AP Calculus BC Quiz: Disc Method Revolving Around Xy Axes

What this quiz covers

This quiz focuses on Disc Method Revolving Around Xy Axes, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\pi\int_{0}^{1} e^y\,dy$
2. $\pi\int_{0}^{1} (e^y)^2\,dy$ (correct answer)
3. $\pi\int_{0}^{1} (1-e^y)^2\,dy$
4. $\pi\int_{0}^{e} y^2\,dy$
5. $\pi\int_{0}^{1} (e^y)^2\,dx$

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 $\pi$.

Question 2

1. $\pi\int_{0}^{2} (y^2)^2\,dy$ (correct answer)
2. $\pi\int_{0}^{2} y^2\,dy$
3. $\pi\int_{0}^{2} (y^2)^2\,dx$
4. $\pi\int_{0}^{4} y^2\,dy$
5. $\pi\int_{0}^{2} (2-y^2)^2\,dy$

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.

Question 3

1. $\pi\int_{0}^{1/2} \frac{1}{1-x}\,dx$
2. $\pi\int_{0}^{1/2} \left(\frac{1}{1-x}\right)^2\,dx$ (correct answer)
3. $\pi\int_{0}^{1} \left(\frac{1}{1-x}\right)^2\,dx$
4. $\pi\int_{0}^{1/2} \left(\frac{1}{1-x}\right)^2\,dy$
5. $\pi\int_{0}^{1/2} (1-\tfrac{1}{1-x})^2\,dx$

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.

Question 4

1. $\pi\int_{0}^{1} e^x\,dx$
2. $\pi\int_{0}^{1} (e^x)^2\,dx$ (correct answer)
3. $\pi\int_{0}^{1} (1-e^x)^2\,dx$
4. $\pi\int_{0}^{e} y^2\,dy$
5. $\pi\int_{0}^{1} (e^x)^2\,dy$

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.

Question 5

1. $\pi\int_{0}^{1} x^3\,dx$
2. $\pi\int_{0}^{1} (x^3)^2\,dx$ (correct answer)
3. $\pi\int_{0}^{3} x^2\,dx$
4. $\pi\int_{0}^{1} (x^3)^2\,dy$
5. $\pi\int_{0}^{1} (1-x^3)^2\,dx$

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.

Question 6

1. $\pi\int_{-1}^{3} \sqrt{x+1}\,dx$
2. $\pi\int_{-1}^{3} (\sqrt{x+1})^2\,dx$ (correct answer)
3. $\pi\int_{0}^{3} (\sqrt{x+1})^2\,dx$
4. $\pi\int_{-1}^{3} (\sqrt{x+1})^2\,dy$
5. $\pi\int_{-1}^{3} (3-\sqrt{x+1})^2\,dx$

Explanation: This problem applies the disc method when the region bounded by y = √(x+1) and the x-axis on -1 ≤ x ≤ 3 is revolved about the x-axis. Since the region is solid (bounded by the curve and the axis of rotation), we form discs with radius R(x) = √(x+1). The volume is π∫₋₁³ [R(x)]² dx = π∫₋₁³ [√(x+1)]² dx = π∫₋₁³ (x+1) dx. Choice A omits the crucial squaring operation, and the key principle is: disc method requires squaring the radius function, so [√(x+1)]² = x+1.

Question 7

1. $\pi\int_{-1}^{1} (1-x^2)^2\,dx$ (correct answer)
2. $\pi\int_{-1}^{1} (1-x^2)\,dx$
3. $\pi\int_{0}^{1} (1-x^2)^2\,dx$
4. $\pi\int_{-1}^{1} (x^2-1)^2\,dy$
5. $\pi\int_{-1}^{1} (1-(1-x^2))^2\,dx$

Explanation: This problem applies the disc method for revolving the region bounded by y = 1 - x² and the x-axis for -1 ≤ x ≤ 1 about the x-axis. The region forms solid discs with radius R(x) = 1 - x² extending from the axis of rotation to the curve. The volume integral is π∫₋₁¹ [R(x)]² dx = π∫₋₁¹ (1 - x²)² dx. Choice B omits the crucial squaring of the radius, which would calculate area under the curve rather than volume of revolution. The key rule: disc method requires π times the radius function squared.

Question 8

1. $\pi\int_{0}^{1} (3y)^2\,dy$ (correct answer)
2. $\pi\int_{0}^{1} 3y\,dy$
3. $\pi\int_{0}^{3} y^2\,dy$
4. $\pi\int_{0}^{1} (1-3y)^2\,dy$
5. $\pi\int_{0}^{1} (3y)^2\,dx$

Explanation: This problem applies the disc method when the region bounded by x = 3y and the y-axis from y = 0 to y = 1 is revolved about the y-axis. Since the region is solid (bounded by the curve and the axis of rotation), we form discs with radius R(y) = 3y. The volume integral is π∫₀¹ [R(y)]² dy = π∫₀¹ (3y)² dy. Choice B omits the essential squaring of the radius, which would calculate area instead of volume. The core rule: disc method about the y-axis always requires π times the square of the x-function.

Question 9

1. $\pi\int_{0}^{5}(2x)^2\,dx$ (correct answer)
2. $\pi\int_{0}^{5}2x\,dx$
3. $\pi\int_{0}^{10}x^2\,dx$
4. $\pi\int_{0}^{5}(5-2x)^2\,dx$
5. $\pi\int_{0}^{5}(2x)^2\,dy$

Explanation: This problem involves the disc method for finding volumes of solids of revolution. The disc method applies as the region is between y = 2x and the x-axis from 0 to 5, revolved about the x-axis, with no holes. The radius is y = 2x at each x. The volume integral is π∫(2x)² dx = π∫4x² dx from 0 to 5. Choice B, π∫2x dx, is tempting but incorrect since it omits squaring the radius, treating it like area. Remember, use the disc method when revolving a region directly against the axis of rotation without any gap; switch to washers if there's a hole in the solid.

Question 10

1. $\pi\int_{0}^{4} x\,dx$
2. $\pi\int_{0}^{4} (\sqrt{x})^2\,dx$ (correct answer)
3. $\pi\int_{0}^{2} (y^2)^2\,dy$
4. $2\pi\int_{0}^{4} \sqrt{x}\,dx$
5. $\pi\int_{0}^{4} \sqrt{x}\,dx$

Explanation: This problem requires the disc method to find the volume of a solid of revolution. The region is bounded by the curve and the x-axis, and when revolved about the x-axis, it forms a solid without holes, making the disc method appropriate rather than washers. The radius of each disc is the distance from the x-axis to the curve, which is given by y = √x. To set up the integral, we square this radius and integrate with respect to x from 0 to 4, yielding π ∫ from 0 to 4 of (√x)^2 dx. A tempting distractor like choice D includes a 2π factor, which would apply to the shell method instead, but that's incorrect here since we're revolving around the x-axis using discs. Use the disc method when revolving a region between a curve and the axis of rotation with no gap creating a hole; opt for washers if there's an inner and outer radius.

Question 11

1. $\pi\int_{-2}^{2} |y|\,dy$
2. $\pi\int_{-2}^{2} (|y|)^2\,dy$ (correct answer)
3. $\pi\int_{0}^{2} (|y|)^2\,dy$
4. $\pi\int_{-2}^{2} (|y|)^2\,dx$
5. $\pi\int_{-2}^{2} (2-|y|)^2\,dy$

Explanation: This problem uses the disc method when the region bounded by $x = |y|$ and the y-axis for $-2 \le y \le 2$ is revolved about the y-axis. Since we have solid discs with radius $R(y) = |y|$ extending from the y-axis to the curve, the volume is $\pi \int_{-2}^{2} [R(y)]^2 \, dy = \pi \int_{-2}^{2} (|y|)^2 \, dy = \pi \int_{-2}^{2} y^2 \, dy$ (since $|y|^2 = y^2$). Choice A incorrectly omits the squaring of the radius function, giving area rather than volume. The fundamental principle: disc method about the y-axis always involves $\pi$ times the square of the x-function.

Question 12

1. $\pi\int_{0}^{\pi/2} (2\cos x)^2\,dx$ (correct answer)
2. $\pi\int_{0}^{\pi/2} 2\cos x\,dx$
3. $\pi\int_{0}^{\pi} (2\cos x)^2\,dx$
4. $\pi\int_{0}^{\pi/2} (\cos x)^2\,dx$
5. $\pi\int_{0}^{\pi/2} (2\cos x)^2\,dy$

Explanation: This problem applies the disc method when the region bounded by y = 2cos x and the x-axis on 0 ≤ x ≤ π/2 is revolved about the x-axis. Since the region is solid (bounded by the curve and the axis of rotation), we form discs with radius R(x) = 2cos x. The volume is π∫₀^(π/2) [R(x)]² dx = π∫₀^(π/2) (2cos x)² dx. Choice B omits the essential squaring of the radius, which would yield area instead of volume. The core rule: disc method about the x-axis always involves π times the y-function squared.

Question 13

1. $\pi\int_{0}^{e-1} (\ln(y+1))^2\,dy$ (correct answer)
2. $\pi\int_{0}^{e-1} \ln(y+1)\,dy$
3. $\pi\int_{0}^{1} (\ln(y+1))^2\,dy$
4. $\pi\int_{0}^{e-1} (e-1-\ln(y+1))^2\,dy$
5. $\pi\int_{0}^{e-1} (\ln(y+1))^2\,dx$

Explanation: This problem uses the disc method for revolving the region bounded by x = ln(y+1) and the y-axis on 0 ≤ y ≤ e-1 about the y-axis. The region creates solid discs with radius R(y) = ln(y+1) extending from the axis of rotation to the curve. Each disc has area π[R(y)]² = π[ln(y+1)]², giving volume π∫₀^(e-1) [ln(y+1)]² dy. Choice B incorrectly omits the squaring of the radius function, yielding area rather than volume. The fundamental principle: disc method about the y-axis always requires π times the square of the x-function.

Question 14

1. $\pi\int_{0}^{2} (\tfrac{1}{2}x+1)^2\,dx$ (correct answer)
2. $\pi\int_{0}^{2} (\tfrac{1}{2}x+1)\,dx$
3. $\pi\int_{0}^{2} (1-\tfrac{1}{2}x)^2\,dx$
4. $\pi\int_{0}^{2} (\tfrac{1}{2}x+1)^2\,dy$
5. $\pi\int_{1}^{2} (\tfrac{1}{2}x+1)^2\,dx$

Explanation: This problem uses the disc method for revolving the region under $y = \frac{1}{2}x + 1$ above the x-axis from $x = 0$ to $x = 2$ about the x-axis. The region creates solid discs with radius $R(x) = \frac{1}{2}x + 1$ extending from the axis of rotation to the curve. Each disc has area $\pi [R(x)]^2 = \pi \left( \frac{1}{2}x + 1 \right)^2$, giving volume $\pi \int_0^2 \left( \frac{1}{2}x + 1 \right)^2 \, dx$. Choice B incorrectly omits the squaring of the radius function, calculating area instead of volume. The fundamental principle: disc method about the x-axis always requires $\pi$ times the square of the y-function.

Question 15

1. $\pi\int_{0}^{4} x\,dx$
2. $\pi\int_{0}^{2} (\sqrt{x})^2\,dx$
3. $\pi\int_{0}^{4} (\sqrt{x})^2\,dx$ (correct answer)
4. $\pi\int_{0}^{4} (4-\sqrt{x})^2\,dx$
5. $\pi\int_{0}^{4} (\sqrt{x})^2\,dy$

Explanation: This problem uses the disc method since we're revolving the region between $y = \sqrt{x}$ and the x-axis around the x-axis, creating solid discs. The radius of each disc at position x is $R(x) = \sqrt{x}$, extending from the x-axis up to the curve. The area of each disc is $\pi [R(x)]^2 = \pi (\sqrt{x})^2 = \pi x$, and integrating from x = 0 to x = 4 gives the volume $\pi \int_0^4 (\sqrt{x})^2 \, dx$. Choice A incorrectly omits the $\pi$ factor and doesn't square the radius. Remember: disc method always requires squaring the radius function and including the $\pi$ factor.

Question 16

1. $\pi\int_{0}^{2} (2-y)^2\,dy$ (correct answer)
2. $\pi\int_{0}^{2} (2-y)\,dy$
3. $\pi\int_{0}^{2} (y-2)^2\,dx$
4. $\pi\int_{0}^{2} (2-(2-y))^2\,dy$
5. $\pi\int_{0}^{2} y^2\,dy$

Explanation: This problem uses the disc method since the region bounded by x = 2 - y and the y-axis for 0 ≤ y ≤ 2 is revolved about the y-axis, creating solid discs. The radius of each disc at height y is R(y) = 2 - y, extending from the y-axis to the curve. The volume is π∫₀² [R(y)]² dy = π∫₀² (2 - y)² dy. Choice B incorrectly omits the squaring of the radius function, yielding area under the curve rather than volume of revolution. The key principle: disc method about the y-axis requires π times the square of the x-function.

Question 17

1. $\pi\int_{0}^{\pi} \sin y\,dy$
2. $\pi\int_{0}^{\pi} (\sin y)^2\,dy$ (correct answer)
3. $\pi\int_{0}^{\pi} (\sin y)^2\,dx$
4. $\pi\int_{0}^{1} y^2\,dy$
5. $\pi\int_{0}^{\pi} (\pi-\sin y)^2\,dy$

Explanation: This problem applies the disc method for revolving the region bounded by $x = \sin y$ and the y-axis on $0 \le y \le \pi$ about the y-axis. The region forms solid discs with radius $R(y) = \sin y$ extending from the axis of rotation to the curve. Each disc has area $\pi [R(y)]^2 = \pi (\sin y)^2$, so the volume integral is $\pi \int_0^\pi (\sin y)^2 \, dy$. Choice A omits the crucial squaring of the radius, which would give area instead of volume. The fundamental rule: disc method about the y-axis always involves $\pi$ times the square of the x-function.

Question 18

1. $\pi\int_{0}^{\pi} \sin x\,dx$
2. $\pi\int_{0}^{\pi} (\sin x)^2\,dx$ (correct answer)
3. $\pi\int_{0}^{\pi} (\pi-\sin x)^2\,dx$
4. $\pi\int_{0}^{1} x^2\,dx$
5. $\pi\int_{0}^{\pi} (\sin x)^2\,dy$

Explanation: This problem requires the disc method since the region bounded by $y = \sin x$ and the x-axis on $0 \le x \le \pi$ is revolved about the x-axis, creating solid discs. The radius of each disc is $R(x) = \sin x$, extending from the x-axis up to the curve. The volume is $\pi \int_0^\pi [R(x)]^2 \, dx = \pi \int_0^\pi (\sin x)^2 \, dx$. Choice A fails to square the radius function, which would give the area under the curve rather than the volume of revolution. The key rule: disc method always involves $\pi$ times the square of the radius function.

Question 19

1. $\pi\int_{0}^{2}(y^2)^2\,dy$ (correct answer)
2. $\pi\int_{0}^{2}y^2\,dx$
3. $\pi\int_{0}^{2}(2-y^2)^2\,dy$
4. $\pi\int_{0}^{2}(y^2)\,dy$
5. $\pi\int_{0}^{2}(y)^2\,dy$

Explanation: This problem involves the disc method for finding volumes of solids of revolution. Discs are used because the region is bounded by x = y² and the y-axis, revolving around the y-axis, resulting in a solid without holes. The radius is the distance from the y-axis to the curve, which is x = y². Integrate π times the square of the radius along y from 0 to 2, yielding π∫(y²)² dy = π∫y⁴ dy. Choice D, π∫y² dy, is a common distractor but incorrect as it omits squaring the radius, treating it like an area integral. Remember, use the disc method when revolving a region directly against the axis of rotation without any gap; switch to washers if there's a hole in the solid.

Question 20

1. $\pi\int_{0}^{4}x\,dx$
2. $\pi\int_{0}^{4}(\sqrt{x})\,dx$
3. $\pi\int_{0}^{4}(\sqrt{x})^2\,dx$ (correct answer)
4. $\pi\int_{0}^{2}x^2\,dx$
5. $\pi\int_{0}^{4}(4-\sqrt{x})^2\,dx$

Explanation: This problem involves the disc method for finding volumes of solids of revolution. The disc method is appropriate since the region is between $y = \sqrt{x}$ and the $x$-axis, revolving around the $x$-axis, forming a solid with no holes. The radius of each disc is the function value $y = \sqrt{x}$ at each $x$. The volume is given by $\pi \int_{0}^{4} (\sqrt{x})^2 \, dx$, which simplifies to $\pi \int_{0}^{4} x \, dx$. Choice A is tempting as it matches the simplified integral, but it fails to show the explicit squaring of the radius, missing the conceptual setup. Remember, use the disc method when revolving a region directly against the axis of rotation without any gap; switch to washers if there's a hole in the solid.