Chapter 8.3, Problem 10E

To determine

To find : the volume of the solid.

Answer to Problem 10E

The volume of the solid is $\frac{{\pi }^2}{16}$ .

Explanation of Solution

Given information : The shaded region about the x -axis.

  

Calculation :

The shaded region of the y -axis, the area of these cross sectional disks with radius

  $\sin \left(x\right)\cos \left(x\right)$ and the area is given by:

  $A\left(x\right)=\pi {\sin }^2\left(x\right){\cos }^2\left(x\right)$

Hence, the volume of the solid is given by solving the integral.

  $\begin{array}{l}V=\pi {\sin }^2\left(x\right){\cos }^2\left(x\right)\\ V={\displaystyle \underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}\pi {\sin }^2\left(x\right){\cos }^2\left(x\right)dx}\\ V=\pi {\displaystyle \underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}{\left(\sin x\cos x\right)}^{2}dx}\\ V=\pi {\displaystyle \underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}\left({\sin }^{2}x{\cos }^{2}x\right)dx}\\ V=\pi {\displaystyle \underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}\left(\frac{1-\cos 2x}{2}\right)\left(\frac{1+\cos 2x}{2}\right)dx}\\ V=\pi {\displaystyle \underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}\frac{\left(1-\cos 2x\right)\left(1+\cos 2x\right)}{4}dx}\\ V=\frac{\pi }{4}{\displaystyle \underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}\left(1-{\left(\cos 2x\right)}^2\right)}dx\\ V=\frac{\pi }{4}\left[\left(\frac{\pi }{2}-\frac{1}{4}\left(\pi +0\right)\right)-\left(0\right)\right]\\ V=\frac{\pi }{4}\left(\frac{2\pi }{4}-\frac{\pi }{4}\right)\\ V=\frac{\pi }{4}\left(\frac{\pi }{4}\right)\\ V=\frac{{\pi }^2}{16}\end{array}$