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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?

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

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  • $\begingroup$ 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 $\endgroup$ – Milan Dec 25 '18 at 21:08
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$y=\sqrt x,y=x^2$ 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 $$\displaystyle\int_0^1\pi\Big(\frac{\sqrt x-x}{\sqrt2}\Big)^2\sqrt2dx=\frac\pi{\sqrt2}\int_0^1(x^2+x-2x\sqrt x)dx=\frac{\pi}{30\sqrt2}$$

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$x + w/\sqrt{2} = \sqrt{x - w/\sqrt{2}}$

means $w = \frac{1}{2} \left(\sqrt{2} \sqrt{8 x+1}-\sqrt{2} (2 x+1)\right)$.

Volume:

$$\int\limits_{x=0}^1 \sqrt{2} \pi \left(\sqrt{2} \sqrt{8 x+1}-\sqrt{2} (2 x+1)\right)^2\ dx = {\pi \over 30 \sqrt{2}}$$

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