I need to calculate this integral:
$$\int\int\arctan{\frac{y}{x}}dxdy$$ on the area of the circle $x^2+y^2=ax$, $a>0$
here's what I did:
$x=r\cos\theta$
$y=r\sin\theta$
Substituting, we have
$r = a\cos\theta$
Now, the boundaries for $r$ are $r \in [0,a\cos\theta]$
As for the boundaries of the angle, from our condition we have that the left side is always $\ge 0$, because it's a sum of two squares. The right side must be $\ge 0$, too. Since $a>0$, then $\cos\theta \ge 0$, which is $\theta \in [-\pi/2, \pi/2]$
Now, when I plug in my substitution into the function $f(x,y)$, I get that it evaluates to only $\theta$, since it's $\arctan{\tan{\theta}}$
I've got the integral:
$$ \large \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_0^{a\cos\theta}r\theta dr d\theta$$
The integral evaluates to zero. I think I have made a mistake somewhere. Can someone point it out?