Question: The region bounded by $y=x^2$ and $y=x$ is rotated about $y=x$. Find the volume of the solid of the revolution.
My answer: I rotated the region $45^\circ$ clockwise and obtained the curve $(x,y)=(\large\frac{t^2+t}{\sqrt2},\frac{t^2-t}{\sqrt2})$ where $0\leq t\leq 1$. Then the volume can be found by the disk method formula: $$V=\int_0^1\pi y^2dx=\pi\int_0^1\frac{t^2(t-1)^2(2t+1)}{2\sqrt2}dt=\frac{\pi}{30\sqrt2}\approx 0.074.$$
Is my answer correct? Can you give another solution? Thanks in advance.