Q. 26 - 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 $r=\theta for 0\le \theta \le 2\pi$

Step-by-Step Solution

Answer

The integral can be given as ${\int }_0^{2\pi }\sqrt{1+{\theta }^2}d\theta$ and its length is $21.25units$

Step 1: Given information

We are given a spiral represented by a function $r=\theta for 0\le \theta \le 2\pi$

Step 2: We find the definite integral and evaluate it

We know that length of polar curve can be given as 

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

We are given $$\begin{gathered} r=\theta \\ \frac{dr}{d\theta }=1 \end{gathered}$$

Substituting the above values in the integral we get,

${\int }_0^{2\pi }\sqrt{1+{\theta }^2}d\theta$

On solving the integral we get the value as

${\int }_0^{2\pi }\sqrt{1+{\theta }^2}d\theta =21.25units$