Find the volume of the solid generated by revolving the region about the given line.
-The region bounded above by the line $y = 3$, below by the curve $y = 3 \cos ( \pi x )$, on the left by the line $x = - 0.5$, and on the right by the line $x = 0.5$, about the line $y = 3$
A) $\frac { 27 } { 2 } \pi - 36$
B) $27 \pi - 18$
C) $\frac { 9 } { 2 } \pi$
D) $\frac { 27 } { 2 } \pi$

Question

Quizplus · Accepted Answer

Explanation:The volume of the solid generated by revolving the region about the line $y = 3$ is given by the integral:$$V = \pi \int_{-0.5}^{0.5} (3 - 3\cos(\pi x))^2 dx$$$$= \pi \int_{-0.5}^{0.5} (9 - 18\cos(\pi x) + 9\cos^2(\pi x)) dx$$$$= \pi \int_{-0.5}^{0.5} (9 - 18\cos(\pi x) + \frac{9}{2}(1 + \cos(2\pi x))) dx$$$$= \pi \int_{-0.5}^{0.5} (13.5 - 18\cos(\pi x) + \frac{9}{2}) dx$$$$= \pi \int_{-0.5}^{0.5} (22.5 - 18\cos(\pi x)) dx$$$$= \pi \left[22.5x - 18\frac{\sin(\pi x)}{\pi}\right]_{-0.5}^{0.5}$$$$= \pi \left[22.5(0.5) - 18\frac{\sin(\pi (0.5))}{\pi} - 22.5(-0.5) + 18\frac{\sin(\pi (-0.5))}{\pi}\right]$$$$= \pi \left[11.25 - 9 - 11.25 + 9\right]$$$$= \pi (0) = 0$$Therefore, the volume of the solid is 0.

Find the Volume of the Solid Generated by Revolving the Region