The area of the region bounded by the curve \( y=\sqrt{9-x^2} \) and \( x \)-axis is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{9\pi}{2} \) Sq.units
\( y=\sqrt{9-x^2} \) gives \( x^2+y^2=9 \) with \( y\ge 0 \): the upper half of the circle of radius 3. It meets the x-axis at x = −3 and x = 3.
\( \displaystyle \text{Area}=\int_{-3}^{3}\sqrt{9-x^2}\,dx \)
\( \displaystyle =2\int_{0}^{3}\sqrt{9-x^2}\,dx \)
\( \displaystyle =2\left[\frac{x}{2}\sqrt{9-x^2}+\frac{9}{2}\sin^{-1}\frac{x}{3}\right]_{0}^{3} \)
\( \displaystyle =2\cdot\frac{9}{2}\cdot\frac{\pi}{2}=\frac{9\pi}{2} \)
Shortcut: The region is a semicircle of radius 3, so its area is \( \frac{1}{2}\pi(3)^2=\frac{9\pi}{2} \).
Why the other options are incorrect
Option (1) \( 9\pi \) Sq.units — Incorrect. \( 9\pi \) is the area of the full circle; the square root gives only \( y\ge 0 \).
Option (2) \( \displaystyle\frac{9\pi}{4} \) Sq.units — Incorrect. \( \frac{9\pi}{4} \) is a quarter circle, the integral from 0 to 3 only.
Option (3) \( 2\pi \) Sq.units — Incorrect. The semicircle of radius 3 has area \( \frac{9\pi}{2} \), not \( 2\pi \).