Mathematics - Area Under Curves Question with Solution | TestHub

Given equation of circle is $x^2+y^2=16$ $\therefore y^2=16-x^2\therefore y=\sqrt{16-x^2}$ Required area is shaded. $A=2\mathrm{A}(\mathrm{OABCO})$ $\begin{array}{rl}\text{ Area }& =2{\int }_0^2\sqrt{16-x^2}\mathrm{dx}\\ & =2{\left[\frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}{\sin }^{-1}\frac{x}{4}\right]}_0^2\\ & =2\left\{\frac{2}{2}\sqrt{12}+8{\sin }^{-1}\frac{1}{2}-0\right\}=2\left[2\sqrt{3}+8\left(\frac{\pi }{6}\right)\right]=4\sqrt{3}+\frac{8\pi }{3}\end{array}$