Home / AP Calculus AB 8.11 Volume with Washer Method: Revolving Around the x- or y-Axis – MCQs

AP Calculus AB 8.11 Volume with Washer Method: Revolving Around the x- or y-Axis - MCQs - Exam Style Questions

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No-Calc Question


Let \(R\) be the shaded region bounded by the graph of \(y=\sqrt{x}\), the graph of \(y=x-2\), and the \(x\)-axis, as shown. Which of the following gives the volume of the solid generated when \(R\) is revolved about the \(x\)-axis?

(A) \(\displaystyle \pi\!\int_{0}^{4}\!\big(x-(x-2)^{2}\big)\,dx\)
(B) \(\displaystyle \pi\!\int_{0}^{4}\!\big(\sqrt{x}-(x-2)\big)^{2}dx\)
(C) \(\displaystyle \pi\!\int_{0}^{2}\!x\,dx+\pi\!\int_{2}^{4}\!\big(x-(x-2)^{2}\big)\,dx\)
(D) \(\displaystyle \pi\!\int_{0}^{2}\!x\,dx+\pi\!\int_{2}^{4}\!\big(\sqrt{x}-(x-2)\big)^{2}dx\)

▶️ Answer/Explanation

Revolve around the \(x\)-axis ⟹ washers.
For \(0\le x\le 2\): outer radius \(=\sqrt{x}\), inner radius \(=0\) ⇒ area \(=\pi(\sqrt{x})^{2}=\pi x\).
For \(2\le x\le 4\): outer radius \(=\sqrt{x}\), inner radius \(=x-2\) ⇒ area \(=\pi\big(x-(x-2)^{2}\big)\).
Volume is the sum: \(\displaystyle \pi\!\int_{0}^{2}\!x\,dx+\pi\!\int_{2}^{4}\!\big(x-(x-2)^{2}\big)\,dx\).
✅ Answer: (C)

Question

A region in the first quadrant is enclosed by the graphs of  \(y=e^{2x}\),x = 1, and the coordinate axes.If the region is rotated about the y-axis, the volume of the solid that is generated is represented by which of the following integrals?

(A) \(2\pi\int_{0}^{1}xe^{2x}dx\)
(B)\(2\pi\int_{0}^{1}e^{2x}dx\)
(C)\(\pi\int_{0}^{1}e^{4x}dx\)
(D)\(\pi\int_{0}^{e}y ln ydy\)
(E)\(\frac{\pi }{4}\int_{0}^{e}ln^{2}ydy\)

▶️Answer/Explanation

Ans:A

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