In a problem i am working on, equation of the circle is given to be
$x^2+y^2-2ax=0,z=0$
So the polar equation taking become $r=2a\cos\theta$
So for the purpose of surface integral, $r$ varies from $0$ to $2a\cos\theta$ and it is given that $\theta$ varies from $0$ to $\pi$ should it not be till $2\pi$ ?? The integral is to be determined above the plane $z=0$
You have the circle
$$0=x^2+y^2-2ax=(x-a)^2+y^2-a^2\implies (x-a)^2+y^2=a^2$$
a circle centered at $\,(a,0)\;$ and with radius $\;a\;$ , so it has part of $\,x-$axis as one of its diameters and the $\,y-$ axis is tangent to it, and thus I'd say that we have
$$-\frac\pi2\le \theta\le\frac\pi2$$
as you can see if you draw a little sketch of the given circle...