Question

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

Answer 1

We have, $y^2=2pxand{x}^2=2py$

$\therefore y=\sqrt{2px}$

$\Rightarrow x^2=2p.\sqrt{2px}$

$\Rightarrow x^4=4p^2.\left(2px\right)$

$\Rightarrow x^4=8p^{3}x$

$\Rightarrow x^4-8p^{3}x=0$

$\Rightarrow x^3(x-8p^3)=0$

$\Rightarrow x=0,2p$

$\therefore Requiredarea={\int }_0^{2p}\sqrt{2px}dx-{\int }_0^{2p}\frac{x^2}{2p}dx=\sqrt{2p}{\int }_0^{2p}{x}^{\frac{1}{2}}dx-\frac{1}{2p}{\int }_0^{2p}{x}^{2}dx=\sqrt{2p}{\left[\frac{2(x{)}^{\frac{3}{2}}}{3}\right]}_0^{2p}-\frac{1}{2p}{\left[\frac{x^3}{3}\right]}_0^{2p}=\sqrt{2p}\left[\frac{2}{3}\right(2p{)}^{\frac{3}{2}}-0]-\frac{1}{2p}\left[\frac{1}{3}\right(2p{)}^3-0]=\sqrt{2p}\left(\frac{2}{3}.2\sqrt{2}{p}^{\frac{3}{2}}\right)-\frac{1}{2p}\left(\frac{1}{3}8p^3\right)=\sqrt{2p}\left(\frac{4\sqrt{2}}{3}{p}^{\frac{3}{2}}\right)-\frac{1}{2p}\left(\frac{8}{3}{p}^3\right)=\frac{4\sqrt{2}}{3}.\sqrt{2}{p}^2-\frac{8}{6}{p}^2=\frac{(16-8)p^2}{6}=\frac{8p^2}{6}=\frac{4p^2}{3}sq.units$