Question

Area is the region bounded by the circle $x^2+y^2=4$ and lines x=0, x=2 in the first quadrant is ?

• A. $\pi$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{4}$

Answer 1

Answer: (A)

$\pi$

$\begin{array}{c}Weknowthat,\\ x^2+y^2=4\\ y^2=4-x^2\\ \therefore y=\pm \sqrt{4-x^2}\\ As,OBAisin1stquadrant\\ \therefore y=\sqrt{4-x^2}\\ \therefore Requiredarea={\int }_0^{2}y.dx\\ ={\int }_0^2\sqrt{4-x^2}dx\\ ={\int }_0^2\sqrt{{\left(2\right)}^2-x^2}dx\\ ={\left[\frac{x}{2}\sqrt{{\left(2\right)}^2-x^2}+2{\sin }^{-1}\frac{x}{2}\right]}_0^2\\ =\left[\frac{2}{2}\sqrt{{\left(2\right)}^2-2^2}+2{\sin }^{-1}\frac{2}{2}-\frac{0}{2}\sqrt{{\left(2\right)}^2-{\left(0\right)}^2}-2{\sin }^{-1}\frac{0}{2}\right]\\ =\left[0+2{\sin }^{-1}\left(1\right)-2{\sin }^{-1}\left(0\right)-0\right]\\ =2\left[{\sin }^{-1}\left(1\right)-{\sin }^{-1}\left(0\right)\right]\\ =2\left[\frac{\pi }{2}-0\right]\\ =2.\frac{\pi }{2}\\ =\pi \end{array}$

Hence, Option $A$ is correct answer.