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$r=\theta for0\le \theta \le 2\pi$

Short answer

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

Step-by-step solution

Given information

We are given a spiral represented by a function$r=\theta for0\le \theta \le 2\pi$

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

role="math" localid="1649471467926" ${\int }_0^{2\pi }\sqrt{1+{\theta }^2}d\theta =21.25units$