Disc Method: Revolving Around x/y Axes - AP Calculus AB

The region bounded by $x=\frac{1}{2}y^2$ and the y-axis on $0\le y\le 4$ is revolved about the y-axis. Which integral gives the volume?
Answer: $\pi\int_{0}^{4} \left(\frac{1}{2}y^2\right)^2\,dy$ This problem uses the disc method for revolution about the y-axis. Since the region is bounded by x = (1/2)y² and the y-axis with no inner boundary, discs are the appropriate method. The radius of each disc at position y is the distance from the y-axis to the curve: r(y) = (1/2)y². The volume formula becomes V = π∫[a to b] [r(y)]² dy = π∫[0 to 4] ((1/2)y²)² dy. Choice B fails because it doesn't square the radius function, which is essential in the disc method. Apply the disc method when there's a single boundary curve extending from the axis of revolution.
A region bounded by $x=y^2$ and the y-axis for $0 \le y \le 2$ is revolved about the y-axis. Which integral gives the volume?
Answer: $\pi\int_{0}^{2} (y^2)^2\,dy$ This problem applies the disc method for revolution about the y-axis. The region bounded by $x = y^2$ and the y-axis creates a solid with no hole, so discs rather than washers are appropriate. The radius of each disc at position y equals the distance from the y-axis to the curve: $r(y) = y^2$ . Using the disc method formula: $V = \pi \int_{a}^{b} [r(y)]^2 \, dy = \pi \int_{0}^{2} (y^2)^2 \, dy$ . Choice B is incorrect because it omits the required squaring of the radius in the disc method formula. Use discs when the solid extends from the axis of revolution to a single boundary curve.
The region bounded by $y=1+x^3$ and the x-axis on $0\le x\le 1$ is revolved about the x-axis. Which integral gives the volume?
Answer: $\pi\int_{0}^{1} (1+x^3)^2\,dx$ This problem uses the disc method for revolution about the x-axis. Since the region is bounded by y = 1+x³ and the x-axis with no inner boundary, discs are appropriate rather than washers. The radius of each disc at position x is the distance from the x-axis to the curve: r(x) = 1+x³. The volume formula becomes V = π∫[a to b] [r(x)]² dx = π∫[0 to 1] (1+x³)² dx. Choice B fails because it doesn't square the radius function, which is essential in the disc method. Apply discs when there's exactly one boundary curve extending from the axis of revolution.
The region bounded by $y=2- x$ and the x-axis on $0\le x\le 2$ is revolved about the x-axis. Which integral gives the volume?
Answer: $\pi\int_{0}^{2} (2-x)^2\,dx$ This problem uses the disc method for revolution about the x-axis. The region bounded by y = 2-x and the x-axis creates a solid with no hole, so discs are appropriate rather than washers. The radius of each disc at position x equals the distance from the x-axis to the curve: r(x) = 2-x. Using the disc method formula: V = π∫[a to b] [r(x)]² dx = π∫[0 to 2] (2-x)² dx. Choice A fails because it doesn't square the radius function, which is essential in the disc method. Use discs when the region has a single boundary extending from the axis of revolution to one curve.
The region bounded by $x=\sqrt{1-y}$ and the y-axis on $0\le y\le 1$ is revolved about the y-axis. Which integral gives the volume?
Answer: $\pi\int_{0}^{1} (\sqrt{1-y})^2\,dy$ This problem uses the disc method for revolution about the y-axis. Since the region is bounded by x = √(1-y) and the y-axis with no inner boundary, discs are the appropriate method. The radius of each disc at position y is the distance from the y-axis to the curve: r(y) = √(1-y). The volume formula becomes V = π∫[a to b] [r(y)]² dy = π∫[0 to 1] (√(1-y))² dy. Choice B fails because it doesn't square the radius function, which is essential in the disc method. Apply the disc method when there's a single boundary curve extending from the axis of revolution.
What integral gives the volume when the region under $y=\frac{1}{x}$ from $x=1$ to $x=3$ is revolved about the $x$ -axis?
Answer: $\pi\displaystyle\int_{1}^{3}\left(\frac{1}{x}\right)^{2}\,dx$ This problem applies the disc method for rotation about the x-axis. Revolving the region under y = 1/x from x = 1 to x = 3 around the x-axis creates circular discs perpendicular to the x-axis. The radius of each disc equals the y-value, which is 1/x, making the disc's area π(1/x)². We integrate from x = 1 to x = 3 to find the volume: π∫₁³(1/x)²dx = π∫₁³(1/x²)dx. Choice B incorrectly uses π∫₁³(1/x)dx, missing the crucial squaring step—this would give ln|x| evaluated from 1 to 3, which doesn't represent a volume. Remember: disc method for x-axis rotation always requires π times the square of the function, which accounts for the circular area of each disc.
The region bounded by $y=\ln(x)$ and the x-axis on $1\le x\le e$ is revolved about the x-axis. Which integral gives the volume?
Answer: $\pi\int_{1}^{e} (\ln x)^2\,dx$ This problem uses the disc method for revolution about the x-axis. Since the region is bounded by y = ln(x) and the x-axis with no inner boundary, discs rather than washers apply. The radius of each disc at position x is r(x) = ln(x), the distance from the x-axis to the curve. The volume is V = π∫[a to b] [r(x)]² dx = π∫[1 to e] (ln x)² dx. Choice B fails because it doesn't square the radius function, which is essential in the disc method formula. Use discs when the solid extends from the axis of revolution to a single bounding curve.
A region bounded by $y=1-\cos x$ and the x-axis for $0\le x\le \pi$ is revolved about the x-axis. Which integral gives the volume?
Answer: $\pi\int_{0}^{\pi} (1-\cos x)^2\,dx$ This problem requires the disc method for revolution about the x-axis. The region bounded by y = 1-cos x and the x-axis forms a solid without a hole, making discs appropriate rather than washers. The radius of each disc at position x equals the distance from the x-axis to the curve: r(x) = 1-cos x. Applying the disc method: V = π∫[a to b] [r(x)]² dx = π∫[0 to π] (1-cos x)² dx. Choice B is tempting but incorrect because it omits the essential squaring of the radius required in the disc method formula. Use discs when the region extends from the axis of revolution to a single bounding curve.
The region bounded by $x=5-y$ and the y-axis on $0\le y\le 5$ is revolved about the y-axis. Which integral gives the volume?
Answer: $\pi\int_{0}^{5} (5-y)^2\,dy$ This problem uses the disc method for revolution about the y-axis. Since the region is bounded by x = 5-y and the y-axis with no inner boundary, discs are the correct method. The radius of each disc at position y is the distance from the y-axis to the curve: r(y) = 5-y. The volume formula becomes V = π∫[a to b] [r(y)]² dy = π∫[0 to 5] (5-y)² dy. Choice A fails because it doesn't square the radius function, which is essential in the disc method. Apply discs when there's a single boundary curve extending from the axis of revolution.
A region bounded by $y=3x$ and the x-axis for $0 \le x \le 2$ is revolved about the x-axis. Which integral gives the volume?
Answer: $\pi\int_{0}^{2} (3x)^2\,dx$ This problem applies the disc method for finding volume when revolving about the x-axis. Since the region is bounded by $y = 3x$ and the x-axis with no inner boundary, we use discs rather than washers. The radius of each disc at position x is the distance from the x-axis to the curve: $r(x) = 3x$ . The volume is $V = \pi \int_{a}^{b} [r(x)]^2 \, dx = \pi \int_{0}^{2} (3x)^2 \, dx$ . Choice B is tempting but incorrect because it omits the essential squaring of the radius in the disc method formula. Apply discs when the solid extends from the axis of revolution to exactly one bounding curve.
A region bounded by $y=\sqrt{1-x}$ and the x-axis for $0\le x\le 1$ is revolved about the x-axis. Which integral gives the volume?
Answer: $\pi\int_{0}^{1} (\sqrt{1-x})^2\,dx$ This problem requires the disc method for revolution about the x-axis. The region bounded by y = √(1-x) and the x-axis forms a solid without a hole, making discs appropriate rather than washers. The radius of each disc at position x equals the distance from the x-axis to the curve: r(x) = √(1-x). Applying the disc method: V = π∫[a to b] [r(x)]² dx = π∫[0 to 1] (√(1-x))² dx. Choice B is tempting but incorrect because it omits the essential squaring of the radius required in the disc method formula. Use discs when the region extends from the axis of revolution to a single bounding curve.
Which integral gives the volume when the region under $y=x^2$ from $x=0$ to $x=2$ is revolved about the $x$ -axis?
Answer: $\pi\displaystyle\int_{0}^{2}(x^2)^{2}\,dx$ This problem uses the disc method for revolving around the x-axis. When rotating the region under y = x² from x = 0 to x = 2 around the x-axis, we form circular discs perpendicular to the x-axis. Each disc has radius equal to the y-value at that x-position, which is x², so the disc's area is π(x²)². The volume is found by integrating these disc areas: π∫₀²(x²)²dx = π∫₀²x⁴dx. Choice A incorrectly uses π∫₀²x²dx, forgetting that we need to square the radius function when finding the area of each disc. The key principle: disc method around the x-axis requires π times the square of the function, not just π times the function.
What is the correct disc-method integral for the volume when the region under $y=x^{2}$ from $x=0$ to $x=2$ is revolved about the $x$ -axis?
Answer: $V=\pi\displaystyle\int_{0}^{2}(x^{2})^{2}\,dx$ This problem applies the disc method to revolve y = x² around the x-axis. When revolving around the x-axis, each disc has radius equal to the y-value, which is x². The disc method formula requires squaring this radius: V = π∫₀²(x²)² dx = π∫₀² x⁴ dx. The bounds are from x = 0 to x = 2 as given. Choice A incorrectly uses x² without squaring it again, confusing the function value with the squared radius needed for the disc's area. Remember: even if the function is already squared, you must square it again in the disc method to get [radius]².
The region bounded by $y=\cos x$ and the x-axis on $0\le x\le \frac{\pi}{2}$ is revolved about the x-axis. Which integral gives the volume?
Answer: $\pi\int_{0}^{\pi/2} (\cos x)^2\,dx$ This problem uses the disc method for revolution about the x-axis. Since the region is bounded by y = cos x and the x-axis with no inner boundary, we apply discs rather than washers. The radius of each disc at position x is the distance from the x-axis to the curve: r(x) = cos x. The volume formula becomes V = π∫[a to b] [r(x)]² dx = π∫[0 to π/2] (cos x)² dx. Choice B is tempting but fails because it doesn't square the radius function, which is required by the disc method. Use the disc method when the region extends from the axis of revolution to exactly one bounding curve.
Which integral gives the volume when the region under $y=\sqrt{1-x^{2}}$ on $0\le x\le 1$ is revolved about the $x$ -axis?
Answer: $V=\pi\displaystyle\int_{0}^{1}(1-x^{2})\,dx$ This problem uses the disc method for y = √(1 - x²) revolved around the x-axis. The function represents the upper semicircle of radius 1, and when revolving around the x-axis, each disc has radius √(1 - x²). The disc method requires V = π∫₀¹[√(1 - x²)]² dx = π∫₀¹(1 - x²) dx. The bounds are from x = 0 to x = 1, representing the first quadrant portion of the semicircle. Choice A incorrectly uses √(1 - x²) without squaring, which would give a linear measure rather than the disc's area. For disc method, always square the radius function—even when it contains a square root, the squaring simplifies the expression.
Which integral gives the volume when the region under $y=\cos x$ from $x=0$ to $x=\tfrac{\pi}{2}$ is revolved about the $x$ -axis?
Answer: $\pi\displaystyle\int_{0}^{\pi/2}(\cos x)^{2}\,dx$ This problem applies the disc method for revolving around the x-axis. When rotating the region under y = cos x from x = 0 to x = π/2 around the x-axis, we form circular discs perpendicular to the x-axis. Each disc has radius equal to the y-value at that x-position, which is cos x, so the disc's area is π(cos x)². The volume is found by integrating these disc areas: π∫₀^(π/2)(cos x)²dx. Choice B incorrectly uses π∫₀^(π/2) cos x dx, which fails to square the radius—this fundamental error would give a result that doesn't represent volume at all. The disc method for x-axis rotation always requires squaring: use π∫[f(x)]²dx to account for the circular area of each disc.
The region bounded by $x=\frac{y}{3}$ and the y-axis on $0\le y\le 9$ is revolved about the y-axis. Which integral gives the volume?
Answer: $\pi\int_{0}^{9} \left(\frac{y}{3}\right)^2\,dy$ This problem uses the disc method for revolution about the y-axis. Since the region is bounded by x = y/3 and the y-axis with no inner boundary, discs are the correct approach. The radius of each disc at position y is r(y) = y/3, the distance from the y-axis to the curve. The volume is V = π∫[a to b] [r(y)]² dy = π∫[0 to 9] (y/3)² dy. Choice B fails because it doesn't square the radius function, which is essential in the disc method. Apply discs when there's exactly one boundary curve extending from the axis of revolution.
A region bounded by $y=4-x^2$ and the x-axis for $-2\le x\le 2$ is revolved about the x-axis. Which integral gives the volume?
Answer: $\pi\int_{-2}^{2} (4-x^2)^2\,dx$ This problem requires the disc method for revolution about the x-axis. The region bounded by y = 4-x² and the x-axis forms a solid without a hole, making discs appropriate rather than washers. The radius of each disc at position x equals the distance from the x-axis to the curve: r(x) = 4-x². Applying the disc method: V = π∫[a to b] [r(x)]² dx = π∫[-2 to 2] (4-x²)² dx. Choice B is tempting but incorrect because it omits the essential squaring of the radius required in the disc method formula. Use discs when the region extends from the axis of revolution to a single bounding curve.
The region bounded by $x=\frac{y}{y+1}$ and the y-axis on $0\le y\le 2$ is revolved about the y-axis. Which integral gives the volume?
Answer: $\pi\int_{0}^{2} \left(\frac{y}{y+1}\right)^2\,dy$ This problem uses the disc method for revolution about the y-axis. Since the region is bounded by x = y/(y+1) and the y-axis with no inner boundary, discs are the correct approach. The radius of each disc at position y is r(y) = y/(y+1), the distance from the y-axis to the curve. The volume is V = π∫[a to b] [r(y)]² dy = π∫[0 to 2] (y/(y+1))² dy. Choice B fails because it doesn't square the radius function, which is essential in the disc method. Apply discs when there's exactly one boundary curve extending from the axis of revolution.
The region bounded by $x=1-\cos y$ and the y-axis on $0\le y\le \pi$ is revolved about the y-axis. Which integral gives the volume?
Answer: $\pi\int_{0}^{\pi} (1-\cos y)^2\,dy$ This problem uses the disc method for revolution about the y-axis. Since the region is bounded by x = 1-cos y and the y-axis with no inner boundary, discs are the appropriate method. The radius of each disc at position y is the distance from the y-axis to the curve: r(y) = 1-cos y. The volume formula becomes V = $\pi \int_{a}^{b} [r(y)]^2 \, dy$ = $\pi \int_{0}^{\pi} (1 - \cos y)^2 \, dy$ . Choice B fails because it doesn't square the radius function, which is essential in the disc method. Apply the disc method when there's a single boundary curve extending from the axis of revolution.

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