The area (in sq. units) bounded by the curve $y=\sqrt{16-x^2}$ and x-axis is :
Answer & explanation
Correct answer: option 1
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