Disc Method: Revolving Around x/y Axes
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AP Calculus BC › Disc Method: Revolving Around x/y Axes
The region bounded by $x=e^y$ and the y-axis for $0 \le y \le 1$ is revolved about the y-axis; which integral represents the volume?
$\pi\int_{0}^{1} (e^y)^2,dy$
$\pi\int_{0}^{1} (e^y)^2,dx$
$\pi\int_{0}^{1} (1-e^y)^2,dy$
$\pi\int_{0}^{e} y^2,dy$
$\pi\int_{0}^{1} e^y,dy$
Explanation
This problem applies the disc method since the region bounded by $x = e^y$ and the y-axis for $0 \le y \le 1$ is revolved about the y-axis, forming solid discs. The radius of each disc at height y is $R(y) = e^y$, extending from the y-axis to the curve. The volume integral is $\pi \int_0^1 [R(y)]^2 , dy = \pi \int_0^1 (e^y)^2 , dy$. Choice A omits the essential squaring of the radius, which would yield area instead of volume. The core rule for disc method about the y-axis: always square the x-function and multiply by $\pi$.
Revolve the region bounded by $y=\frac{x}{2}$ and the x-axis from $x=0$ to $x=4$ about the x-axis; select the correct integral.
$\pi\int_{0}^{4} \left(\frac{x}{2}\right)^2,dy$
$\pi\int_{0}^{2} \left(\frac{x}{2}\right)^2,dx$
$\pi\int_{0}^{4} \frac{x}{2},dx$
$\pi\int_{0}^{4} (4-\tfrac{x}{2})^2,dx$
$\pi\int_{0}^{4} \left(\frac{x}{2}\right)^2,dx$
Explanation
This problem uses the disc method since the region bounded by y = x/2 and the x-axis from x = 0 to x = 4 is revolved about the x-axis, creating solid discs. The radius of each disc at position x is R(x) = x/2, extending from the x-axis to the curve. The volume integral is π∫₀⁴ [R(x)]² dx = π∫₀⁴ (x/2)² dx. Choice B incorrectly omits the squaring of the radius function, giving area under the curve rather than volume of revolution. The fundamental principle: disc method about the x-axis always requires π times the square of the y-function.
Revolve the region bounded by $x=2\sin y$ and the y-axis on $0\le y\le \frac{\pi}{2}$ about the y-axis; select the volume setup.
$\pi\int_{0}^{\pi/2} (\sin y)^2,dy$
$\pi\int_{0}^{\pi/2} (2\sin y)^2,dy$
$\pi\int_{0}^{\pi/2} 2\sin y,dy$
$\pi\int_{0}^{\pi} (2\sin y)^2,dy$
$\pi\int_{0}^{\pi/2} (2\sin y)^2,dx$
Explanation
This problem uses the disc method since the region bounded by x = 2sin y and the y-axis on 0 ≤ y ≤ π/2 is revolved about the y-axis, creating solid discs. The radius of each disc at height y is R(y) = 2sin y, extending from the y-axis to the curve. The volume integral is π∫₀^(π/2) [R(y)]² dy = π∫₀^(π/2) (2sin y)² dy. Choice B incorrectly omits the squaring of the radius function, giving area under the curve rather than volume of revolution. The fundamental principle: disc method about the y-axis always requires π times the square of the x-function.
Revolve the region bounded by $x=y^2$ and the y-axis for $0\le y\le 2$ about the y-axis; which disc-method integral is correct?
$\pi\int_{0}^{4} y^2,dy$
$\pi\int_{0}^{2} (y^2)^2,dy$
$\pi\int_{0}^{2} y^2,dy$
$\pi\int_{0}^{2} (y^2)^2,dx$
$\pi\int_{0}^{2} (2-y^2)^2,dy$
Explanation
This problem applies the disc method when the region bounded by x = y² and the y-axis for 0 ≤ y ≤ 2 is revolved about the y-axis. Since we're revolving around the y-axis, the radius at height y is R(y) = y², extending from the y-axis to the curve. Each disc has area π[R(y)]² = π(y²)², and the volume integral is π∫₀² (y²)² dy. Choice B omits the crucial squaring of the radius, yielding area instead of volume. The essential rule: for disc method about the y-axis, square the x-function and integrate with respect to y.
A region bounded by $y=x^2$ and the x-axis from $x=0$ to $x=2$ is revolved about the x-axis; which integral gives the volume?
$\pi\int_{0}^{2} (2-x^2)^2,dx$
$\pi\int_{0}^{2} (x^2)^2,dx$
$\pi\int_{0}^{4} y^2,dy$
$\pi\int_{0}^{2} x^2,dy$
$\pi\int_{0}^{2} x^2,dx$
Explanation
This problem requires the disc method to find the volume when the region bounded by y = x² and the x-axis from x = 0 to x = 2 is revolved about the x-axis. Since we're revolving around the x-axis and the region extends from the curve down to the axis of rotation, we use discs with radius R(x) = x². Each disc has area π[R(x)]² = π(x²)² = πx⁴, so the volume integral is π∫₀² (x²)² dx. Choice A omits the squaring of the radius, which would give area rather than volume. The key principle is: for disc method, always square the radius function in the integrand.
The region under $y=\frac{1}{1-x}$ above the x-axis on $0\le x\le \frac{1}{2}$ is revolved about the x-axis; which setup is correct?
$\pi\int_{0}^{1/2} \frac{1}{1-x},dx$
$\pi\int_{0}^{1/2} \left(\frac{1}{1-x}\right)^2,dy$
$\pi\int_{0}^{1/2} (1-\tfrac{1}{1-x})^2,dx$
$\pi\int_{0}^{1} \left(\frac{1}{1-x}\right)^2,dx$
$\pi\int_{0}^{1/2} \left(\frac{1}{1-x}\right)^2,dx$
Explanation
This problem applies the disc method since the region under y = 1/(1-x) above the x-axis on 0 ≤ x ≤ 1/2 is revolved about the x-axis, forming solid discs. The radius of each disc at position x is R(x) = 1/(1-x), extending from the x-axis to the curve. The volume integral is π∫₀^(1/2) [R(x)]² dx = π∫₀^(1/2) [1/(1-x)]² dx. Choice A omits the crucial squaring of the radius function, yielding area under the curve rather than volume of revolution. The key rule: disc method about the x-axis always requires π times the square of the y-function.
The region between $y=e^x$ and the x-axis from $x=0$ to $x=1$ is revolved about the x-axis; which integral represents the volume?
$\pi\int_{0}^{1} (e^x)^2,dx$
$\pi\int_{0}^{e} y^2,dy$
$\pi\int_{0}^{1} (1-e^x)^2,dx$
$\pi\int_{0}^{1} (e^x)^2,dy$
$\pi\int_{0}^{1} e^x,dx$
Explanation
This problem applies the disc method when the region between y = eˣ and the x-axis from x = 0 to x = 1 is revolved about the x-axis. The region creates solid discs with radius R(x) = eˣ extending from the axis of rotation to the curve. Each disc has area π[R(x)]² = π(eˣ)², giving volume π∫₀¹ (eˣ)² dx. Choice A omits the essential squaring of the radius, yielding area rather than volume. The core rule for disc method: always square the radius function and multiply by π to obtain volume.
The region under $y=2- x$ above the x-axis on $0\le x\le 2$ is revolved about the x-axis; which integral gives the volume?
$\pi\int_{0}^{2} (x-2)^2,dy$
$\pi\int_{0}^{2} (2-x)^2,dx$
$\pi\int_{0}^{2} (2-x),dx$
$\pi\int_{0}^{2} x^2,dx$
$\pi\int_{0}^{2} (2-(2-x))^2,dx$
Explanation
This problem uses the disc method to find volume when the region under y = 2 - x above the x-axis on 0 ≤ x ≤ 2 is revolved about the x-axis. The region forms solid discs with radius R(x) = 2 - x extending from the x-axis to the curve. Each disc has area π[R(x)]² = π(2 - x)², so the volume integral is π∫₀² (2 - x)² dx. Choice A omits the crucial squaring of the radius, giving area rather than volume. The essential principle: disc method always requires squaring the distance from the axis of rotation.
Revolve the region under $y=x^3$ above the x-axis from $x=0$ to $x=1$ about the x-axis; select the correct integral.
$\pi\int_{0}^{1} (x^3)^2,dx$
$\pi\int_{0}^{1} (x^3)^2,dy$
$\pi\int_{0}^{3} x^2,dx$
$\pi\int_{0}^{1} x^3,dx$
$\pi\int_{0}^{1} (1-x^3)^2,dx$
Explanation
This problem uses the disc method for revolving the region under y = x³ above the x-axis from x = 0 to x = 1 about the x-axis. The region creates solid discs with radius R(x) = x³ extending from the axis of rotation to the curve. Each disc has area π[R(x)]² = π(x³)², giving volume π∫₀¹ (x³)² dx. Choice A incorrectly omits the squaring of the radius, which would calculate area instead of volume. The essential principle: disc method about the x-axis requires π times the square of the y-function for proper volume calculation.
The region under $y=\frac{2}{1+x^2}$ above the x-axis on $0\le x\le 1$ is revolved about the x-axis; which setup is correct?
$\pi\int_{0}^{1} (1-\tfrac{2}{1+x^2})^2,dx$
$\pi\int_{0}^{1} \frac{2}{1+x^2},dx$
$\pi\int_{0}^{2} \left(\frac{2}{1+x^2}\right)^2,dx$
$\pi\int_{0}^{1} \left(\frac{2}{1+x^2}\right)^2,dy$
$\pi\int_{0}^{1} \left(\frac{2}{1+x^2}\right)^2,dx$
Explanation
This problem applies the disc method for revolving the region under y = 2/(1+x²) above the x-axis on 0 ≤ x ≤ 1 about the x-axis. The region forms solid discs with radius R(x) = 2/(1+x²) extending from the axis of rotation to the curve. Each disc has area π[R(x)]² = π[2/(1+x²)]², so the volume integral is π∫₀¹ [2/(1+x²)]² dx. Choice A omits the crucial squaring of the radius, which would calculate area instead of volume. The key rule: disc method about the x-axis requires π times the radius function squared for proper volume calculation.