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

$\frac{4p^2}{3}$

The correct answer is Option (3) → $\frac{4p^2}{3}$

We have,

$ y^2 = 2px \quad \dots(i)$

$\text{and} \quad x^2 = 2py \quad \dots(ii)$

On solving Eqs. (i) and (ii), we get

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

On solving both sides, we get

$x^4 = 4p^2 \cdot (2px)$

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

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

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

$\Rightarrow x = 0, 2p$

Let $A_1 = \text{Area under the curve } y^2 = 2px = \text{Area of OABDO}$

$A_2 = \text{Area under the curve } x^2 = 2py = \text{Area OCBDO}$

$∴\text{Required area} = A_1 - A_2$

$= \int_{0}^{2p} \sqrt{2px} \, dx - \int_{0}^{2p} \frac{x^2}{2p} \, dx$

$= \sqrt{2p} \int_{0}^{2p} x^{1/2} \, dx - \frac{1}{2p} \int_{0}^{2p} x^2 \, dx$

$= \sqrt{2p} \left[ \frac{2(x)^{3/2}}{3} \right]_{0}^{2p} - \frac{1}{2p} \left[ \frac{x^3}{3} \right]_{0}^{2p} \qquad \left[ ∵\int x^n \, dx = \frac{x^{n+1}}{n+1} \right]$

$= \sqrt{2p} \left[ \frac{2}{3} \cdot (2p)^{3/2} - 0 \right] - \frac{1}{2p} \left[ \frac{1}{3} (2p)^3 - 0 \right]$

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

$= \sqrt{2p} \left( \frac{4\sqrt{2}}{3} p^{3/2} \right) - \frac{1}{2p} \left( \frac{8}{3} p^3 \right)$

$= \frac{4\sqrt{2}}{3} \cdot \sqrt{2}p^2 - \frac{8}{6} p^2$

$= \frac{(16 - 8)p^2}{6} = \frac{8p^2}{6}$

$= \frac{4p^2}{3} \text{ sq. units}$