The area (in sq. units) of the region bounded by the curve \(y = \sqrt{36-4x^{2}}\) and the x-axis is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(9\pi\)
Since \(36-4x^{2} = 4(9-x^{2})\), the curve is
\(y = 2\sqrt{9-x^{2}}\)
It meets the x-axis where \(9-x^{2} = 0\), that is x = ±3, and it lies above the x-axis in between. The curve is symmetric about the y-axis, so
\( \displaystyle A = 2\int_{0}^{3}2\sqrt{9-x^{2}}\,dx \)
\( \displaystyle = 4\left[\frac{x}{2}\sqrt{9-x^{2}}+\frac{9}{2}\sin^{-1}\frac{x}{3}\right]_{0}^{3} \)
\( \displaystyle = 4\times\frac{9}{2}\times\frac{\pi}{2} = 9\pi \)
Shortcut: Squaring gives \(4x^{2}+y^{2} = 36\), that is \(\displaystyle\frac{x^{2}}{9}+\frac{y^{2}}{36} = 1\), with \(y \ge 0\). So the curve is the upper half of an ellipse with semi-axes 3 and 6, and the area is \(\displaystyle\frac{1}{2}\pi\times3\times6 = 9\pi\).
Why the other options are incorrect
Option (2) \(18\pi\) — Incorrect. This is the area of the whole ellipse; a square root is never negative, so the curve is only its upper half.
Option (3) \(\displaystyle\frac{9\pi}{2}\) — Incorrect. This is the area under \(y = \sqrt{9-x^{2}}\); the factor 2 taken out of the square root is lost.
Option (4) \(36\pi\) — Incorrect. This is the area of a full circle of radius 6, not of the half-ellipse.