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AP Calculus AB 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

Let R be the region in the first quadrant bounded by the graph of y = √x, the x-axis, and the line x = 4. 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)

Let R be the region in the first quadrant bounded by the graph of y = √x, the x-axis, and the line x = 4. Which of the following definite integrals gives the volume of the solid generated when R is revolved about the line y = -1?

A) π ∫[0, 4] (√x + 1)² dx

B) π ∫[0, 4] (√x - 1)² dx

C) π ∫[0, 4] (x + 1) dx

D) π ∫[0, 4] ((√x)² - 1²) dx

Correct Answer: A

The solid is formed by revolving a region about a horizontal line, so the disc method with integration with respect to x is appropriate. The radius R(x) of a disc at a given x is the distance from the axis of revolution (y = -1) to the outer edge of the region (y = √x). This distance is R(x) = √x - (-1) = √x + 1. The volume is given by the definite integral V = π ∫[a, b] (R(x))² dx. The bounds of integration are given as x = 0 (since it's bounded by the x-axis) to x = 4. Therefore, the integral is π ∫[0, 4] (√x + 1)² dx.

Let R be the region enclosed by the graph of y = x² and the line y = 4. What is the volume of the solid generated when R is revolved about the line y = 4?

A) π ∫[-2, 2] (4 - x²)² dx

B) π ∫[-2, 2] (x² - 4)² dx

C) π ∫[0, 4] (16 - x⁴) dx

D) π ∫[0, 4] (√y)² dy

Correct Answer: A

The axis of revolution is the horizontal line y = 4, which is a boundary of the region. We use the disc method with respect to x. The radius R(x) is the distance from the axis of revolution (y = 4) to the curve (y = x²), which is R(x) = 4 - x². To find the bounds of integration, we set the functions equal: x² = 4, which gives x = ±2. The volume is V = π ∫[a, b] (R(x))² dx, so the integral is π ∫[-2, 2] (4 - x²)² dx.

Let R be the region in the first quadrant bounded by the graph of x = y², the y-axis, and the line y = 2. Which of the following definite integrals represents the volume of the solid formed by revolving R about the vertical line x = -3?

A) π ∫[0, 2] (y² + 3)² dy

B) π ∫[0, 2] (y² - 3)² dy

C) π ∫[0, 4] (√x + 3)² dx

D) π ∫[0, 2] ((y²)² - 3²) dy

Correct Answer: A

Since the region is revolved around a vertical line, we integrate with respect to y. The radius R(y) of a disc is the distance from the axis of revolution (x = -3) to the outer edge of the region (x = y²). The distance is R(y) = (right function) - (left axis) = y² - (-3) = y² + 3. The region is bounded by y = 0 (the y-axis) and y = 2. The volume is V = π ∫[c, d] (R(y))² dy. Therefore, the integral is π ∫[0, 2] (y² + 3)² dy.

Let R be the region bounded by the graph of y = ln(x), the x-axis, and the line x = e. The region R is revolved about the vertical line x = e. Which definite integral gives the volume of the resulting solid?

A) π ∫[0, 1] (e - eʸ)² dy

B) π ∫[1, e] (e - ln(x))² dx

C) π ∫[0, 1] (eʸ - e)² dy

D) π ∫[0, e] (e - eʸ)² dy

Correct Answer: A

To revolve around a vertical line, we must integrate with respect to y. First, express x in terms of y: if y = ln(x), then x = eʸ. Next, determine the bounds for y. The region is bounded by the x-axis (y = 0). The upper bound is where x = e, so y = ln(e) = 1. The bounds are from y = 0 to y = 1. The radius R(y) is the distance from the axis of revolution (x = e) to the curve (x = eʸ). This is R(y) = (right axis) - (left curve) = e - eʸ. The volume is V = π ∫[c, d] (R(y))² dy, which gives π ∫[0, 1] (e - eʸ)² dy.

The region R is bounded by the graph of a positive function y = f(x), the x-axis, and the lines x = a and x = b. The solid generated by revolving R about the horizontal line y = k, where k is a negative constant, is found using the disc method. Which expression represents the volume of this solid?

A) π ∫[a, b] (f(x) - k)² dx

B) π ∫[a, b] (f(x) + k)² dx

C) π ∫[a, b] ((f(x))² - k²) dx

D) π ∫[a, b] (k - f(x))² dx

Correct Answer: A

The volume of a solid of revolution using the disc method is V = π ∫[a, b] (R(x))² dx. The radius R(x) is the distance between the axis of revolution and the curve. In this case, the axis is y = k and the curve is y = f(x). The distance is the 'top' value minus the 'bottom' value. Since f(x) is positive and k is negative, f(x) is always above k. Therefore, the radius is R(x) = f(x) - k. Squaring this radius gives the integrand (f(x) - k)², leading to the integral π ∫[a, b] (f(x) - k)² dx.

Let R be the region in the first quadrant enclosed by the graph of y = x³, the y-axis, and the line y = 8. Which of the following gives the volume of the solid generated when R is revolved about the vertical line x = 2?

A) π ∫[0, 8] (2 - ³√y)² dy

B) π ∫[0, 2] (8 - x³)² dx

C) π ∫[0, 8] (4 - (³√y)²) dy

D) π ∫[0, 2] (2 - x³)² dx

Correct Answer: A

The revolution is about a vertical line, so we integrate with respect to y. We must express the curve as x in terms of y: x = ³√y. The bounds for y are given as 0 to 8. The axis of revolution is x = 2. The curve x = ³√y is the left boundary of the solid's cross-section, and the axis x=2 is the right boundary. The radius is R(y) = (right) - (left) = 2 - ³√y. The volume is given by V = π ∫[c, d] (R(y))² dy. Substituting the radius and bounds gives π ∫[0, 8] (2 - ³√y)² dy.

The region in the first quadrant bounded by the graph of y = x, the x-axis, and the line x = 3 is revolved about the line y = -2. The volume of the resulting solid is given by which definite integral?

A) π ∫[0, 3] (x + 2)² dx

B) π ∫[0, 3] (x - 2)² dx

C) π ∫[0, 3] x² dx

D) π ∫[0, 3] (y + 2)² dy

Correct Answer: A

The solid is generated by revolving a region about a horizontal line, so we use the disc method with respect to x. The radius R(x) is the distance from the axis of revolution (y = -2) to the function (y = x). The distance is R(x) = (top function) - (bottom axis) = x - (-2) = x + 2. The bounds of integration are from x = 0 to x = 3. The volume formula is V = π ∫[a, b] (R(x))² dx, which results in the integral π ∫[0, 3] (x + 2)² dx.