Mathematics - Area Under Curves Question with Solution | TestHub

Solving for $x$ : the curves are $x=\frac{a^3}{a^2+y^2}$ and $x=\frac{a{y}^2}{a^2+y^2}$, meeting at $y=\pm a$. Integrating along the y-axis: area $=2{\int }_0^a\frac{a\left(a^2-y^2\right)}{a^2+y^2}dy=2a{\int }_0^a\left(\frac{2a^2}{a^2+y^2}-1\right)dy=2a{\left[2a{\tan }^{-1}\frac{y}{a}-y\right]}_0^a=2a\left(\frac{\pi a}{2}-a\right)=(\pi -2)a^2$.