Find the area of a figure bounded by curves$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,x^2+y^2=ab,x^2+y^2\geq ab,a>b$$
I called $x^2=ab-y^2$ and substituted the ellipse into the equation, which gave me the equation
$$b^2\left ( ab-y^2 \right )+a^2y^2=a^2b^2$$
I wanted to solve it with respect to $y=\pm\frac{b\sqrt{a^2-ab}}{\sqrt{a^2-b^2}}$ and substitute it into the condition $x^2+y^2=ab$ to find the intersection points
It takes a long time and is very ugly. My question is, is there a simpler and more efficient way to find the area of a given shape?