# 007B Sample Midterm 3, Problem 4 Detailed Solution

Find the volume of the solid obtained by rotating the region bounded by  ${\displaystyle y={\sqrt {\sin x}},~0\leq x\leq \pi ,}$  and  ${\displaystyle y=0}$  about the  ${\displaystyle x-}$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   ${\displaystyle f(x),g(x),}$

by setting  ${\displaystyle f(x)=g(x)}$  and solving for  ${\displaystyle x.}$

2. The volume of a solid obtained by rotating an area around the  ${\displaystyle x}$-axis using the washer method is given by

${\displaystyle \int \pi (r_{\text{outer}}^{2}-r_{\text{inner}}^{2})~dx,}$

where  ${\displaystyle r_{\text{inner}}}$  is the inner radius of the washer and  ${\displaystyle r_{\text{outer}}}$  is the outer radius of the washer.

Solution:

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

${\displaystyle x=0,\pi .}$

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
${\displaystyle {\begin{array}{rcl}\displaystyle {V}&=&\displaystyle {\int _{0}^{\pi }\pi ({\sqrt {\sin x}})^{2}~dx}\\&&\\&=&\displaystyle {\int _{0}^{\pi }\pi \sin x~dx}\\&&\\&=&\displaystyle {-\pi \cos x{\bigg |}_{0}^{\pi }}\\&&\\&=&\displaystyle {-\pi \cos(\pi )+\pi \cos(0)}\\&&\\&=&\displaystyle {2\pi .}\end{array}}}$

${\displaystyle V=2\pi }$