Disc Method: Revolving Around Other Axes - AP Calculus BC

Let $R$ be bounded by $y=x$ and $y=0$ on $0\le x\le1$ ; revolving about $y=3$ , which setup is correct?
Answer: $V=\pi\displaystyle\int_{0}^{1}\big((3-0)^2-(3-x)^2\big)\,dx$ This problem uses the disc method with revolution around y = 3, a horizontal line above the triangular region. The region R is bounded by y = x and y = 0 on 0 ≤ x ≤ 1, forming a right triangle. When revolving around y = 3, the outer radius from the axis down to y = 0 is (3 - 0) = 3, while the inner radius from the axis down to y = x is (3 - x). Using the washer method, V = π∫[0 to 1][(3)² - (3 - x)²]dx = π∫[0 to 1][9 - (3 - x)²]dx. Choice B incorrectly computes (3 - x)² - 9, which would give negative values throughout the interval. The key insight is that when revolving around an axis outside the region, you create a washer (annulus) in each cross-section perpendicular to the axis.
Choose the correct volume setup for revolving the region bounded by $y=x^3$ , $y=0$ , $x=0$ , $x=1$ about $y=2$ .
Answer: $V=\pi\int_{0}^{1}\big(2^2-(2-x^3)^2\big)\,dx$ 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 $y=1-x^2$ from $x=-1$ to $x=1$ about $y=-2$ .
Answer: $\pi\displaystyle\int_{-1}^{1}(3-x^2)^2\,dx$ This problem involves the disc method for finding volumes of revolution around a shifted axis, specifically around y = -2 instead of the x-axis. When the axis is shifted down to y = -2, the radius of each disc is the distance from the curve y = 1 - x² to the axis, which is (1 - x²) - (-2) = 3 - x². This adjustment adds the shift magnitude, resulting in a larger radius. The integrand is π(3 - x²)², integrated from -1 to 1. A tempting distractor is choice D, π ∫ (1 - x² - 2)² dx, which simplifies to (-1 - x²)² and might stem from subtracting instead of adding the shift. 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.
Let $R$ be bounded by $y=\ln(x)$ , $y=0$ , and $x=e$ . Revolve $R$ about $y=1$ . Which setup gives the volume?
Answer: $V=\pi\int_{1}^{e}\Big(\big(1-0\big)^2-\big(1-\ln x\big)^2\Big)\,dx$ This problem requires the disc method with revolution around y = 1, above the region. The region R is bounded by y = ln(x), y = 0, and x = e. When revolving around y = 1, the outer radius is from y = 1 down to y = 0: R_outer = 1 - 0 = 1, and the inner radius is from y = 1 down to y = ln(x): R_inner = 1 - ln(x). The volume is V = π∫₁ᵉ[1² - (1-ln x)²]dx. Choice A incorrectly uses only the inner radius without the washer structure. For revolution around a horizontal line above the region, the x-axis creates the outer radius and the curve creates the inner radius of the washer.
Let $R$ be bounded by $y=x^2$ and $y=4$ ; revolving $R$ about $y=5$ , which integral gives the volume?
Answer: $V=\pi\displaystyle\int_{-2}^{2}\big((5-x^2)^2-(5-4)^2\big)\,dx$ 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 $y=3-x$ and $y=1$ for $0\le x\le2$ ; revolving about $y=0$ , which volume integral is correct?
Answer: $V=\pi\displaystyle\int_{0}^{2}\big((3-x)^2-1^2\big)\,dx$ 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.
Let $R$ be bounded by $y=x^2$ and $y=4$ for $0\le x\le2$ . Revolve $R$ about $y=5$ . Which setup gives the volume?
Answer: $V=\pi\int_{0}^{2}\Big(\big(5-x^2\big)^2-\big(5-4\big)^2\Big)\,dx$ This problem requires the disc method with revolution around a horizontal line y = 5, which is above the region. The region R is bounded by y = x² (bottom curve) and y = 4 (top curve) for 0 ≤ x ≤ 2. When revolving around y = 5, the outer radius is the distance from y = 5 to the bottom curve: R_outer = 5 - x², and the inner radius is the distance from y = 5 to the top curve: R_inner = 5 - 4 = 1. The volume is V = π∫[R_outer² - R_inner²]dx = π∫₀²[(5-x²)² - 1²]dx. Choice A incorrectly uses only the outer radius without subtracting the hole. When revolving around a line outside the region, always use (axis - curve) for distances and apply the washer formula when there's a hole.
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?
Answer: $V=\pi\int_{0}^{\pi}\Big(\big(\sin x+1\big)^2-1^2\Big)\,dx$ 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.
Choose the correct volume setup when the region between $y=x$ and $y=0$ for $0\le x\le 2$ is revolved about $y=3$ .
Answer: $V=\pi\int_{0}^{2}\big(3^2-(3-x)^2\big)\,dx$ This problem involves the disc method for volumes of revolution around a shifted axis, specifically around y=3. The shift to y=3 alters the radii to distances from this axis to the region bounds. The outer radius is 3 - 0 =3, to y=0, and the inner is 3 - x, to y=x. The volume integral is π times ( $3^2$ - $(3 - x)^2$ ) from 0 to 2. A tempting distractor is choice A, which omits the washer subtraction, treating it as a solid disc. A transferable axis-shift strategy is to subtract the axis value from the curve functions to find the effective radii.
Which integral represents the volume when the region bounded by $y=3x-x^2$ and $y=0$ is revolved about $y=4$ ?
Answer: $V=\pi\int_{0}^{3}\big(4^2-(4-(3x-x^2))^2\big)\,dx$ 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.
Select the correct volume setup when the region bounded by $y=\ln x$ , $y=0$ , $x=1$ , $x=e$ is revolved about $y=-2$ .
Answer: $V=\pi\int_{1}^{e}\big((2+\ln x)^2-2^2\big)\,dx$ This problem involves the disc method for volumes of revolution around a shifted axis, specifically around y=-2. Shifting the axis to y=-2 modifies the radii to distances from this line. The outer radius is ln x + 2, to y=ln x, and inner is 2, to y=0. The volume integral is π times ( $(ln x + 2)^2$ - $2^2$ ) from 1 to e. A tempting distractor is choice C, which subtracts instead of adding the shift for a below-axis rotation. A transferable axis-shift strategy is to add or subtract the shift value to the curve functions depending on the direction.
Let $R$ be bounded by $y=\frac{1}{x}$ , $y=0$ , $x=1$ , and $x=2$ . Revolve $R$ about $y=3$ . Which setup is correct?
Answer: $V=\pi\int_{1}^{2}\Big(\big(3-0\big)^2-\big(3-\tfrac{1}{x}\big)^2\Big)\,dx$ This problem uses the disc method with revolution around y = 3, above the region. The region R is bounded by y = 1/x, y = 0, x = 1, and x = 2. When revolving around y = 3, the outer radius is from y = 3 down to y = 0: R_outer = 3 - 0 = 3, and the inner radius is from y = 3 down to y = 1/x: R_inner = 3 - 1/x. The volume is V = π∫₁²[3² - (3-1/x)²]dx. Choice E incorrectly computes the radii as distances from the curves to the axis rather than from the axis to the curves. For revolution around a line above the region, always measure distances downward from the axis to find positive radii.
Region $R$ is bounded by $y=1-x$ , $y=0$ , and $x=0$ . Revolve $R$ about $y=-2$ . Which setup gives the volume?
Answer: $V=\pi\int_{0}^{1}\Big(\big((1-x)+2\big)^2-2^2\Big)\,dx$ 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.
Region $R$ is bounded by $y=2x$ , $y=0$ , and $x=3$ . Revolve $R$ about $y=4$ . Which setup gives the volume?
Answer: $V=\pi\int_{0}^{3}\Big(\big(4-0\big)^2-\big(4-2x\big)^2\Big)\,dx$ This problem requires the disc method with revolution around y = 4, above the region. The region R is bounded by y = 2x, y = 0, and x = 3, forming a triangle. When revolving around y = 4, the outer radius is from y = 4 down to y = 0: R_outer = 4 - 0 = 4, and the inner radius is from y = 4 down to y = 2x: R_inner = 4 - 2x. The volume is V = π∫₀³[4² - (4-2x)²]dx. Choice A incorrectly uses only the inner radius without accounting for the washer shape. When the axis is above a triangular region, the horizontal boundary (y = 0) creates the outer radius while the slanted boundary creates the inner radius.
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?
Answer: $V=\pi\int_{0}^{\pi/2}\Big(\big(2-0\big)^2-\big(2-\cos x\big)^2\Big)\,dx$ 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.
Select the correct disc/washer setup for revolving the region bounded by $y=\sin x$ , $y=0$ , $x=0$ , $x=\pi$ about $y=-1$ .
Answer: $V=\pi\int_{0}^{\pi}\big((1+\sin x)^2-1^2\big)\,dx$ 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.
Region $R$ is bounded by $x=y^2$ , $x=4$ , and $y=0$ ; revolving about $x=5$ , which disc-method integral is correct?
Answer: $V=\pi\displaystyle\int_{0}^{2}\big((5-y^2)^2-(5-4)^2\big)\,dy$ This problem requires the disc method with revolution around the vertical line x = 5, which is to the right of the region. The region R is bounded by the parabola x = y², the vertical line x = 4, and y = 0, with the parabola and line intersecting at y = 2. When revolving around x = 5, the outer radius from the axis leftward to x = y² is (5 - y²), while the inner radius from the axis to x = 4 is (5 - 4) = 1. Using washers with dy integration, V = π∫[0 to 2][(5 - y²)² - 1²]dy. Choice E incorrectly reverses the subtraction as 1² - (5 - y²)², which would yield negative values. For vertical axes of revolution, integrate with respect to y and measure horizontal distances from the axis.
Find the disc-method setup for the volume when the region between $y=x^2$ and $y=4$ is revolved about $y=1$ .
Answer: $V=\pi\displaystyle\int_{-2}^{2}\big[(4-1)^2-(x^2-1)^2\big]dx$ 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.
Find the correct volume setup when the region bounded by $y=x^2$ and $y=4$ is revolved about $y=5$ .
Answer: $V=\pi\int_{-2}^{2}\big((5-x^2)^2-(5-4)^2\big)\,dx$ This problem involves the disc method for volumes of revolution around a shifted axis, specifically around y=5. When the axis is shifted to y=5, the radii are measured as distances from this line to the boundaries of the region. The outer radius is 5 - $x^2$ , the distance to the lower curve y= $x^2$ , and the inner radius is 5 - 4 =1, the distance to the upper line y=4. Therefore, the volume integral is π times the integral of ( $(5 - x^2)^2$ - $1^2$ ) from -2 to 2. A tempting distractor is choice A, which omits the inner radius, thus calculating the volume without the hole. 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 $y=x^3$ , $y=0$ , and $x=2$ . Revolve $R$ about $y=-3$ . Which setup gives the volume?
Answer: $V=\pi\int_{0}^{2}\Big(\big(x^3+3\big)^2-3^2\Big)\,dx$ 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.

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