<s>[INST] Evaluate $\iint_{\{(x,y)\mid (x-1)^2+y^2\leq1\}}\sqrt{x^2+y^2} dxdy$ [Background]: I'm trying to find the volume of the region bounded by the $xy$-plane, the cone $z^2=x^2+y^2$ and the cylinder $(x-1)^2+y^2=1$.
[Attempt]: I tried to use the polar coordinate:
\begin{align*}
\iint_{\{(x,y)\mid (x-1)^2+y^2\leq1\}}\sqrt{x^2+y^2} dxdy &= \int_0^{2\pi}\int_0^1 \sqrt{(1+r\cos{\theta})^2+(r\sin{\theta})^2}rdrd\theta\\
&= \int_0^{2\pi}\int_0^1 \sqrt{1+2r\cos\theta+r^2}r drd\theta
\end{align*}
Then I cannot continue. Could you give me a hint?
 [/INST] By using the usual polar coordinates $x=r\cos(\theta)$ and $y=r\sin(\theta)$ then  the domain is
$$1\geq (x-1)^2+y^2=(r\cos(\theta)-1)^2+r^2\sin^2(\theta)\Leftrightarrow r\leq 2\cos(\theta),$$
and the integral becomes
$$\iint_{\{(x,y)\mid (x-1)^2+y^2\leq 1\}}\sqrt{x^2+y^2} dxdy=\int_{\theta=-\pi/2}^{\pi/2} \int_{r=0}^{2\cos(\theta)}r^2\,dr\,d\theta.$$
Can you take it from here?
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