The area (in sq. units) of the region bounded by the curve \(x = \sqrt{9-y^{2}}\) and the y-axis is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{9\pi}{2}\)
Squaring \(x = \sqrt{9-y^{2}}\) gives \(x^{2}+y^{2} = 9\) with \(x \ge 0\), so the curve is the right half of the circle of radius 3. It meets the y-axis at y = −3 and y = 3, so, integrating with respect to y,
\( \displaystyle A = \int_{-3}^{3}\sqrt{9-y^{2}}\,dy \)
\( \displaystyle = \left[\frac{y}{2}\sqrt{9-y^{2}}+\frac{9}{2}\sin^{-1}\frac{y}{3}\right]_{-3}^{3} \)
\( \displaystyle = \frac{9}{2}\left(\frac{\pi}{2}+\frac{\pi}{2}\right) = \frac{9\pi}{2} \)
Shortcut: The region is half of a circle of radius 3, so its area is \(\frac{1}{2}\pi(3)^{2} = \frac{9\pi}{2}\).