Question

Find the area enclosed between $y^2=4ax$ and $x^2=y$.

Answer 1

Given, $y=x^2,y^2=4ax$

$(x^2{)}^2=4ax$

$⟹x(x^3-4a)=0$

$x=0,x=(4a{)}^{1/3}$

$y=0,y=(4a{)}^{2/3}$

Area$={\int }_0^{(4a{)}^{1/3}}(2a^{1/2}{x}^{1/2}-x^2)dx$

$={\left[\frac{4a^{1/2}{x}^{3/2}}{3}\right]}_0^{(4a{)}^{1/3}}$$+{\left[\frac{x^3}{3}\right]}_0^{(4a{)}^{1/3}}$

$=\frac{8a}{3}-\frac{4a}{3}=\frac{4a}{3}$ sq units