Question

3 $r = e^{\theta}, 0 \leq \theta \leq \pi$ $r = e^{\theta}$ $r = e^{\theta}$

Explanation:

Step1: Recall the formula for the arc - length of a polar curve

The formula for the arc - length \(L\) of a polar curve \(r = r(\theta)\) from \(\theta=a\) to \(\theta = b\) is \(L=\int_{a}^{b}\sqrt{r^{2}+(r')^{2}}d\theta\). Given \(r = e^{\theta}\), then \(r'=\frac{dr}{d\theta}=e^{\theta}\).

Step2: Substitute \(r\) and \(r'\) into the arc - length formula

Substitute \(r = e^{\theta}\) and \(r'=e^{\theta}\) into \(L=\int_{a}^{b}\sqrt{r^{2}+(r')^{2}}d\theta\). We get \(L=\int_{0}^{\pi}\sqrt{(e^{\theta})^{2}+(e^{\theta})^{2}}d\theta=\int_{0}^{\pi}\sqrt{2e^{2\theta}}d\theta\). Since \(\sqrt{2e^{2\theta}}=\sqrt{2}e^{\theta}\), the integral becomes \(L = \sqrt{2}\int_{0}^{\pi}e^{\theta}d\theta\).

Step3: Evaluate the integral

We know that \(\int e^{\theta}d\theta=e^{\theta}+C\). Using the fundamental theorem of calculus \(\int_{0}^{\pi}e^{\theta}d\theta=e^{\theta}\big|_{0}^{\pi}=e^{\pi}-e^{0}\).

Answer:

\(L=\sqrt{2}(e^{\pi}- 1)\)