I need to find the area of the surface $x^2+y^2+z^2 = a^2$ for $y^2 \ge a(a+x)$.
I know that $A = 4a \int_{-a}^0 dx \int_{\sqrt{a^2+ax}}^{\sqrt{a^2-x^2}} \frac{dy}{\sqrt{a^2-x^2-y^2}}$, but I have trouble finding the limits of integration when converting the integral into polar coordinates:
$A = 4a \int_{\frac{\pi}{2}}^{\pi} d \phi \int_{r(\phi)}^{a} \frac{rdr}{\sqrt{a^2-r^2}}$
where $r(\phi)$ solves $r^2 \sin^2 \phi - a \cos \phi r - a^2 =0$.
Is there anything I'm doing wrong? Thank you.
You have to consider the projection of your surface on $xy$-plane, show the above figure for $a=2$. So you have to divide the region into two equal regions.
The polar arm changes from $$r=r(\phi)\to r=a$$ So you are completely right. The main problem now is how to get the limits of $r$. I have checked your equation it is also true, use the general rule for the second order equation, you will get a long root in terms of $\csc , \cot$.