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

\(8\pi\) sq.units

\(y=\sqrt{16-x^2}\) and equation of x-axis is y = 0.

$\sqrt{16-x^2}=0⇒x=±4$

Point of intersection between $\sqrt{16-x^2}$ & x - axis is (-4, 0) & (4, 0).

Area = $\int\limits_{-4}^{4}\sqrt{16-x^2}dx=\int\limits_{-4}^{4}\sqrt{4^2-x^2}dx$

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

$=8\pi$ sq. units.