Area lying in first quadrant and bounded by the circle $x^2+y^2=9$ and the lines $x=1$ and $x=3$ is :

$\left|\frac{9\pi }{4}-\sqrt{2}-\frac{9}{2}sin^{-1}\frac{1}{3}\right|$sq.units

The correct answer is Option (4) → $\left|\frac{9\pi }{4}-\sqrt{2}-\frac{9}{2}sin^{-1}\frac{1}{3}\right|$sq.units

$x^2+y^2=9⇒y=\sqrt{9-x^2}$

so area = $\int\limits_1^3\sqrt{9-x^2}dx$

$=\left[\frac{x}{2}\sqrt{9-x^2}+\frac{9}{2}\sin^{-1}\frac{x}{3}\right]_1^3$

$=\frac{9π}{4}-\sqrt{2}-\frac{9}{2}\sin^{-1}\frac{1}{3}$