So I have to compute the following integral:
$$ \iint_D \sqrt{(x-1)^2+y^2}\,dx\,dy $$
where $D=\{(x,y)\enspace :\enspace x^2 +y^2 \le 1; \enspace y\ge 0 \}$
So, as stated on the title, this is basically the volume of the object above half a circumference with radius $1,$ centered at the origin, but below the cone which is centered at $(1,0)$.
As it is natural, I changed to cylindrical coordinates to solve the problem more easily, only to find I still couldn't calculate the integral. The integral I came across after changing the coordinates was
$$ \int_0^\pi \int_0^1 r\sqrt{r^2 - 2r\cos(\theta)+1} \, dr\,d\theta $$
I can't seem to progress from there. Is there a substitution I can't see? I would appreciate any help! Thank you in advance!
If follows your method, please note the following and you can get the result(maybe need more calculation): $$\sqrt{r^2 - 2r\cos(\theta)+1} =\sqrt{(r-\cos(\theta))^2+\sin^2(\theta)}.$$
$$r\sqrt{r^2 - 2r\cos(\theta)+1} =(r-\cos(\theta))\sqrt{r^2 - 2r\cos(\theta)+1}\\ +\sqrt{(r-\cos(\theta))^2+\sin^2(\theta)}.$$
But I think the better method is "Coordinate transformation". Let $$X=x-1,Y=y,$$ then $D$ changes to $$D:(X+1)^2+Y^2\leq 1,Y\geq0,$$ and also we have $dXdY=dxdy$, so( also use Polar coordinate transformation): $$\iint\limits_{x^2 +y^2 \le 1} \sqrt{(x-1)^2+y^2}\,dx\,dy =\iint\limits_{(X+1)^2+Y^2\leq 1}\sqrt{X^2+Y^2}dXdY\\ =\int_{\pi/2}^{\pi}\int_{0}^{2\cos(\pi-\theta)}r^2drd\theta =\frac{-8}{3}\int_{\pi/2}^{\pi}\cos^3(\theta)d\theta=\frac{16}{9}.$$