set up as π ∫ from 1 to 5 of (√x + 2)² dx. The distractor choice D fails because it uses (2 - √x)², which would be appropriate if revolving around y=2 above the curve, but not for y=-2 below. 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 $y=\frac{x^2}{9}$ from $x=0$ to $x=3$ is revolved about $y=2$?
$V=\pi\int_{0}^{3}\left(2+\frac{x^2}{9}\right)^2,dx$
$V=\pi\int_{0}^{3}\left(\frac{x^2}{9}-2\right)^2,dx$
$V=\pi\int_{0}^{3}\left(2-\frac{x^2}{9}\right)^2,dx$
$V=\pi\int_{0}^{3}\left(\frac{x^2}{9}\right)^2,dx$
$V=\pi\int_{0}^{2}(2-y)^2,dy$
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=x²/9 to the axis y=2, given by 2 - x²/9 since x²/9 ≤ 1 < 2 throughout [0,3]. This expression is always positive in the interval, directly providing the radius. Therefore, the volume is correctly set up as π ∫ from 0 to 3 of (2 - x²/9)² dx. The distractor choice A fails because it uses (x²/9)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 $y=\frac{2}{x+1}$ from $x=0$ to $x=3$ is revolved about $y=2$?
$V=\pi\int_{0}^{3}\left(\frac{2}{x+1}\right)^2,dx$
$V=\pi\int_{0}^{3}\left(2-\frac{2}{x+1}\right)^2,dx$
$V=\pi\int_{0}^{3}\left(\frac{2}{x+1}-2\right)^2,dx$
$V=\pi\int_{0}^{2}(2-y)^2,dy$
$V=\pi\int_{0}^{3}\left(2+\frac{2}{x+1}\right)^2,dx$
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=2/(x+1) to the axis y=2, given by |2 - 2/(x+1)|. Since the curve is below y=2 throughout [0,3], 2 - 2/(x+1) is positive and directly gives the radius. Therefore, the volume is correctly set up as π ∫ from 0 to 3 of (2 - 2/(x+1))² dx. The distractor choice A fails because it uses (2/(x+1))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 setup for the volume when the region under $y=x^2+1$ from $x=0$ to $x=2$ is revolved about $y=0$?
$V=\pi\int_{0}^{2}(1-x^2)^2,dx$
$V=\pi\int_{0}^{2}(x^2+1-2)^2,dx$
$V=\pi\int_{0}^{2}(x^2+1)^2,dx$
$V=\pi\int_{0}^{2}(x^2-1)^2,dx$
$V=\pi\int_{0}^{2}(0-(x^2+1))^2,dx$
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 setup for the volume when the region under $y=\sqrt{x+1}$ from $x=0$ to $x=3$ is revolved about $y=2$?
$V=\pi\int_{0}^{3}(\sqrt{x+1}-2)^2,dx$
$V=\pi\int_{0}^{3}(2+\sqrt{x+1})^2,dx$
$V=\pi\int_{0}^{3}(2-\sqrt{x+1})^2,dx$
$V=\pi\int_{1}^{4}(2-y)^2,dy$
$V=\pi\int_{0}^{3}(\sqrt{x+1})^2,dx$
Explanation
This problem involves the disc method for finding volumes of solids of revolution when revolving around a shifted axis, specifically y=2. The standard disc method uses the radius as the function value when revolving around y=0, but here the axis is shifted up to y=2. Since the function y = √(x+1) is below y=2, the radius becomes 2 - √(x+1) to account for the distance. This reversal in subtraction reflects that the axis is above the curve. A tempting distractor is choice D, which uses (2 + $√(x+1))^2$, but this adds instead of subtracting, overstating the radius and volume. To generalize for revolutions around y = k, always compute the radius as |f(x) - k| and square it in the integral setup.