The area of the region bounded by the curve $y = \sqrt{16 - x^2}$ and X-axis is

$8\pi$ sq units

The correct answer is Option (1) → $8\pi$ sq units

Given equation of curve is $y = \sqrt{16 - x^2}$ and the equation of X-axis i.e., $y = 0$.

$∴\sqrt{16 - x^2} = 0 \quad \dots(i)$

$\Rightarrow 16 - x^2 = 0$

$\Rightarrow x^2 = 16$

$\Rightarrow x = \pm 4$

So, the intersection points are $(4, 0)$ and $(-4, 0)$.

$∴\text{Required area of curve} = \int_{-4}^{4} (16 - x^2)^{1/2} dx$

$= \int_{-4}^{4} \sqrt{4^2 - x^2} dx$

$= \left[ \frac{x}{2} \sqrt{4^2 - x^2} + \frac{4^2}{2} \sin^{-1} \frac{x}{4} \right]_{-4}^{4}$

$= \left[ \frac{4}{2} \sqrt{4^2 - 4^2} + 8 \sin^{-1} \left( \frac{4}{4} \right) \right] - \left[ -\frac{4}{2} \sqrt{4^2 - (-4)^2} + 8 \sin^{-1} \left( \frac{-4}{4} \right) \right]$

$= [2 \times 0 + 8 \sin^{-1} (1)] - [-2 \times 0 + 8 \sin^{-1} (-1)]$

$= 8 \sin^{-1} \left( \sin \frac{\pi}{2} \right) - \left[ 8 \sin^{-1} \left( -\sin \frac{\pi}{2} \right) \right]$

$= 8 \times \frac{\pi}{2} - 8 \sin^{-1} \left[ \sin \left( -\frac{\pi}{2} \right) \right]$

$= 4\pi + 8 \times \frac{\pi}{2} = 8\pi \text{ sq. units.}$