Q. 27 - 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

the spiral $r=e^{\theta } for 0\le \theta \le 2\pi$

Step-by-Step Solution

Answer

The integral can be given as $\sqrt{2}{\int }_0^{2\pi }{e}^{\theta }d\theta$ and the length of the polar curve can be given as $40.36units$

Step 1: Given information

We are given a spiral with $r=e^{\theta } for 0\le \theta \le 2\pi$

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 we get,

$$\begin{gathered} {\int }_0^{2\pi }\sqrt{e^{2\theta }+e^{2\theta }}d\theta \\ ={\int }_0^{2\pi }\sqrt{2e^{2\theta }}d\theta \\ =\sqrt{2}{\int }_0^{2\pi }\sqrt{e^{2\theta }}d\theta \\ =\sqrt{2}{\int }_0^{2\pi }{e}^{\theta }d\theta \\ =\sqrt{2}\left(e^{2\pi }-1\right) \end{gathered}$$