Question

Find the area of the region bounded by the parabola $y^2=2px,x^2=2py$

Answer 1

We have $y^2=2px\cdots \left(1\right)$
& $x^2=2py\cdots \left(2\right)$
Substitute the value of $y$ from equation (2) to equation (1)
$\Rightarrow \frac{x^4}{4p^2}=2px$
$\Rightarrow x^4-8p^{3}x=0$
$\Rightarrow x(x^3-(2p{)}^3)=0)$
$\Rightarrow x(x-2p)(x^2+2px+4p^2)=0$
$\Rightarrow x=0,2p$
$[\because x^2+2px+4p^2=0]$has no real roots
When $x=0\Rightarrow t=0$and when $x=2p$
$\Rightarrow y=2p$
Hence $(0,0)\text{and}(2p,2p)$are the points of intersection.
So, the area bounded by curves is shaded in the diagram below:

Area $={\int }_{x_1}^{x_2}(y_2-y_1)dx$
Area $={\int }_0^{2p}(\sqrt{2px}-\frac{x^2}{2p})dx$
$[\because x\text{varies from}0\text{to}2p]$
$=\sqrt{2p}[\frac{2}{3}{x}^{\frac{3}{2}}{]}_0^{2p}-\frac{1}{2p}[\frac{x^3}{3}{]}_0^{2p}$
$[\because {\int }_a^b{x}^{n}dx=[\frac{x^{n+1}}{n+1}{]}_a^b]$
$=\sqrt{2p}\left(\frac{2}{3}2p\sqrt{2p}\right)-\frac{1}{2p}\frac{8p^3}{3}$
$=\frac{8}{3}{p}^2-\frac{4}{3}{p}^2$
$=\frac{4p^2}{3}sq.units$
Hence the required area is $\frac{4p^2}{3}sq.units$