# 009B Sample Final 2, Problem 3

Find the volume of the solid obtained by rotating the region bounded by the curves  $y=x$ and  $y=x^{2}$ about the line  $y=2.$ 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 an area around the  $x$ -axis using the washer method is given by

$\int \pi (r_{\text{outer}}^{2}-r_{\text{inner}}^{2})~dx,$ where  $r_{\text{inner}}$ is the inner radius of the washer and  $r_{\text{outer}}$ is the outer radius of the washer.

Solution:

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