# 009B Sample Final 3, Problem 4

Find the volume of the solid obtained by rotating about the  ${\displaystyle x}$-axis the region bounded by  ${\displaystyle y={\sqrt {1-x^{2}}}}$  and  ${\displaystyle y=0.}$

Foundations:
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 a region around the  ${\displaystyle x}$-axis using disk method is given by

${\displaystyle \int \pi r^{2}~dx,}$  where  ${\displaystyle r}$  is the radius of the disk.

Solution:

Step 1:
We start by finding the intersection points of the functions  ${\displaystyle y={\sqrt {1-x^{2}}}}$  and  ${\displaystyle y=0.}$
We need to solve
${\displaystyle 0={\sqrt {1-x^{2}}}.}$
If we square both sides, we get
${\displaystyle 0=1-x^{2}.}$
The solutions to this equation are  ${\displaystyle x=-1}$  and  ${\displaystyle x=1.}$
Hence, we are interested in the region between  ${\displaystyle x=-1}$  and  ${\displaystyle x=1.}$
Step 2:
Using the disk method, the radius of each disk is given by
${\displaystyle r={\sqrt {1-x^{2}}}.}$
Therefore, the volume of the solid is
${\displaystyle {\begin{array}{rcl}\displaystyle {V}&=&\displaystyle {\int _{-1}^{1}\pi ({\sqrt {1-x^{2}}})^{2}~dx}\\&&\\&=&\displaystyle {\int _{-1}^{1}\pi (1-x^{2})~dx}\\&&\\&=&\displaystyle {\pi {\bigg (}x-{\frac {x^{3}}{3}}{\bigg )}{\bigg |}_{-1}^{1}}\\&&\\&=&\displaystyle {\pi {\bigg (}1-{\frac {1}{3}}{\bigg )}-\pi {\bigg (}-1+{\frac {1}{3}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {4\pi }{3}}.}\end{array}}}$

${\displaystyle {\frac {4\pi }{3}}}$