How to solve a triple integral with a circle not centered at origin? My task is this:
Use cylinder coordinates to calculate:$$\iiint\limits_{A}z\sqrt{x^2 + y^2}dA, \enspace A = \left\{(x,y,z):x^2 + (y - 1)^2 \leq 1,\: 0 \leq z \leq 2\right\}.$$
My works so far is this;
Switching to cylindrical coordinates we get:$$A = \left\{(r,\theta,z):0\leq r\leq 1,\: 0\leq \theta \leq 2\pi,\: 0\leq z \leq 2\right\}.$$
Now my book tells me that if you want the center in another point $(a,b,c)$, you should use the substitution: $$x = a + r\cos(\theta),\: y = b + r\sin(\theta),\: z = c + z.$$
With this in mind we change to cylindrical and add the bounderies (don't forget the jacobian).$$\int\limits_{0}^{2\pi}\int\limits_{0}^{1}\int\limits_{0}^{2}zr\sqrt{r^2\cos^2(\theta) + (1 + r\sin(\theta))^2}dz\:dr\:d\theta \:=\:2\int\limits_{0}^{2\pi}\int\limits_{0}^{1} r\sqrt{r^2\cos^2(\theta) + (1 + r\sin(\theta))^2}dr \:d\theta.$$
Now this is the part where i get stuck, if i did my calculations right teh expression under the root becomes $r^2 + 1 + 2r\sin(\theta).$ I'm not sure where to go from here so any tips and tricks would be appreciated. I would very much like to see how this is done with this substitution, but alternative solution that leads me to the right answer would also be of great value. Finally, don't show me the calculations down to the answer as i would like to do that myself.
Thanks in advance!
 A: Your can use a simpler (?) transformation with cylindrical coordinates. Given that $x^2+(y-1)^2=1$ is a cylinder shifted 1 unit along the $y$ axis, you should not consider points in the $y<0$ plane ($\pi \le \theta\le 2\pi$), and the polar equation of the cylinder is not $r=1$ but $r=2\sin\theta$. Therefore, it should be
$$
A=\{(r,\theta,z)\;|\; 0 \le \theta \le  \pi, 0 \le r \le 2\sin\theta,   0\le z \le 2\}
$$
Once you have done that, the integral is easy and equals:
$$
\int_0^{\pi} \int_0^{2\sin\theta} \int_0^{2} z\, r\, r \; dz drd\theta = \frac{64}{9}
$$
A: It's actually easier to evaluate the integral as it stands.  I entered
$\int_{0}^{2}\int_{0}^{2}\int_{-\sqrt{2y-y^2}}^{\sqrt{2y-y^2}}z\sqrt{x^2+y^2} dxdydz$
in my calculator.  It returned $64/9.$  You may want to look at this post where another person was having a very similar problem with a double integral.  
As a general rule, a change of variables in a multiple integral is most effective when the same (or very similar) expressions occur in both the integrand and the region of integration. 
