Find the area of the region bounded by the curve $y^2 = 4x$ and $x^2 = 4y$.

$\frac{16}{3}$

The correct answer is Option (1) → $\frac{16}{3}$

Given equation of curves are

$y^2 = 4x \quad \dots(i)$

$\text{and} \quad x^2 = 4y \quad \dots(ii)$

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

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

$⇒\frac{x^4}{16} = 4x$

$⇒x^4 = 64x$

$⇒x^4 - 64x = 0$

$⇒x(x^3 - 4^3) = 0$

$⇒x = 4, 0$

Let $A_1 =$ Area under the curve $y^2 = 4x$

$\text{Area of } OABDO = \int_{0}^{4} \sqrt{4x} \, dx$

and $A_2 =$ Area under the curve $x^2 = 4y$

$= \text{Area of } OCBDO = \int_{0}^{4} \frac{x^2}{4} \, dx$

$∴$ Required area of shaded region $= A_1 - A_2$

$= \int_{0}^{4} \left( \sqrt{4x} - \frac{x^2}{4} \right) dx$

$= \int_{0}^{4} \left( 2\sqrt{x} - \frac{x^2}{4} \right) dx = \left[ \frac{2x^{3/2} \cdot 2}{3} - \frac{1}{4} \cdot \frac{x^3}{3} \right]_{0}^{4}$

$= \frac{2 \cdot 2}{3} \cdot 8 - \frac{1}{4} \cdot \frac{64}{3} - 0$

$= \frac{32}{3} - \frac{16}{3} = \frac{16}{3} \text{ sq. units}$