[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?
On
Hint Writing the domain in polar coordinates gives $$(r \cos \theta - 1)^2 + (r \sin \theta)^2 \leq 1 .$$ Expanding and using the Pythagorean identity gives $$r^2 - 2 r \cos \theta + 1 \leq 1,$$ rearranging gives $$r^2 \leq 2 r \cos \theta ,$$ and for $r \neq 0$, dividing by $r$ gives $$r \leq 2 \cos \theta .$$ So, in polar coordinates the integral becomes $$\int_{-\pi / 2}^{\pi / 2} \int_0^{2 \cos \theta} \underbrace{r}_{\sqrt{x^2 + y^2}} \cdot \underbrace{r \,dr \,d\theta}_{dx\,dy} = \int_{-\pi / 2}^{\pi / 2} \int_0^{2 \cos \theta} r^2 \,dr \,d\theta .$$
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?