Area lying in first quadrant and bounded by the circle $x^2+y^2=9$ and the lines $x=1$ and $x=3$ is :
Answer & explanation
Correct answer: option 4
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 = $\int\limits_1^3\sqrt{9-x^2}dx$
$=\left[\frac{x}{2}\sqrt{9-x^2}+\frac{9}{2}\sin^{-1}\frac{x}{3}\right]_1^3$
$=\frac{9π}{4}-\sqrt{2}-\frac{9}{2}\sin^{-1}\frac{1}{3}$