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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle r={\sqrt {1-x^{2}}}.}
Therefore, the volume of the solid is
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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}}}

Final Answer:
${\frac {4\pi }{3}}$

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