AP Calculus AB: Volumes: Disk and Washer Method (Drill 1)

Question 1. The region bounded by \( y = \sqrt{x} \), the x-axis, and the line \( x = 4 \) is revolved around the x-axis. Which of the following gives the volume of the resulting solid?

• A) \( \pi \int_0^4 x \, dx \) ✓
• B) \( \pi \int_0^4 \sqrt{x} \, dx \)
• C) \( 2\pi \int_0^4 \sqrt{x} \, dx \)
• D) \( \pi \int_0^4 x^2 \, dx \)

Explanation: Choice A is correct. The disk method gives \( V = \pi \int_a^b [f(x)]^2 \, dx \). Since \( f(x) = \sqrt{x} \), we have \( [\sqrt{x}]^2 = x \), so the volume is \( \pi \int_0^4 x \, dx \). Choice B is incorrect because the student forgot to square the radius, the integrand should be \( [\sqrt{x}]^2 = x \), not \( \sqrt{x} \). Choice C is incorrect because it both omits the squaring and applies the \( 2\pi \) coefficient from the shell method, which does not apply here. Choice D is incorrect because the student squared the function twice, arriving at \( x^2 \) instead of \( x \).

Question 2. The region enclosed by \( y = x \) and \( y = x^2 \) on \( [0, 1] \) is revolved around the x-axis. Which of the following gives the volume of the resulting solid?

• A) \( \pi \int_0^1 (x - x^2)^2 \, dx \)
• B) \( \pi \int_0^1 (x - x^2) \, dx \)
• C) \( \pi \int_0^1 (x^4 - x^2) \, dx \)
• D) \( \pi \int_0^1 (x^2 - x^4) \, dx \) ✓

Explanation: Choice D is correct. The washer method gives \( V = \pi \int_a^b \left([R(x)]^2 - [r(x)]^2\right) dx \). On \( [0,1] \), \( x \geq x^2 \), so the outer radius is \( R(x) = x \) and the inner radius is \( r(x) = x^2 \). Thus \( V = \pi \int_0^1 \left(x^2 - x^4\right) dx \). Choice A is incorrect because the student subtracted the functions before squaring, \( (x - x^2)^2 \neq x^2 - x^4 \). This is a very common washer-method error. Choice B is incorrect because the student forgot to square the radii entirely, computing an area between curves rather than a volume. Choice C is incorrect because the student reversed the outer and inner radii, writing \( x^4 - x^2 \), which produces a negative integrand.

Question 3. The region bounded by \( x = \sqrt{y} \), the y-axis, and the line \( y = 9 \) is revolved around the y-axis. What is the volume of the resulting solid?

• A) \( 18\pi \)
• B) \( 81\pi \)
• C) \( \dfrac{81\pi}{2} \) ✓
• D) \( \dfrac{81\pi}{4} \)

Explanation: Choice C is correct. Revolving around the y-axis, the disk radius at height \( y \) is \( x = \sqrt{y} \), so \( V = \pi \int_0^9 (\sqrt{y})^2 \, dy = \pi \int_0^9 y \, dy = \pi \left[\dfrac{y^2}{2}\right]_0^9 = \dfrac{81\pi}{2} \). Choice A is incorrect because the student forgot to square the radius, integrating \( \pi \int_0^9 \sqrt{y} \, dy = \pi \cdot \dfrac{2}{3}(9)^{3/2} = 18\pi \). Choice B is incorrect because the student dropped the \( \dfrac{1}{2} \) from the antiderivative of \( y \), evaluating \( \pi \cdot 9^2 = 81\pi \) instead of \( \pi \cdot \dfrac{81}{2} \). Choice D is incorrect because the student applied an extra \( \dfrac{1}{2} \) factor as if computing a semicircle area rather than a full disk, arriving at \( \dfrac{\pi}{2}\int_0^9 y \, dy = \dfrac{81\pi}{4} \).

Question 4. A student attempts to find the volume of the solid formed by revolving the region between \( y = 4 \) and \( y = x^2 \) on \( [-2, 2] \) around the x-axis. The student sets up \( \pi \int_{-2}^{2} (4 - x^2)^2 \, dx \). Which of the following correctly identifies the error?

• A) The limits of integration are wrong; to integrate with respect to \( y \) instead, the interval would become \( [0, 4] \).
• B) The student omitted \( \pi \); the correct setup has no \( \pi \) coefficient for washer problems.
• C) The student should have used \( \pi \int_{-2}^{2} (4 + x^2)^2 \, dx \) because the two radii must be added, not subtracted.
• D) The student subtracted the functions before squaring; the correct integrand is \( 16 - x^4 \). ✓

Explanation: Choice D is correct. The washer method requires \( \pi \int_a^b \left([R(x)]^2 - [r(x)]^2\right) dx \). The outer radius is \( R(x) = 4 \) and the inner radius is \( r(x) = x^2 \), giving integrand \( 16 - x^4 \). The student incorrectly subtracted the functions first and then squared, \( (4 - x^2)^2 = 16 - 8x^2 + x^4 \), which is an entirely different expression. Choice A is incorrect because \( [-2, 2] \) is the correct interval for integrating with respect to \( x \); the note about \( [0, 4] \) describes an alternative method (integrating with respect to \( y \)) that would require a completely different setup. Choice B is incorrect because \( \pi \) is always required in the disk and washer methods. Choice C is incorrect because the washer method subtracts the squared inner radius from the squared outer radius, the two terms are never added.

Question 5. A solid is formed by revolving the region bounded by \( x = \sqrt{y} \) and the y-axis around the y-axis for \( 0 \leq y \leq 4 \). A cylindrical hole of radius 1 is drilled through the center of the solid along the y-axis. Which integral gives the volume of material remaining?

• A) \( \pi \int_0^4 (\sqrt{y} - 1)^2 \, dy \)
• B) \( \pi \int_1^4 (y - 1) \, dy \) ✓
• C) \( \pi \int_0^4 (y - 1) \, dy \)
• D) \( \pi \int_0^4 (y + 1) \, dy \)

Explanation: Choice B is correct. The solid is a paraboloid with outer radius \( \sqrt{y} \) at height \( y \). The cylindrical hole has radius 1, which equals the outer radius when \( \sqrt{y} = 1 \), i.e., at \( y = 1 \). For \( 0 \leq y < 1 \), the drill removes the entire cross section, no material remains there. For \( 1 \leq y \leq 4 \), the washer area is \( \pi\left[(\sqrt{y})^2 - 1^2\right] = \pi(y-1) \). Therefore the volume is \( \pi \int_1^4 (y - 1) \, dy \). Choice A is incorrect because the student subtracted the radii before squaring: \( (\sqrt{y} - 1)^2 \neq y - 1 \), the same subtract-then-square error seen in Q2 and Q4. Choice C is incorrect because integrating \( \pi(y-1) \) from 0 to 4 includes the interval \( [0,1] \) where \( y - 1 < 0 \); a cross-sectional area cannot be negative, and extending the lower limit to 0 produces an invalid setup that underestimates the true volume. Choice D is incorrect because the student added the squared radii instead of subtracting them, writing \( y + 1 \) instead of \( y - 1 \), which corresponds to a solid with no hole, a fundamental misapplication of the washer formula.