How to change the limits of a double integral to polar coordinates limits? Problem:
Use polar coordinates to evaluate the following integral:
$$\int_{0}^{2}\int_{0}^{\sqrt{2x-x^{2}}}xdydy$$
Solution:
First, this is the graph I manually plotted to define the new limits:

So I set up the new integral with these new limits in polar coordinates: 
$$1\leqslant r\leqslant 2$$
$$0\leqslant \theta\leqslant\pi$$
But the integral gives me $0$ as a result. This is the integral to evaluate:
$$\int_{0}^{\pi}\int_{1}^{2}r\cos(\theta)rdrd\theta$$
 A: The problem is your range for $r$. In this case, the circle has equation $$(x-1)^2+y^2=1,$$from where you take $$x^2-2x+1+y^2=1 \Leftrightarrow (x^2+y^2)=2x \Leftrightarrow r^2=2r\cos(\theta) \Leftrightarrow r=2\cos(\theta)$$ so actually $r\in[0,2\cos(\theta)]$.
A: The problem is that polar coordinates are centered at the origin, and your original region of integration is not.
To fix this, I'd start by making the change of variables $w = x-1$ in the original integral, which results in 
$$ \int_{-1}^1 \int_{0}^\sqrt{1-w^2} (w + 1) \; dy \; dw. $$
Now convert to polar to get
$$ \int_0^\pi \int_0^1 (r \cos(\theta) + 1) r \; dr \; d\theta$$ 
A: Notice that
$$2x - x^2 = (1 - 1) + 2x - x^2 = 1 - (x-1)^2 ,$$
so $0 \le y \le \sqrt {2x - x^2}$ implies $y^2 \le 1 - (x-1)^2$ (this is the equation of a circle!), which in turn implies $y^2 + (x-1)^2 \le 1$, so you have to choose polar coordinates relative to the circle of radius $1$ and center $(1,0)$. This means
$$x = \color{red} {1 + } r \cos \theta \\
y = r \sin \theta$$
with $0 \le r \le 1$ and $0 \le \theta \le \pi$. With this, you should be able to find your way by yourself now.
A: Your conversion from cartesian bounds to polar is not quite correct. Your bounds describe an area that looks like this:

One way to go about this is to substitute $x$ with a variable $u = x - 1$ and then integrate over the bounds $\int_0^\pi\int_0^1$ after converting $\int_0^2\int_0^{\sqrt{2(u+1) - (u+1)^2}}(u+1)u\mathrm{d}u\mathrm{d}y = \int_0^2\int_0^{\sqrt{1 - u^2}}(u+1)\mathrm{d}u\mathrm{d}y$
A: The region in your sketch is half a tangent circle.  The equation is not $r=2$, but $r=2\cos \theta$.  Integrate $\theta$ from $0$ to $\pi/2$ and $r$ from 0 to $2\cos \theta.$
