Mathematics - Area Under Curves Question with Solution | TestHub

Given curves, $$\begin{array}{rl}{x}^2+y^2& =\mathrm{8...}(i)\\ and{y}^2& =2x...(ii)\end{array}$$
On solving Eqs. (i) and (ii), we get $\therefore$ $$\begin{array}{rl}{x}^2+2x-8& =0\\ x^2+4x-2x-8& =0\\ x(x+4)-2(x+4)& =0\\ (x-2)(x+4)& =0\\ x=2\text{ and }y& =\pm 2\end{array}$$ $\therefore$ Required area $$\begin{array}{l}=2[\text{ Area of }OAP+\text{ Area of }PAB]\\ =2\left[{\int }_0^2\sqrt{2x}dx+{\int }_2^{2\sqrt{2}}\sqrt{8-x^2}dx\right]\\ =2[\sqrt{2}{\left(x^{3/2}\cdot \frac{2}{3}\right)}_0^2+(\frac{x}{2}\sqrt{8-x^2}\\ {+\frac{8}{2}{\sin }^{-1}\frac{x}{2\sqrt{2}})}_2^{2\sqrt{2}}]\\ =2\left[\frac{2\sqrt{2}}{3}\left(2^{3/2}\right)+4\times \frac{\pi }{2}-2-4\times \frac{\pi }{4}\right]\end{array}$$ $=2\left[\frac{2\sqrt{2}}{3}\cdot 2\sqrt{2}+2\pi -2-\pi \right]$ $=2\left[\frac{8}{3}-2+\pi \right]=2\left(\frac{2}{3}+\pi \right)=2\pi +\frac{4}{3}$