Question

Area of the region enclosed by circle $x^2+y^2=16$ and lines $x=0$ and $x=4$ in the first quadrant is __________.

• A. $4\pi$
• B. $16\pi$
• C. $8\pi$
• D. $2\pi$

Answer 1

The correct option is A $4\pi$

Here $x^2+y^2=16$
$\therefore y^2=16-x^2$
$y=\sqrt{4^2-x^2}$
Area $=\stackrel{\pi /2}{\underset{0}{\int }}\sqrt{4^2-x^2}dx$
$={[\frac{x\left(\sqrt{4^2-x^2}\right)}{2}+\frac{4^2}{2}{\sin }^{-1}\left(\frac{\pi }{4}\right)]}_0^{\pi /2}$
$=(\frac{4\left(0\right)}{2}+\frac{16}{2}{\sin }^{-1}(1\left)\right)-(0+\frac{16}{2}{\sin }^{-1}(0\left)\right)$
$=0+8\left(\frac{\pi }{2}\right)-0$
$\therefore$ Required area $=4\pi$