Q. 28 - Calculus

Question

In exercise 26-30 Find a definite integral that represents the length of the specified polar curve and then find the exact value of integral 

$r=e^{\theta } for 2k\pi \le \theta \le 2(k+1)\pi$

Step-by-Step Solution

Answer

The definite integral can be given as ${\int }_0^{2\pi }\sqrt{2}{e}^{\theta }d\theta$ and length of the polar curve is $\sqrt{2}\left[e^{2k\pi }\right(e^{2\pi }-1\left)\right]$

Step 1: Given information

We are given a polar curve $r=e^{\theta } for 2k\pi \le \theta \le 2(k+1)\pi$

Step 2: We find the definite integral and evaluate it

We know that length of polar curve can be given as 

${\int }_0^{2\pi }\sqrt{(f{\left(\theta \right))}^2+(f\text{'}{\left(\theta \right))}^2}d\theta$

We are given 

$$\begin{gathered} r=e^{\theta } \\ r\text{'}=e^{\theta } \end{gathered}$$

On substituting the values

$$\begin{gathered} {\int }_{2k\pi }^{2(k+1)\pi }\sqrt{e^{2\theta }+e^{2\theta }}d\theta \\ ={\int }_{2k\pi }^{2(k+1)\pi }\sqrt{2e^{2\theta }}d\theta \\ =\sqrt{2}{\int }_{2k\pi }^{2(k+1)\pi }\sqrt{e^{2\theta }}d\theta \\ =\sqrt{2}{\int }_{2k\pi }^{2(k+1)\pi }{e}^{\theta }d\theta \\ =\sqrt{2}{{\left[e^{\theta }\right]}^{2(k+1)\pi }}_{2k\pi }=\sqrt{2}[e^{2(k+1)\pi }-e^{2k\pi }] \\ =\sqrt{2}\left[e^{2k\pi }\right(e^{2\pi }-1\left)\right] \end{gathered}$$