Question

Prove that the area enclosed between two parabolas $y^2=4ax$ and $x^2=4ay$ is $\frac{16a^2}{3}$

Answer 1

To Prove that the area enclosed between two parabolas $y^2=4ax$ and $x^2=4ay$ is $\frac{16a^3}{3}$

Given curves are

$y^2=4ax$ and

$x^2=4ay$

First we have to find the area of Intersection of the two curves

Point of Intersection of the two curves are

${\left(\frac{x^2}{4a}\right)}^2=4ax$

$\left(\frac{x^4}{16a^2}\right)=4ax$

$x^4=64a^{3}x$

$x^4-64a^{3}x=0$

$x(x^3-64a^3)=0$

$x=0,x=4a$

Also $y=0,y=4a$

The Point of Intersection of these 2 curves are (0,0) and (4a, 4a )

The Area of the two region between the two curves

$=$ Area of the shaded region

${\int }_0^{4a}[y_2-y_1]dx$

${\int }_0^{4a}[\sqrt{4ax}-\frac{x^2}{4a}]dx$

On Integrating this ,we get

${[4a^{1/2}\frac{x^{3/2}}{3}-\frac{x^3}{12a}]}_0^{4a}$

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

$=\frac{16a^2}{3}$

Hence , Area $=\frac{16a^2}{3}$