Q. 30 - 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 cardoid $r=2-2\sin 2\theta for 0\le \theta \le 2\pi$

Step-by-Step Solution

Answer

The integral can be given as $2\sqrt{2}{\int }_0^{2\pi }\sqrt{1-\sin \theta }d\theta$ and length of the polar curve can be given as $8\sqrt{2}units$

Step 1: Given information

We are given an equation of cardioid

$r=2-2\sin 2\theta for 0\le \theta \le 2\pi$

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

$$\begin{gathered} r=2-2\sin \theta \\ r\text{'}=-2\cos \theta \end{gathered}$$

Substituting the values we get,

$$\begin{gathered} \mathrm{I}={\int }_0^{2\pi }\sqrt{{(2-2\sin \theta )}^2+{\left(2\cos \theta \right)}^2}d\theta \\ ={\int }_0^{2\pi }\sqrt{4-8\sin \theta +4{\sin }^2\theta +4{\cos }^2\theta }d\theta \\ ={\int }_0^{2\pi }\sqrt{8-8\sin \theta }d\theta \\ =2\sqrt{2}{\int }_0^{2\pi }\sqrt{1-\sin \theta }d\theta \\ =2\sqrt{2}{\int }_0^{2\pi }(-\cos \frac{\theta }{2}+\sin \frac{\theta }{2})d\theta \\ =4\sqrt{2}{{[-\sin \frac{\theta }{2}-\cos \frac{\theta }{2}]}^{2\pi }}_0=4\sqrt{2}\left(2\right) \\ =8\sqrt{2}units \end{gathered}$$