Application of Integrals — Class 12 Maths Solution

We have, ${y^2} = 2px$ and ${x^2} = 2py$
Solving curves, we get
$\therefore {x^4} = 4{p^2}{y^2}$

$\Rightarrow$ ${x^4} = 4{p^2}(2px)$

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

$\Rightarrow$ $x\left( {{x^3} - 8{p^3}} \right) = 0$

$\Rightarrow$ $x = 0,20$

When $x = 0$, $y = 0$ and when $x = 2p$, $y = 2p$
So, points of intersection are (0,0) and $(2p,2p)$
Graph of both the parabolas is as shown in the following figure.

From the figure, Area of shaded region,
$A = \int_0^{2p} {\left( {\sqrt {2px} - \frac{{{x^2}}}{{2p}}} \right)} dx$

$= \sqrt {2p} \int_0^{2p} {{x^{1/2}}} dx - \frac{1}{{2p}}\int_0^{2p} {{x^2}} dx$

$= \sqrt {2p} \left[ {\frac{2}{3}{x^{3/2}}} \right]_0^{2p} - \frac{1}{{2p}}\left[ {\frac{{{x^3}}}{3}} \right]_0^{2p}$

$= \sqrt {2p} \left( {\frac{2}{3} \cdot 2\sqrt 2 {p^{3/2}}} \right) - \frac{1}{{2p}}\left( {\frac{1}{3} - 8{p^3}} \right)$

$= \frac{8}{3}{p^2} - \frac{4}{3}{p^2} = \frac{{4{p^2}}}{3}sq,$ units