Consider a circle parametrized as $(r\cos (t), r \sin (t))$ and an ellipse parametrized as $(a\cos (t), b \sin (t))$. Assuming that $a>r>b$, you find the area of region of intersection of circle and elipse by setting up line integral and using greens theorem.
I tried to parametrize those four curves (boundary of the region). For the left and right region, $r$ is fixed but couldn't find $θ$. For upper and lower curves, both $r$ and $θ$ are varying. Are there other ways to approach this problem?
Using Green's theorem area bounded by curve $C$ (as I guess $b<r<a$) can be calculated as: $$S=\int\limits_Cxdy=-\int\limits_Cydx=\frac{1}{2}\int\limits_C(xdy-ydx)$$
1)
Intersection points can be calculated as $$\begin{array}{} x^2=a^2\frac{r^2-b^2}{a^2-b^2} \\ y^2=b^2\frac{a^2-r^2}{a^2-b^2} \end{array} $$ So we have 4 symmetrical point with argument in first quadrant $$\tan (\phi_0) = \frac{b}{a}\sqrt{\frac{a^2-r^2}{r^2-b^2}}$$ From obtained curve $C$ is ellipse on $[\phi_0, \pi-\phi_0]$ and $[\pi+\phi_0, 2\pi-\phi_0]$ and is circle on remained part of $[0,2\pi]$. Now integral can be divided in 4 summands and each part separately can be parametrized by polar or extended polar coordinates.
2) Sweet in the end: possibly it is more easy to calculate ellipse parts outside of circle, then subtract from ellipse area. Obtained subtract from circle area. So you'll need calculate only integral on $[-\phi_0, \phi_0]$ more then ellipse and less then circle. As suggested by Charlie Chang it's more easy by double integral, but as you insist on Green, then everything needed you have.