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.
The outer radius is
$r_{\text{outer}}=2-x^{2}$
and the inner radius is
$r_{\text{inner}}=2-x.$
Therefore, the volume of the solid is
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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}}

Step 3:
Now, we integrate to get
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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}}

Final Answer:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{8\pi}{15}}

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