EDIT: Let's say the three vertices are at $[0,0]$, $[R,0]$ and $[a,b]$ with $-R < a < R$, $b > 0$ and $a^2 + b^2 < R^2$. Then the integration can be set up as $$ 2 \int_0^b dy \int_{ay/b}^{R - (R-a)y/b} dx \; \sqrt{R^2 - x^2 - y^2} $$
This corresponds to the picture
= (R-a)y$. The extent of the intersection of the sphere in the $z$ direction over the point $(x,y)$ in the plane is $2 \sqrt{R^2 - x^2 - y^2}$.