Question

Find the area of the region lying between the parabolas $y^2=4ax$ and $x^2=4ay$

Answer 1

Given curves are $y^2=4ax$ and $x^2=4ay$

The point of intersections of given curves are $A(0,0)$ and $B(4a,4a)$

The area enclosed by arc $AB$ of curve $y^2=4ax$ with $x$ axis is $P={\int }_0^{4a}2\sqrt{ax}$

$\Rightarrow P={\int }_0^{4a}2\sqrt{ax}=2\sqrt{a}{\int }_0^{4a}\sqrt{x}=\frac{4\sqrt{a}}{3}{\left(4a\right)}^{\frac{3}{2}}=\frac{32a^2}{3}$

The area enclosed by arc $AB$ of curve $x^2=4ay$ with $x$ axis is $Q={\int }_0^{4a}\frac{x^2}{4a}$

$\Rightarrow Q={\int }_0^{4a}\frac{x^2}{4a}=\frac{1}{4a}{\int }_0^{4a}{x}^2=\frac{1}{12a}\left(64a^3\right)=\frac{16a^2}{3}$

The area of region lying between them is $P-Q=\frac{16a^2}{3}$