Draw a rough sketch of the region $\{(x, y) : y^2 \leq 6ax \text{ and } x^2 + y^2 \leq 16a^2\}$. Also, find the area of the region sketched using method of integration.

$\frac{4a^2}{3}(4\pi + \sqrt{3})$ square units

The correct answer is Option (2) → $\frac{4a^2}{3}(4\pi + \sqrt{3})$ square units

We have, $y^2 = 6ax$ and $x^2 + y^2 = 16a^2$

On solving both equations, we get

$x^2 + 6ax = 16a^2$

$\Rightarrow x^2 + 6ax - 16a^2 = 0$

$\Rightarrow x^2 + 8ax - 2ax - 16a^2 = 0$

$\Rightarrow x(x + 8a) - 2a(x + 8a) = 0$

$\Rightarrow (x - 2a)(x + 8a) = 0$

$\Rightarrow x = 2a, -8a$

Let $A_1 = \text{Area under the curve (parabola) in first quadrant}$

and $A_2 = \text{Area under the curve (circle) in first quadrant}$

As, $A_1$ and $A_2$ symmetrical to X-axis on both sides

Therefore required area $= 2(A_1 + A_2)$

where $A_1 = \int\limits_{0}^{2a} \sqrt{6ax} \, dx$ and $A_2 = \int\limits_{2a}^{4a} \sqrt{(4a)^2 - x^2} \, dx$

$∴$ Area of required region $= 2 \left[ \int\limits_{0}^{2a} \sqrt{6ax} \, dx + \int\limits_{2a}^{4a} \sqrt{(4a)^2 - x^2} \, dx \right]$

$= 2 \left[ \int\limits_{0}^{2a} \sqrt{6a} \cdot x^{1/2} \, dx + \int\limits_{2a}^{4a} \sqrt{(4a)^2 - x^2} \, dx \right]$

$= 2 \left[ \sqrt{6a} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{2a} + \left( \frac{x}{2} \sqrt{(4a)^2 - x^2} + \frac{(4a)^2}{2} \sin^{-1} \frac{x}{4a} \right)_{2a}^{4a} \right]$

$= 2 \left[ \sqrt{6a} \cdot \frac{2}{3} ((2a)^{3/2} - 0) + \frac{4a}{2} \cdot 0 + \frac{16a^2}{2} \cdot \frac{\pi}{2} - \frac{2a}{2} \sqrt{16a^2 - 4a^2} - \frac{16a^2}{2} \sin^{-1} \frac{2a}{4a} \right]$

$= 2 \left[ \sqrt{6a} \cdot \frac{2}{3} \cdot 2\sqrt{2} \cdot a^{3/2} + 0 + 4\pi a^2 - \frac{2a}{2} \cdot 2\sqrt{3}a - 8a^2 \cdot \frac{\pi}{6} \right]$

$= 2 \left[ \sqrt{12} \cdot \frac{4}{3} a^2 + 4\pi a^2 - 2\sqrt{3} a^2 - \frac{4a^2 \pi}{3} \right]$

$= 2 \left[ \frac{8\sqrt{3} a^2 + 12\pi a^2 - 6\sqrt{3} a^2 - 4a^2 \pi}{3} \right]$

$= \frac{2}{3} a^2 [8\sqrt{3} + 12\pi - 6\sqrt{3} - 4\pi]$

$= \frac{2}{3} a^2 [2\sqrt{3} + 8\pi] = \frac{4}{3} a^2 [\sqrt{3} + 4\pi] \text{ sq. units}$