I want to write the following double integral
$$\int_0^2 \int_{0}^{\sqrt{1-(x-1)^2}} \frac{x+y}{x^2+y^2} \, \mathrm d x \mathrm d y$$
in polar coordinates. The region is a circle centered at $(1,0)$ and with radius $1$.
I am having problems with finding the bounds for $r$. When we are on the circle,
$$(r \cos\theta -1 )^2 + r^2 \sin^2 \theta = 1$$
which implies that $r^2 -2 r \cos \theta +1 =0$. At his point I am stuck. Can you please explain how to find the polar equivalent of this integral? Thank you.
Best Answer
Guide:
$\theta$ goes from $-\frac{\pi}2$ to $\frac{\pi}2$.
We have $$r^2-2r\cos \theta = 0$$
(You forgot to cancel out the $1$ on both sides).
That is the upper limit of $r$ is $r= 2\cos \theta$.
That is
$$\int_{0}^\frac{\pi}{2}\int_0^{2\cos \theta} \frac{r\cos \theta + r\sin \theta}{r^2}\cdot r\, \, drd\theta$$