The area of region bounded by the curve $y^2 = 4ax$ and the straight line $x = 2a, a > 0$ in the first quardant is:

$\frac{8\sqrt{2}a^2}{3}$ sq. units

The correct answer is Option (2) → $\frac{8\sqrt{2}a^2}{3}$ sq. units

Given the parabola: $y^2 = 4ax$

Given the vertical line: $x = 2a$

The required area lies in the first quadrant, so only the upper half of the parabola is considered.

From $y^2 = 4ax$, the positive root gives: $y = 2\sqrt{ax}$

Area under the curve from $x = 0$ to $x = 2a$ is given by:

$A = \int_{0}^{2a} 2\sqrt{ax} \, dx$

Factor out constants:

$A = 2\sqrt{a} \int_{0}^{2a} \sqrt{x} \, dx$

Using the identity: $\int x^{n} dx = \frac{x^{n+1}}{n+1}$

$A = 2\sqrt{a} \left[ \frac{2}{3} x^{3/2} \right]_0^{2a}$

Evaluate the definite integral:

$= 2\sqrt{a} \cdot \frac{2}{3} \cdot (2a)^{3/2}$

$= 2\sqrt{a} \cdot \frac{2}{3} \cdot 2\sqrt{2} \cdot a^{3/2}$

$= \frac{8\sqrt{2}}{3} a^2$