The area (in sq. units) of the region \(\left\{(x, y): \displaystyle\frac{x^{2}}{36}+\frac{y^{2}}{16}\le 1,\ y \ge 0\right\}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(12\pi\)
The region is the upper half of the ellipse with a = 6 and b = 4. On the upper half, \(y = \frac{4}{6}\sqrt{36-x^{2}} = \frac{2}{3}\sqrt{36-x^{2}}\):
\( \displaystyle A = \frac{2}{3}\int_{-6}^{6}\sqrt{36-x^{2}}\,dx \)
\( \displaystyle = \frac{2}{3}\left[\frac{x}{2}\sqrt{36-x^{2}}+18\sin^{-1}\frac{x}{6}\right]_{-6}^{6} \)
\( \displaystyle = \frac{2}{3}\times18\left(\frac{\pi}{2}+\frac{\pi}{2}\right) = 12\pi \)
Shortcut: The whole ellipse encloses \(\pi ab = \pi\times6\times4 = 24\pi\), and the region is half of it, so the area is \(12\pi\).