Chapter 7, Problem 46RE

To determine

To find: The volume of the solid generated by resolving the region bounded by the curves.

Answer to Problem 46RE

The volume of the solid generated is $V={\pi }^2\text{ cu}\text{. unit}$ .

Explanation of Solution

Given information:

  $\begin{array}{c}y=\sin x\\ y=0\end{array}$

Formula used:

The formula for cylindrical shell is,

  $V={\displaystyle {\int }_a^{b}2\pi xf\left(x\right)dx}$

Calculation:

Let's graph thegiven area and draw some conclusions based on it.

  

Given that the interval of the integration is $\left[0,\pi \right]$ , because the region revolving about the $x$ -axis.

According to the shell approach, the volume of the solid is determined by the integral because, from the image seen above, the shell height is $x$ and the shell radius is $\sin x$ .

  $V={\displaystyle {\int }_0^{\pi }2\pi \cdot x\cdot \sin \left(x\right)dx}$

The above integral is calculated by using integration by parts,

  $\begin{array}{c}u=x\Rightarrow du=dx\\ dv=\sin xdx\\ v=-\cos x\end{array}$

Thus,

  $\begin{array}{c}V=2\pi \left({\left.-x\cos \left(x\right)\right|}_0^{\pi }+{\displaystyle \underset{0}{\overset{\pi }{\int }}\cos \left(x\right)dx}\right)\\ =\pi \left(\pi +{\left.\sin \left(x\right)\right|}_0^{\pi }\right)\\ V={\pi }^2\end{array}$

Therefore, the volume of the solid generated is $V={\pi }^2\text{ cu}\text{. unit}$ .