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:
$$\int_{0}^{\pi}\int_{1}^{2}r\cos(\theta)rdrd\theta$$
Best Answer
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 $r\in[0,2\cos(\theta)]$.