9. $[\boldsymbol{7}]$ let $\mathcal{R}$ be the region in the plane bounded by the graphs of $y=x$ and $y=x^{2}$. Compute the volume of the region formed by revolving $\mathcal{R}$ around the line $y=x$.
Answer:
$\frac{\sqrt{2} \pi}{60}$
Solution: We integrate from 0 to 1 using the method of washers. Fix $d$ between 0 and 1 . Let the line $x=d$ intersect the graph of $y=x^{2}$ at $Q$, and let the line $x=d$ intersect the graph of $y=x$ at $P$. Then $P=(d, d)$, and $Q=\left(d, d^{2}\right)$. Now drop a perpendicular from $Q$ to the line $y=x$, and let $R$ be the foot of this perpendicular. Because $P Q R$ is a $45-45-90$ triangle, $Q R=\left(d-d^{2}\right) / \sqrt{2}$. So the differential washer has a radius of $\left(d-d^{2}\right) / \sqrt{2}$ and a height of $\sqrt{2} d x$. So we integrate (from 0 to 1 ) the expression $\left[\left(x-x^{2}\right) / \sqrt{2}\right]^{2} \sqrt{2} d x$, and the answer follows.