Mathematics - Area Under Curves Question with Solution | TestHub

$\frac{x^2}{18}+\frac{y^2}{6}=1$

$\begin{array}{rl}& \frac{x^2}{18}+\frac{3x^2}{18}=1\Rightarrow 4x^2=18\Rightarrow x^2=\frac{9}{2}\\ & {\int }_{\frac{3}{\sqrt{2}}}^{\sqrt[3]{2}}\frac{\sqrt{18-x^2}}{\sqrt{3}}dx\\ & =\frac{1}{\sqrt{3}}{\left(\frac{x\sqrt{18-x^2}}{2}+\frac{18}{2}{\sin }^{-1}\frac{x}{3\sqrt{2}}\right)}_{\frac{3}{\sqrt{2}}}^{3\sqrt{2}}\\ & =\frac{1}{\sqrt{3}}\left(9\times \frac{\pi }{2}-\frac{3}{2\sqrt{2}}\times \frac{3\sqrt{3}}{\sqrt{2}}-9\times \frac{\pi }{6}\right)\end{array}$
Required Area $\begin{array}{rl}& =\frac{1}{2}\times \frac{9}{2}+\left(\frac{18\pi }{6}-\frac{9\sqrt{3}}{4}\right)\frac{1}{\sqrt{3}}\\ & =\sqrt{3}\pi \end{array}$