PrepGo

AP Calculus BC Practice Quiz: Volume with Disc Method: Revolving Around Other Axes

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 7 questions to check your progress.

Question 1 of 7

The region R is bounded by the graph of y = x², the x-axis, and the line x = 2. Which of the following definite integrals gives the volume of the solid generated when R is revolved about the line y = -1?

All Questions (7)

The region R is bounded by the graph of y = x², the x-axis, and the line x = 2. Which of the following definite integrals gives the volume of the solid generated when R is revolved about the line y = -1?

A) π ∫₀² (x² + 1)² dx

B) π ∫₀² (x² - 1)² dx

C) π ∫₀² (x²)² dx

D) π ∫₀⁴ (√y + 1)² dy

Correct Answer: A

When revolving around a horizontal line, the disc method uses an integral with respect to x. The radius R(x) of a representative disc is the distance from the axis of revolution (y = -1) to the outer edge of the region (y = x²). This distance is calculated as 'top curve - bottom line', so R(x) = x² - (-1) = x² + 1. The volume formula is V = π ∫[a,b] (R(x))² dx. Therefore, the integral is π ∫₀² (x² + 1)² dx.

Let R be the region enclosed by the graph of y = √x, the y-axis, and the line y = 3. Which definite integral represents the volume of the solid formed by revolving R about the vertical line x = -2?

A) π ∫₀³ (y²)² dy

B) π ∫₀³ (y² - 2)² dy

C) π ∫₀³ (y² + 2)² dy

D) π ∫₀⁹ (√x + 2)² dx

Correct Answer: C

To revolve around a vertical line, the disc method requires integration with respect to y. The function must be expressed as x in terms of y: y = √x becomes x = y². The radius R(y) is the distance from the axis of revolution (x = -2) to the outer edge of the region (x = y²). This distance is 'right curve - left line', so R(y) = y² - (-2) = y² + 2. The bounds for y are given as 0 to 3. The volume is V = π ∫[c,d] (R(y))² dy, which gives π ∫₀³ (y² + 2)² dy.

The region in the first quadrant bounded by the graph of y = 4 - x², the x-axis, and the y-axis is revolved about the line y = 5. The volume of the resulting solid is given by which integral?

A) π ∫₀² (5 - (4 - x²))² dx

B) π ∫₀² ((4 - x²) - 5)² dx

C) π ∫₀⁴ (5 - √ (4 - y))² dy

D) π ∫₀² (4 - x²)² dx

Correct Answer: A

The problem describes revolving a region around a horizontal line (y=5) that is above the region. The radius R(x) is the distance from the axis of revolution to the curve. This is calculated as 'top line - bottom curve'. Here, the top is the axis y = 5 and the bottom is the function y = 4 - x². So, R(x) = 5 - (4 - x²). The region is bounded by the x-axis (y=0) and y=4-x², so the x-bounds are from x=0 to x=2. The volume integral is V = π ∫[a,b] (R(x))² dx, which is π ∫₀² (5 - (4 - x²))² dx.

Let R be the region bounded by the graph of x = eʸ, the y-axis, and the lines y = 0 and y = 1. What is the volume of the solid generated when R is revolved about the line x = 4?

A) π ∫₀¹ (4 - eʸ)² dy

B) π ∫₀¹ (eʸ - 4)² dy

C) π ∫₁ᵉ (4 - ln(x))² dx

D) π ∫₀¹ (eʸ)² dy

Correct Answer: A

The revolution is about a vertical line (x = 4), so we must integrate with respect to y. The radius R(y) is the distance from the axis of revolution (x=4) to the curve (x=eʸ). Since the axis is to the right of the region, the radius is 'right line - left curve', which is R(y) = 4 - eʸ. The bounds for y are given as 0 to 1. The volume is V = π ∫[c,d] (R(y))² dy, resulting in the integral π ∫₀¹ (4 - eʸ)² dy.

A solid is created by revolving the region bounded by y = f(x), the x-axis, and the lines x = a and x = b about the horizontal line y = c, where c is a constant such that c < 0. The radius of a representative disc for this solid is given by:

A) f(x) + c

B) f(x) - c

C) c - f(x)

D) f(x)

Correct Answer: B

The radius of a disc is the distance from the axis of revolution to the edge of the region. For a horizontal axis of revolution y = c, the radius R(x) is the vertical distance. This is calculated as 'top y-value - bottom y-value'. The top of the region is defined by y = f(x) and the bottom is the axis of revolution y = c. Therefore, the radius is R(x) = f(x) - c. Since c is negative, this results in a value larger than f(x), which is correct.

The region enclosed by the graph of y = x, the line x = 4, and the x-axis is revolved about the line x = 4. Which of the following integrals gives the volume of the solid?

A) π ∫₀⁴ (4 - y)² dy

B) π ∫₀⁴ (4 - x)² dx

C) π ∫₀⁴ (y - 4)² dy

D) π ∫₀⁴ (x)² dx

Correct Answer: A

Since the region is revolved around a vertical line (x = 4), the integration must be with respect to y. The function y = x is rewritten as x = y. The region is a triangle with vertices at (0,0), (4,0), and (4,4), so the y-bounds are from 0 to 4. The radius R(y) is the horizontal distance from the axis of revolution (x = 4) to the function (x = y). This is calculated as 'right boundary - left boundary', so R(y) = 4 - y. The volume is given by V = π ∫[c,d] (R(y))² dy, which is π ∫₀⁴ (4 - y)² dy.

Let R be the region bounded by y = 2x, y = 6, and the y-axis. Which definite integral represents the volume of the solid formed by revolving R about the line y = 6?

A) π ∫₀³ (6 - 2x)² dx

B) π ∫₀⁶ (6 - y/2)² dy

C) π ∫₀³ (2x - 6)² dx

D) π ∫₀³ (6)² - (2x)² dx

Correct Answer: A

The region is revolved around a horizontal line (y = 6), so the integration is with respect to x. The region is a triangle bounded by y=2x, y=6, and x=0. The intersection of y=2x and y=6 occurs at 2x=6, or x=3. So, the x-bounds are from 0 to 3. The radius R(x) is the vertical distance from the axis of revolution (y = 6) to the function (y = 2x). This is 'top boundary - bottom boundary', so R(x) = 6 - 2x. The volume is V = π ∫[a,b] (R(x))² dx, which is π ∫₀³ (6 - 2x)² dx.