Question

The area of the smaller segment cut off from the circle $x^2+y^2=9$ by $x=1$ is

• A. $\frac{1}{2}\left(9{\sec }^{-1}3-\sqrt{8}\right)$ sq unit
• B. $\left(9{\sec }^{-1}3-\sqrt{8}\right)$ sq unit
• C. $\left(\sqrt{8}-9{\sec }^{-1}3\right)$ sq unit
• D. None of the above

Answer 1

Answer: (B)

Given equation of the circle is $x^2+y^2=9$.
$\therefore$ Area of the smaller segment cut off from the circle
$x^2+y^2=9$ by $x=1$, is given by
$A=2\underset{1}{\overset{3}{\int }}\sqrt{9-x^2}dx$
$=2\cdot \frac{1}{2}{\left[x\sqrt{9-x^2}+9{\sin }^{-1}\frac{x}{3}\right]}_1^3$

$=\left[3\cdot \sqrt{9-9}+9si{n}^{-1}\left(\frac{3}{3}\right)-1\cdot \sqrt{9-1}-9{\sin }^{-1}\left(\frac{1}{3}\right)\right]$
$=\left[9{\sin }^{-1}\left(\sin \frac{\pi }{2}\right)-\sqrt{8}-9{\sin }^{-1}\left(\frac{1}{3}\right)\right]$
$=\left[9\left(\frac{\pi }{2}-{\sin }^{-1}\left(\frac{1}{3}\right)\right)-\sqrt{8}\right]$
$=\left[9\left({\cos }^{-1}\left(\frac{1}{3}\right)-\sqrt{8}\right)\right]$
$=\left[9{\sec }^{-1}\left(3\right)-\sqrt{8}\right]$ sq unit