Answer: (i) The area is \(\mathrm{e}^{\pi}-\mathrm{e}^{\pi/3}\).

(ii) The length of \(C\) is \(2\sqrt{2}\left(\mathrm{e}^{\pi/2}-\mathrm{e}^{\pi/6}\right)\).

For a polar curve \(r=f(\theta)\), the area swept out between two angles is \(\frac{1}{2}\int r^2\,\mathrm{d}\theta\), and the arc length is \(\int \sqrt{r^2+\left(\frac{\mathrm{d}r}{\mathrm{d}\theta}\right)^2}\,\mathrm{d}\theta\).

(i) Here \(r=2\mathrm{e}^{\theta}\), so \(r^2=4\mathrm{e}^{2\theta}\). Therefore

\(\text{Area}=\frac{1}{2}\int_{\pi/6}^{\pi/2}4\mathrm{e}^{2\theta}\,\mathrm{d}\theta=2\int_{\pi/6}^{\pi/2}\mathrm{e}^{2\theta}\,\mathrm{d}\theta\).

Integrating,

\(2\int \mathrm{e}^{2\theta}\,\mathrm{d}\theta=2\cdot \frac{1}{2}\mathrm{e}^{2\theta}=\mathrm{e}^{2\theta}\), so

\(\text{Area}=\left[\mathrm{e}^{2\theta}\right]_{\pi/6}^{\pi/2}=\mathrm{e}^{\pi}-\mathrm{e}^{\pi/3}.\)

(ii) Differentiate \(r\):

\(\frac{\mathrm{d}r}{\mathrm{d}\theta}=2\mathrm{e}^{\theta}.\)

So

\(\sqrt{r^2+\left(\frac{\mathrm{d}r}{\mathrm{d}\theta}\right)^2}=\sqrt{4\mathrm{e}^{2\theta}+4\mathrm{e}^{2\theta}}=\sqrt{8\mathrm{e}^{2\theta}}=2\sqrt{2}\,\mathrm{e}^{\theta}.\)

Hence the length of \(C\) is

\(\displaystyle \int_{\pi/6}^{\pi/2}2\sqrt{2}\,\mathrm{e}^{\theta}\,\mathrm{d}\theta=2\sqrt{2}\left[\mathrm{e}^{\theta}\right]_{\pi/6}^{\pi/2}=2\sqrt{2}\left(\mathrm{e}^{\pi/2}-\mathrm{e}^{\pi/6}\right).\)