# 007B Sample Midterm 3, Problem 4 Detailed Solution

Find the volume of the solid obtained by rotating the region bounded by  $y={\sqrt {\sin x}},~0\leq x\leq \pi ,$ and  $y=0$ about the  $x-$ axis. Sketch the graph of the region and a typical disk element.

Background Information:

1. You can find the intersection points of two functions, say   $f(x),g(x),$ by setting  $f(x)=g(x)$ and solving for  $x.$ 2. The volume of a solid obtained by rotating an area around the  $x$ -axis using the washer method is given by

$\int \pi (r_{\text{outer}}^{2}-r_{\text{inner}}^{2})~dx,$ where  $r_{\text{inner}}$ is the inner radius of the washer and  $r_{\text{outer}}$ is the outer radius of the washer.

Solution:

Step 1:
First, we need to find the intersection points of  $y={\sqrt {\sin x}}$ and  $y=0.$ To do this, we need to solve
$0={\sqrt {\sin x}}.$ Squaring both sides, we get  $0=\sin x.$ The solutions to this equation in the interval  $[0,\pi ]$ are

$x=0,\pi .$ Now, the graph of the region is below.
Additionally, we are going to be using the washer/disk method.
Below, we show a typically disk element.
(Insert graph)
Step 2:
The volume of the solid using the disk method is
${\begin{array}{rcl}\displaystyle {V}&=&\displaystyle {\int _{0}^{\pi }\pi ({\sqrt {\sin x}})^{2}~dx}\\&&\\&=&\displaystyle {\int _{0}^{\pi }\pi \sin x~dx}\\&&\\&=&\displaystyle {-\pi \cos x{\bigg |}_{0}^{\pi }}\\&&\\&=&\displaystyle {-\pi \cos(\pi )+\pi \cos(0)}\\&&\\&=&\displaystyle {2\pi .}\end{array}}$ $V=2\pi$ 