Disc Method: Revolving Around x/y Axes
Help Questions
AP Calculus AB › Disc Method: Revolving Around x/y Axes
A region bounded by $y=\sqrt{4-x^2}$ and the x-axis for $-2\le x\le 2$ is revolved about the x-axis. Which integral gives the volume?
$\pi\int_{0}^{2} (\sqrt{4-x^2})^2,dx$
$\pi\int_{-2}^{2} (\sqrt{4-x^2})^2,dx$
$\pi\int_{-2}^{2} (\sqrt{4-x^2})^2,dy$
$\pi\int_{-2}^{2} \sqrt{4-x^2},dx$
$\pi\int_{-2}^{2} (4-x^2)^2,dx$
Explanation
This problem applies the disc method for revolution about the x-axis. The region bounded by y = √(4-x²) and the x-axis creates a solid with no hole, so discs rather than washers are appropriate. The radius of each disc at position x equals the distance from the x-axis to the curve: r(x) = √(4-x²). Using the disc method formula: V = π∫[a to b] [r(x)]² dx = π∫[-2 to 2] (√(4-x²))² dx. 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 $x=3\sqrt{y}$ and the y-axis on $0\le y\le 4$ is revolved about the y-axis. Which integral gives the volume?
$\pi\int_{0}^{4} (\sqrt{y})^2,dy$
$\pi\int_{0}^{4} 3\sqrt{y},dy$
$\pi\int_{0}^{4} (3\sqrt{y})^2,dy$
$\pi\int_{0}^{4} (3\sqrt{y})^2,dx$
$\pi\int_{0}^{2} (3\sqrt{y})^2,dy$
Explanation
This problem uses the disc method for revolution about the y-axis. Since the region is bounded by x = 3√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) = 3√y. The volume formula becomes V = π∫[a to b] [r(y)]² dy = π∫[0 to 4] (3√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.
The region bounded by $x=2+y^3$ and the y-axis on $0\le y\le 1$ is revolved about the y-axis. Which integral gives the volume?
$\pi\int_{0}^{1} (2+y^3)^2,dx$
$\pi\int_{0}^{1} (y^3)^2,dy$
$\pi\int_{0}^{1} (2+y^3)^2,dy$
$\pi\int_{0}^{1} (2+y^3),dy$
$\pi\int_{1}^{0} (2+y^3)^2,dy$
Explanation
This problem uses the disc method for revolution about the y-axis. Since the region is bounded by x = 2+y³ and the y-axis with no inner boundary, discs are the correct approach. The radius of each disc at position y is r(y) = 2+y³, the distance from the y-axis to the curve. The volume is V = π∫[a to b] [r(y)]² dy = π∫[0 to 1] (2+y³)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 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?
$\pi\int_{-2}^{2} (x^2-4)^2,dx$
$\pi\int_{0}^{2} (4-x^2)^2,dx$
$\pi\int_{-2}^{2} (4-x^2)^2,dy$
$\pi\int_{-2}^{2} (4-x^2)^2,dx$
$\pi\int_{-2}^{2} (4-x^2),dx$
Explanation
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.
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?
$\pi\int_{0}^{2} (y^2)^2,dx$
$\pi\int_{0}^{4} (y^2)^2,dy$
$\pi\int_{0}^{2} (y^2)^2,dy$
$2\pi\int_{0}^{2} (y^2)^2,dy$
$\pi\int_{0}^{2} y^2,dy$
Explanation
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.
A region bounded by $y=x-x^3$ and the x-axis for $0\le x\le 1$ is revolved about the x-axis. Which integral gives the volume?
$\pi\int_{-1}^{1} (x-x^3)^2,dx$
$\pi\int_{0}^{1} (x-x^3),dx$
$\pi\int_{0}^{1} (x-x^3)^2,dy$
$\pi\int_{0}^{1} (x^3-x)^2,dx$
$\pi\int_{0}^{1} (x-x^3)^2,dx$
Explanation
This problem applies the disc method for revolution about the x-axis. The region bounded by y = x-x³ and the x-axis creates a solid with no hole, so discs rather than washers are appropriate. The radius of each disc at position x equals the distance from the x-axis to the curve: r(x) = x-x³. Using the disc method formula: V = π∫[a to b] [r(x)]² dx = π∫[0 to 1] (x-x³)² dx. 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.
A region bounded by $x=\sin y$ and the y-axis for $0\le y\le \pi$ is revolved about the y-axis. Which integral gives the volume?
$\pi\int_{0}^{\pi} (\sin y)^2,dx$
$\pi\int_{0}^{2\pi} (\sin y)^2,dy$
$2\pi\int_{0}^{\pi} (\sin y)^2,dy$
$\pi\int_{0}^{\pi} (\sin y)^2,dy$
$\pi\int_{0}^{\pi} \sin y,dy$
Explanation
This problem applies the disc method for revolution about the y-axis. The region bounded by x = sin y 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) = sin y. Using the disc method formula: V = π∫[a to b] [r(y)]² dy = π∫[0 to π] (sin y)² dy. Choice B is incorrect because it omits the required squaring of the radius in the disc method formula. Use the disc method when the solid extends from the axis of revolution to a single boundary curve.
A region bounded by $y=e^x$ and the x-axis for $0\le x\le 1$ is revolved about the x-axis. Which integral gives the volume?
$\pi\int_{1}^{0} (e^x)^2,dx$
$2\pi\int_{0}^{1} (e^x)^2,dx$
$\pi\int_{0}^{1} (e^x)^2,dx$
$\pi\int_{0}^{1} e^x,dx$
$\pi\int_{0}^{1} (e^x)^2,dy$
Explanation
This problem applies the disc method for revolution about the x-axis. The region bounded by y = eˣ and the x-axis creates a solid with no hole, so discs are appropriate. The radius of each disc at position x equals the distance from the x-axis to the curve: r(x) = eˣ. Using the disc method formula: V = π∫[a to b] [r(x)]² dx = π∫[0 to 1] (eˣ)² dx. Choice A is incorrect because it omits the essential squaring of the radius function required in the disc method. Apply discs when there's a single boundary curve extending from the axis of revolution.
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?
$\pi\int_{0}^{\pi/2} (\sin x)^2,dx$
$\pi\int_{0}^{\pi/2} \cos x,dx$
$\pi\int_{0}^{\pi} (\cos x)^2,dx$
$\pi\int_{0}^{\pi/2} (\cos x)^2,dx$
$\pi\int_{0}^{\pi/2} (\cos x)^2,dy$
Explanation
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.
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?
$\pi\int_{0}^{2} (2-x)^2,dx$
$\pi\int_{0}^{2} (x-2)^2,dx$
$\pi\int_{0}^{2} (2-x),dx$
$\pi\int_{0}^{2} (2-x)^2,dy$
$\pi\int_{-2}^{0} (2-x)^2,dx$
Explanation
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.