The area (in sq. units) of the region \(\{(x, y): 0 \le x \le 2,\ 0 \le y \le \sqrt{16-x^{2}}\}\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle 2\sqrt{3}+\frac{4\pi}{3}\)
The region lies under the upper half of the circle \(x^{2}+y^{2}=16\), from x = 0 to x = 2. Using the standard integral
\( \displaystyle \int\sqrt{a^{2}-x^{2}}\,dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}+C \)
with a = 4:
\( \displaystyle A = \int_{0}^{2}\sqrt{16-x^{2}}\,dx \)
\( \displaystyle = \left[\frac{x}{2}\sqrt{16-x^{2}}+8\sin^{-1}\frac{x}{4}\right]_{0}^{2} \)
\( \displaystyle = 1\times\sqrt{12}+8\sin^{-1}\frac{1}{2} \)
\( \displaystyle = 2\sqrt{3}+8\times\frac{\pi}{6} \)
\( \displaystyle = 2\sqrt{3}+\frac{4\pi}{3} \)
In shapes, this is a triangle with vertices (0, 0), (2, 0), \((2, 2\sqrt{3})\) of area \(2\sqrt{3}\), together with a sector of radius 4 and angle \(30^{\circ}\) of area \(\frac{4\pi}{3}\).
Why the other options are incorrect
Option (2) \(\displaystyle 4\sqrt{3}+\frac{4\pi}{3}\) — Incorrect. The first term is \(\frac{x}{2}\sqrt{16-x^{2}}\) at x = 2, which is \(\sqrt{12} = 2\sqrt{3}\); dropping the \(\frac{1}{2}\) gives \(4\sqrt{3}\).
Option (3) \(\displaystyle 2\sqrt{3}+\frac{8\pi}{3}\) — Incorrect. \(\sin^{-1}\frac{1}{2} = \frac{\pi}{6}\), so the second term is \(\frac{8\pi}{6} = \frac{4\pi}{3}\); \(\frac{8\pi}{3}\) uses \(\frac{\pi}{3}\).
Option (4) \(\displaystyle \frac{4\pi}{3}\) — Incorrect. This leaves out the first term \(2\sqrt{3}\), which is not zero at x = 2.