The volume of solid of revolution rotated about the line $y=x$ Find the volume of solid of revolution of region between curves $y=\sqrt x$ and $y=x^2$ in $xy-$plane about the line $y=x$.  I know the answer, $\pi/30\sqrt 2$, but how we can obtain it?  Should we rotate axis?


$x + w/\sqrt{2} = \sqrt{x - w/\sqrt{2}}$
means $w = \frac{1}{2} \left(\sqrt{2} \sqrt{8 x+1}-\sqrt{2} (2 x+1)\right)$.
Volume:
$$\int\limits_{x=0}^1 \sqrt{2} \pi \left(\sqrt{2} \sqrt{8 x+1}-\sqrt{2} (2 x+1)\right)^2\ dx = {\pi \over 30 \sqrt{2}}$$
