Find the volume of solid lies under $z=x^2+y^2$ above $x$-$y$ plane and inside the cylinder $x^2+y^2=2x$. I know, for volume we have to us $V=\iiint { \mathrm dx\mathrm dy\mathrm dz}$ but i was not able to proceed further.
Any hint would be sufficient !
\begin{align} & V=\int_{0}^{2}{\int_{-\sqrt{2x-{{x}^{2}}}}^{\sqrt{2x-{{x}^{2}}}}{\int_{0}^{{{x}^{2}}+{{y}^{2}}}{dzdydx=}}}\int_{0}^{2\pi }{\int_{0}^{2\cos \theta }{\int_{0}^{{{r}^{2}}}{r\,dzdrd\theta}}} \\ & V=\int_{0}^{2\pi }{\int_{0}^{2\cos \theta }{{{r}^{3}}}}drd\theta =4\int_{0}^{2\pi }{{{\cos }^{4}}}\theta d\theta =16\int_{0}^{\frac{\pi }{2}}{{{\cos }^{4}}}\theta d\theta \\ \end{align} $$V=16\frac{\Gamma \left( \frac{5}{2} \right)\Gamma \left( \frac{1}{2} \right)}{2\Gamma \left( 3 \right)}$$