# 009B Sample Final 3, Problem 4

Find the volume of the solid obtained by rotating about the  $x$ -axis the region bounded by  $y={\sqrt {1-x^{2}}}$ and  $y=0.$ Foundations:
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 a region around the  $x$ -axis using disk method is given by

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

Solution:

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