AP Calculus BC Practice Quiz: Volume with Washer Method: Revolving Around the $x$- or $y$-Axis
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
All Questions (7)
A) π ∫[1,4] (x - 1) dx
B) π ∫[1,4] (√x - 1)² dx
C) π ∫[1,2] (16 - (y²)²) dy
D) π ∫[1,4] ( (√x)² - 1 ) dx
Correct Answer: A
When revolving around the x-axis, the washer method formula is V = π ∫[a,b] (R(x)² - r(x)²) dx. The region is bounded by x=1 (where y=√x intersects y=1) and x=4. The outer radius R(x) is the distance from the x-axis to the upper curve, so R(x) = √x. The inner radius r(x) is the distance from the x-axis to the lower curve, so r(x) = 1. The integral is π ∫[1,4] ((√x)² - 1²) dx, which simplifies to π ∫[1,4] (x - 1) dx.
A) π ∫[0,1] (y - y⁴) dy
B) π ∫[0,1] (√y - y²)² dy
C) π ∫[0,1] (x² - x⁴) dx
D) π ∫[0,1] (y² - y) dy
Correct Answer: A
To revolve around the y-axis, we use the washer method with functions of y: V = π ∫[c,d] (R(y)² - r(y)²) dy. The curves must be expressed as x in terms of y: x = √y and x = y². The intersection points are (0,0) and (1,1), so the y-bounds are from 0 to 1. In this interval, √y ≥ y², so the outer radius is R(y) = √y and the inner radius is r(y) = y². The integral is π ∫[0,1] ((√y)² - (y²)²) dy, which simplifies to π ∫[0,1] (y - y⁴) dy.
A) π ∫[-1,2] (x² - (x+2))² dx
B) π ∫[-1,2] ((x+2)² - x⁴) dx
C) π ∫[0,4] (y - (y-2)²) dy
D) π ∫[-1,2] (x+2 - x²) dx
Correct Answer: B
First, find the bounds of integration by setting the functions equal: x² = x + 2 → x² - x - 2 = 0 → (x-2)(x+1) = 0. The intersections are at x = -1 and x = 2. For x in [-1, 2], the line y = x + 2 is above the parabola y = x². Using the washer method for revolution about the x-axis, the outer radius is R(x) = x + 2 and the inner radius is r(x) = x². The volume is given by the definite integral V = π ∫[-1,2] (R(x)² - r(x)²) dx = π ∫[-1,2] ((x+2)² - (x²)²) dx, which is π ∫[-1,2] ((x+2)² - x⁴) dx.
A) π ∫[0,1] (e² - (eʸ)²) dy
B) π ∫[1,e] (ln(x))² dx
C) π ∫[0,e] (1 - (ln(x))²) dx
D) π ∫[0,1] (e - eʸ)² dy
Correct Answer: A
For revolution about the y-axis, we integrate with respect to y. The region's y-values range from y = 0 to y = ln(e) = 1. The cross-sections are washers. The outer radius R(y) is the constant distance from the y-axis to the line x = e, so R(y) = e. The inner radius r(y) is the distance from the y-axis to the curve y = ln(x), which must be rewritten as x = eʸ. So, r(y) = eʸ. The volume is V = π ∫[0,1] (R(y)² - r(y)²) dy = π ∫[0,1] (e² - (eʸ)²) dy.
A) The exact volume of the solid.
B) The area of a single ring-shaped cross section (washer) at a specific x-value.
C) The circumference of the outer edge of a washer at a specific x-value.
D) The height of the region at a specific x-value.
Correct Answer: B
The washer method calculates volume by summing the volumes of infinitesimally thin ring-shaped cross sections (washers). The area of one such washer is the area of the outer circle minus the area of the inner circle. The outer radius is R = f(x) and the inner radius is r = g(x). Therefore, the area of the washer is A(x) = πR² - πr² = π(f(x)² - g(x)²). The definite integral sums these areas multiplied by an infinitesimal thickness (dx) to find the total volume.
A) The solid formed by revolving the region bounded by x=√y, x=1, and y=9 about the y-axis.
B) The solid formed by revolving the region bounded by y=x, y=1, x=1, and x=9 about the x-axis.
C) The solid formed by revolving the region bounded by x=y-1 and x=0 from y=1 to y=9 about the y-axis.
D) The solid formed by revolving the region bounded by y=√x and y=1 about the x-axis.
Correct Answer: A
The integral is in the form V = π ∫[c,d] (R(y)² - r(y)²) dy, indicating revolution about the y-axis. We can rewrite the integrand (y - 1) as ((√y)² - 1²). This identifies the outer radius as R(y) = √y (or x = √y) and the inner radius as r(y) = 1 (or x = 1). The bounds of integration are from y = 1 to y = 9. Therefore, the integral represents the volume of the solid generated by revolving the region enclosed by the curves x = √y, x = 1, and the line y=9 about the y-axis.
A) The bounds of integration are incorrect.
B) The student squared the difference of the radii instead of using the difference of the squares of the radii.
C) The student should have integrated with respect to y.
D) The student incorrectly identified the outer and inner radii.
Correct Answer: B
The washer method formula for revolution about the x-axis is V = π ∫[a,b] (R(x)² - r(x)²) dx. In the interval [0, 1], y = x is the upper curve (outer radius, R(x) = x) and y = x² is the lower curve (inner radius, r(x) = x²). The correct setup is π ∫[0,1] (x² - (x²)²) dx. The student's integral, π ∫[0,1] (x - x²)² dx, calculates π ∫[0,1] (R(x) - r(x))² dx. This is a common conceptual error where the difference of the functions is squared, rather than the correct difference of the squares of the functions.