Question

Find the area of region lying between parabolas $y^2=4ax\&x^2=4ay$ where $a>0$

Answer 1

Area of region between parabolas

$y^2=4ax,x^2=4ay$ where $a>0$

These both functions are inverse of each other

So Area bounded by these two curves

$=2$ (Area bounded by arc of there curves and line $y=x$)

$A=2{\int }_0^{4a}(2\sqrt{ax}-x)dx$

$A={[2\sqrt{a}\frac{(x{)}^{3/2}}{3/2}-\frac{x^2}{2}]}_0^{4a}$

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

$A=2[\frac{32}{3}{a}^2-\frac{16a^2}{2}]$

$A=2\left[\frac{16}{6}{a}^2\right]=\frac{16}{3}{a}^2$ sq. units