TITLE: Find integral of a polar function $h(r,\theta)$ over a circle
QUESTION [0 upvotes]: I am studying for my math final and our prof gave us a review but without any solutions or hints. I don't really understand this problem so if anyone could help me out here I would appreciate it.

Find the integral of the polar function $h(r, \theta) = r \cos(\theta) , r \ge 0$, over the circle $r = 2a\cos(\theta)$ .

Again I would appreciate the help.

REPLY [2 votes]: I am going to assume (perhaps incorrectly) that whoever asked the question really meant what they wrote, which is, to integrate over a circle (a curve) and not over the interior of the circle (a disc).  I would find the center and radius of the circle, starting by multiplying both sides of the equation of the circle by $r$, and eventually completing the square.  I'll assume $a > 0$.  I think the circle will have radius $a$ and center $(x,y)=(a,0)$.  Then parametrize the circle, using a letter other than $\theta$ as the angle, and you'll get something like $x(t)=a+a\cos(t),\ y(t)=a+a\sin(t),\ 0 \leq t \leq 2\pi$.  Note that your function $h(r,\theta) = r\cos(\theta) = x$.
Now find $\int_0^{2\pi} x(t)\sqrt{x'(t)^2+y'(t)^2}\,dt$.
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