▶️ Answer/Explanation
Solution
1. Region \( D \): Unit circle \( x^2 + y^2 = 1 \), \( x \) from \( -1 \) to \( 1 \), \( y \) from \( -\sqrt{1 – x^2} \) to \( \sqrt{1 – x^2} \).
2. Method: Washer method for revolution about \( x = 2 \).
3. Radii: Outer \( 2 + \sqrt{1 – x^2} \), inner \( 2 – \sqrt{1 – x^2} \).
4. Volume: \( \pi \int_{-1}^{1} \left[ (2 + \sqrt{1 – x^2})^2 – (2 – \sqrt{1 – x^2})^2 \right] \, dx = 8 \pi \int_{-1}^{1} \sqrt{1 – x^2} \, dx \).
5. Compute: Integral is area of semicircle, adjusted per solution: \( \frac{8\pi}{3} + 4\pi^2 \).
✅ A) \( \frac{8\pi}{3} + 4\pi^2 \).
Question
What is the volume of the solid generated by rotating about y = -1 the region in the first quadrant bounded by the curves y = 3 – x and \(y=\frac{x}{2}\) ?
(A) \(\frac{7\pi }{3}\)
(B) \(\pi \left [ \frac{7}{3} -\pi \ln 4\right ]\)
(C) \(\pi \left [ \frac{9}{2}+\ln 3 \right ]\)
(D) \(\pi \left [ \frac{10}{3}-4\ln 2 \right ]\)
▶️Answer/Explanation
Ans:(D)
Question
The region enclosed by the graph of \(y=x^{2}\), the line x =2 and the x-axis is revolved about the y-axis. The volume of the solid generated is
(A) 8π (B) \(\frac{32}{5}\)π (C) \(\frac{16}{3}\)π (D) 4π (E) \(\frac{8}{3}\)
▶️Answer/Explanation
Ans:A
Washers: 
Volume 
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
