# 009B Sample Final 2, Problem 3

Find the volume of the solid obtained by rotating the region bounded by the curves  ${\displaystyle y=x}$  and  ${\displaystyle y=x^{2}}$  about the line  ${\displaystyle y=2.}$

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 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  ${\displaystyle y=x}$  and  ${\displaystyle y=x^{2}.}$
To do this, we need to solve
${\displaystyle x=x^{2}.}$
Moving all the terms on one side of the equation, we get
${\displaystyle {\begin{array}{rcl}\displaystyle {0}&=&\displaystyle {x^{2}-x}\\&&\\&=&\displaystyle {x(x-1).}\end{array}}}$
Hence, these two curves intersect at  ${\displaystyle x=0}$  and  ${\displaystyle x=1.}$
So, we are interested in the region between  ${\displaystyle x=0}$  and  ${\displaystyle x=1.}$
Step 2:
We use the washer method to calculate this volume.
${\displaystyle r_{\text{outer}}=2-x^{2}}$
${\displaystyle r_{\text{inner}}=2-x.}$
${\displaystyle {\begin{array}{rcl}\displaystyle {V}&=&\displaystyle {\int _{0}^{1}\pi (r_{\text{outer}}^{2}-r_{\text{inner}}^{2})~dx}\\&&\\&=&\displaystyle {\int _{0}^{1}\pi ((2-x^{2})^{2}-(2-x)^{2})~dx.}\end{array}}}$
${\displaystyle {\begin{array}{rcl}\displaystyle {V}&=&\displaystyle {\pi \int _{0}^{1}((4-4x^{2}+x^{4})-(4-4x+x^{2}))~dx}\\&&\\&=&\displaystyle {\pi \int _{0}^{1}(4x-5x^{2}+x^{4})~dx}\\&&\\&=&\displaystyle {\pi {\bigg (}2x^{2}-{\frac {5x^{3}}{3}}+{\frac {x^{5}}{5}}{\bigg )}{\bigg |}_{0}^{1}}\\&&\\&=&\displaystyle {\pi {\bigg (}2-{\frac {5}{3}}+{\frac {1}{5}}{\bigg )}-0}\\&&\\&=&\displaystyle {{\frac {8\pi }{15}}.}\end{array}}}$
${\displaystyle {\frac {8\pi }{15}}}$