Disc Method: Revolving Around Other Axes - AP Calculus AB

What is the correct volume setup when the region under $y=\sqrt{x}$ from $x=1$ to $x=5$ is revolved about $y=-2$ ?
Answer: $V=\pi\int_{1}^{5}(\sqrt{x}+2)^2\,dx$ 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 setup for the volume when the region under $y=\sin x$ from $x=0$ to $x=\pi$ is revolved about $y=2$ ?
Answer: $V=\pi\int_{0}^{\pi}(2-\sin x)^2\,dx$ 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=sin x to the axis y=2, given by 2 - sin x since sin x ≤ 1 < 2 throughout [0,π]. This expression is always positive in the interval, directly providing the radius. Therefore, the volume is correctly set up as π ∫ from 0 to π of (2 - sin x)² dx. The distractor choice A fails because it uses (sin 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 $y=3x^2$ from $x=0$ to $x=2$ is revolved about $y=-2$ ?
Answer: $V=\pi\int_{0}^{2}(3x^2+2)^2\,dx$ 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=3x² to the axis y=-2, given by 3x² - (-2) = 3x² + 2. Since the axis is below the curve throughout [0,2], this expression is always positive and directly gives the radius. Therefore, the volume is correctly set up as π ∫ from 0 to 2 of (3x² + 2)² dx. The distractor choice A fails because it uses (3x²)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 $y=\sqrt{4-x^2}$ from $x=0$ to $x=2$ is revolved about $y=1$ ?
Answer: $V=\pi\int_{0}^{2}(1-\sqrt{4-x^2})^2\,dx$ 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 setup for the volume when the region under $y=1+\cos x$ from $x=0$ to $x=\pi$ is revolved about $y=3$ ?
Answer: $V=\pi\int_{0}^{\pi}(3-(1+\cos x))^2\,dx$ 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=1+cos x to the axis y=3, given by 3 - (1+cos x) since 1+cos x ≤ 2 < 3 throughout [0,π]. This expression is always positive in the interval, directly providing the radius. Therefore, the volume is correctly set up as π ∫ from 0 to π of (3 - (1+cos x))² dx. The distractor choice A fails because it uses (1+cos x)², 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 setup for the volume when the region under $y=2\ln x$ from $x=1$ to $x=e$ is revolved about $y=-1$ ?
Answer: $V=\pi\int_{1}^{e}(2\ln x+1)^2\,dx$ 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 = 2 ln x is above y=-1, the radius becomes 2 ln x + 1 to reflect the distance. This addition accounts for the negative shift of the axis. A tempting distractor is choice B, which uses $(-1 - 2 ln x)^2$ , but while equivalent when squared, it may obscure the positive 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 $y=2-\sin x$ from $x=0$ to $x=\pi$ is revolved about $y=0$ ?
Answer: $V=\pi\int_{0}^{\pi}(2-\sin x)^2\,dx$ 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=2-sin x to the axis y=0, given by 2-sin x - 0 = 2-sin x. Since the axis is at y=0 and the curve is above it throughout [0,π], this expression is positive and directly gives the radius. Therefore, the volume is correctly set up as π ∫ from 0 to π of (2 - sin x)² dx. The distractor choice C fails because it uses (sin x - 2)², which would be equivalent but may confuse if not recognizing the square handles the sign. 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 $y=x^3$ from $x=0$ to $x=1$ is revolved about $y=1$ ?
Answer: $V=\pi\int_{0}^{1}(1-x^3)^2\,dx$ 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=x³ to the axis y=1, given by 1 - x³ since x³ ≤ 1 throughout [0,1]. This expression is non-negative in the interval, directly providing the radius. Therefore, the volume is correctly set up as π ∫ from 0 to 1 of (1 - x³)² dx. The distractor choice B fails because it uses (x³)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 setup for the volume when the region under $y=\sqrt{x}$ from $x=0$ to $x=4$ is revolved about $y=3$ ?
Answer: $V=\pi\int_{0}^{4}(3-\sqrt{x})^2\,dx$ This problem involves the disc method for finding volumes of solids of revolution when revolving around a shifted axis, specifically y=3. 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=3. Since the function y = √x is below y=3, the radius becomes 3 - √x 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 (3 + √x)^2, but this adds instead of subtracting, overstating the 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 $y=4-x^2$ from $x=-1$ to $x=1$ is revolved about $y=6$ ?
Answer: $V=\pi\int_{-1}^{1}(6-(4-x^2))^2\,dx$ 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 setup for the volume when the region under $y=\ln(x+1)$ from $x=0$ to $x=2$ is revolved about $y=-1$ ?
Answer: $V=\pi\int_{0}^{2}(\ln(x+1)+1)^2\,dx$ 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 volume setup when the region under $y=\frac{x^2}{9}$ from $x=0$ to $x=3$ is revolved about $y=2$ ?
Answer: $V=\pi\int_{0}^{3}\left(2-\frac{x^2}{9}\right)^2\,dx$ 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 $y=\frac{x}{x+1}$ from $x=0$ to $x=2$ is revolved about $y=1$ ?
Answer: $V=\pi\int_{0}^{2}\left(1-\frac{x}{x+1}\right)^2\,dx$ 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/(x+1) to the axis y=1, given by 1 - x/(x+1) since x/(x+1) < 1 throughout [0,2]. This expression is always positive in the interval, directly providing the radius. Therefore, the volume is correctly set up as π ∫ from 0 to 2 of (1 - x/(x+1))² dx. The distractor choice A fails because it uses (x/(x+1))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 setup for the volume when the region under $y=\tan x$ from $x=0$ to $x=\frac{\pi}{4}$ is revolved about $y=1$ ?
Answer: $V=\pi\int_{0}^{\pi/4}(1-\tan x)^2\,dx$ 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=tan x to the axis y=1, given by 1 - tan x since tan x ≤ 1 throughout [0,π/4]. This expression is non-negative in the interval, directly providing the radius. Therefore, the volume is correctly set up as π ∫ from 0 to π/4 of (1 - tan x)² dx. The distractor choice A fails because it uses (tan x)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 setup for the volume when the region under $y=x^2$ from $x=0$ to $x=2$ is revolved about $y=3$ ?
Answer: $V=\pi\int_{0}^{2}(3-x^2)^2\,dx$ 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.
Which integral gives the volume when the region under $y=\ln(x+1)$ on $0\le x\le 2$ is revolved about $y=4$ ?
Answer: $V=\pi\int_{0}^{2}(4-\ln(x+1))^2\,dx$ This problem involves the disc method with rotation about a horizontal line well above the curve. When revolving the region under y = ln(x + 1) from x = 0 to x = 2 about y = 4, we need the distance from the curve to y = 4. Since ln(x + 1) < 4 on [0,2] (as ln(3) ≈ 1.1), the radius is 4 - ln(x + 1), giving us π∫₀²(4 - ln(x + 1))² dx. Option C incorrectly uses (ln(x + 1) - 4)², which would result in negative values being squared. For rotation about y = c where c > f(x), always use (c - f(x)) for the radius.
What is the correct volume setup when the region under $y=\ln x$ from $x=1$ to $x=e$ is revolved about $y=4$ ?
Answer: $V=\pi\int_{1}^{e}(4-\ln x)^2\,dx$ 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=4. The radius of each disc is the vertical distance from the curve y=ln x to the axis y=4, given by 4 - ln x since ln x ≤ 1 < 4 throughout [1,e]. This expression is always positive in the interval, directly providing the radius. Therefore, the volume is correctly set up as π ∫ from 1 to e of (4 - ln x)² dx. The distractor choice A fails because it uses (ln x)², which would be correct for rotation around y=0, but ignores the shift to y=4. 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 $y=1-\frac{x}{3}$ from $x=0$ to $x=3$ is revolved about $y=-2$ ?
Answer: $V=\pi\int_{0}^{3}\left(\left(1-\frac{x}{3}\right)+2\right)^2\,dx$ 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 down to y=-2. Since the function y = 1 - x/3 is above y=-2, the radius becomes (1 - x/3) + 2 to reflect the distance. This addition accounts for the negative shift of the axis. A tempting distractor is choice B, which uses $(-2 - (1 - x/3))^2$ , but while mathematically equivalent when squared, it may confuse the conceptual distance. 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 $y=e^x$ from $x=0$ to $x=1$ is revolved about $y=0$ ?
Answer: $V=\pi\int_{0}^{1}(e^x)^2\,dx$ 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 $y=x+2$ from $x=1$ to $x=4$ is revolved about $y=1$ ?
Answer: $V=\pi\int_{1}^{4}((x+2)-1)^2\,dx$ 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.

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