The volume of solid of revolution rotated about the line $y=x$ [closed]

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?

closed as off-topic by Namaste, Paul Frost, Brian Borchers, KReiser, Eevee TrainerDec 26 '18 at 3:34

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• are you familiar with multiple integrals and polar coordinates( if not thats not a problem) thats just my favourite way of solving these kind of problems – Milan Dec 25 '18 at 21:08

$$y=\sqrt x,y=x^2$$ intersect at $$x=0,1$$. The length of segment $$OD$$ is $$\sqrt2$$. The differential segment along $$OD$$ is $$dx/\cos(\pi/4)=\sqrt2dx$$. We have $$AB=\sqrt x-x$$, so that the $$\displaystyle AC=AB\sin(\pi/4)=\frac{\sqrt x-x}{\sqrt2}$$.
The volume of the solid of revolution is given by $$\displaystyle\int_0^1\pi\Big(\frac{\sqrt x-x}{\sqrt2}\Big)^2\sqrt2dx=\frac\pi{\sqrt2}\int_0^1(x^2+x-2x\sqrt x)dx=\frac{\pi}{30\sqrt2}$$
$$x + w/\sqrt{2} = \sqrt{x - w/\sqrt{2}}$$
means $$w = \frac{1}{2} \left(\sqrt{2} \sqrt{8 x+1}-\sqrt{2} (2 x+1)\right)$$.
$$\int\limits_{x=0}^1 \sqrt{2} \pi \left(\sqrt{2} \sqrt{8 x+1}-\sqrt{2} (2 x+1)\right)^2\ dx = {\pi \over 30 \sqrt{2}}$$