Mathematics - Area Under Curves Question with Solution | TestHub

If curve $\mathrm{r}=\mathrm{a}\sin 3\theta$ To trace the curve, we consider the following table : \begin{array}{|c|c|c|c|c|c|c|c|} \hline 3 \theta= & 0 & \frac{\pi}{2} & \pi & \frac{3 \pi}{2} & 2 \pi & \frac{5 \pi}{2} & 3 \pi \\ \hline\theta= & 0 & \frac{\pi}{6} & \frac{\pi}{3} & \frac{\pi}{2} & \frac{2 \pi}{3} & \frac{5 \pi}{6} & \pi \\ \hline \mathrm{r}= & 0 & \mathrm{a} & 0 & -\mathrm{a} & 0 & \mathrm{a} & 0 \\ \hline \end{array} Thus there is a loop between $\theta =0\&\theta =\frac{\pi }{3}$ as r varies from $\mathrm{r}=0$ to $\mathrm{r}=0$.
Hence, the area of the loop lying in the $$\text{ positive quadrant }=\frac{1}{2}{\int }_0^{\frac{\pi }{3}}{r}^2\mathrm{d}\theta$$ $$=\frac{1}{2}{\int }_0^{\frac{\pi }{3}}{\sin }^2\varphi \cdot \frac{1}{3}\mathrm{d}\varphi$$ $[$ On putting, $3\theta =\varphi \Rightarrow \mathrm{d}\theta =\frac{1}{3}\mathrm{d}\varphi ]$ $$\begin{array}{c}=\frac{a^2}{6}{\int }_0^{\frac{\pi }{2}}{\sin }^2\varphi \mathrm{d}\varphi \\ =\frac{{\mathrm{a}}^2}{6}\cdot {\int }_0^{\frac{\pi }{2}}\frac{1-\cos 2\varphi }{2}\mathrm{d}\varphi \left[\because \cos 2\theta =1-2{\sin }^2\theta \right]\end{array}$$ $=\frac{{\mathrm{a}}^2}{12}\cdot {\left[\varphi +\frac{\sin 2\varphi }{2}\right]}_0^{\frac{\pi }{2}}$ $=\frac{{\mathrm{a}}^2}{12}\cdot \left[\frac{\pi }{2}+\sin \pi \right]=\frac{{\mathrm{a}}^2\pi }{24}$