Answer: (i) The area enclosed by the curve and the two half-lines is

\(A=\frac12\int_{\pi/6}^{\pi/2} r^2\,d\theta\).

Since \(r=2e^{\theta}\), we have \(r^2=4e^{2\theta}\). Hence

\(A=\frac12\int_{\pi/6}^{\pi/2}4e^{2\theta}\,d\theta=2\int_{\pi/6}^{\pi/2}e^{2\theta}\,d\theta=[e^{2\theta}]_{\pi/6}^{\pi/2}=e^{\pi}-e^{\pi/3}.\)

So the area is \(e^{\pi}-e^{\pi/3}\approx 20.3\).

(ii) The length of \(C\) is

\(L=\int_{\pi/6}^{\pi/2}\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta.\)

Now \(r=2e^{\theta}\), so \(\frac{dr}{d\theta}=2e^{\theta}\). Therefore

\(L=\int_{\pi/6}^{\pi/2}\sqrt{4e^{2\theta}+4e^{2\theta}}\,d\theta =\int_{\pi/6}^{\pi/2}2\sqrt2\,e^{\theta}\,d\theta =2\sqrt2\,[e^{\theta}]_{\pi/6}^{\pi/2}.\)

Hence

\(L=2\sqrt2\left(e^{\pi/2}-e^{\pi/6}\right)\approx 8.83.\)

(i) For a polar curve, the area swept out between two angles is given by \(\frac12\int r^2\,d\theta\).

Here \(r=2e^{\theta}\), so

\(r^2=(2e^{\theta})^2=4e^{2\theta}.\)

Therefore

\(A=\frac12\int_{\pi/6}^{\pi/2}4e^{2\theta}\,d\theta=2\int_{\pi/6}^{\pi/2}e^{2\theta}\,d\theta.\)

Integrating gives

\(A=2\cdot \frac12 e^{2\theta}\Big|_{\pi/6}^{\pi/2}=[e^{2\theta}]_{\pi/6}^{\pi/2}=e^{\pi}-e^{\pi/3}.\)

So the required area is \(e^{\pi}-e^{\pi/3}\), which is approximately \(20.3\).

(ii) The arc length of a polar curve \(r=f(\theta)\) is

\(L=\int_{\alpha}^{\beta}\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta.\)

With \(r=2e^{\theta}\), we have

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

So

\(L=\int_{\pi/6}^{\pi/2}\sqrt{(2e^{\theta})^2+(2e^{\theta})^2}\,d\theta =\int_{\pi/6}^{\pi/2}\sqrt{8e^{2\theta}}\,d\theta.\)

Since \(e^{\theta}\gt 0\), this simplifies to

\(L=\int_{\pi/6}^{\pi/2}2\sqrt2\,e^{\theta}\,d\theta.\)

Now integrate:

\(L=2\sqrt2\,[e^{\theta}]_{\pi/6}^{\pi/2}=2\sqrt2\left(e^{\pi/2}-e^{\pi/6}\right).\)

Hence the length of \(C\) is \(2\sqrt2\left(e^{\pi/2}-e^{\pi/6}\right)\), which is approximately \(8.83\).