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) → \(8\pi\)
Squaring \(y = \sqrt{16 - x^2}\) gives \(x^2 + y^2 = 16\) with \(y \ge 0\). So the curve is the upper half of the circle of radius 4 centred at the origin, meeting the x-axis at x = −4 and x = 4.
\(\displaystyle \text{Area} = \int_{-4}^{4} \sqrt{16 - x^2}\,dx\)
Using \(\int \sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C\) with a = 4:
\(\displaystyle = \left[\frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\frac{x}{4}\right]_{-4}^{4}\)
\(\displaystyle = 8\cdot\frac{\pi}{2} - 8\left(-\frac{\pi}{2}\right) = 8\pi\)
Shortcut: The region is a semicircle of radius 4, so its area is \(\frac{1}{2}\pi(4)^2 = 8\pi\).