The area (in sq. units) of the region bounded by the curve $y = \sqrt{16-x^2}$ and x-axis is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $8π$
Region is the upper semicircle of radius $4$ (since $y=\sqrt{16-x^{2}}$, $-4\le x\le4$).
Area $=\displaystyle \int_{-4}^{4}\sqrt{16-x^{2}}\,dx=\frac{1}{2}\pi(4)^{2} = 8\pi$