The area (in sq. units) bounded by the curve $y=\sqrt{16-x^2}$ and x-axis is :

$8 \pi $

The curve $y=\sqrt{16-x^{2}}$ represents the upper half of a circle of radius $4$ centered at the origin.

The region bounded by the curve and the x-axis is a semicircle.

Area of a circle $=\pi r^{2}$.

So area of the semicircle:

$=\frac{1}{2}\pi(4)^{2}=8\pi$

final answer: $8\pi$ square units