Q. 36 - Calculus

Question

in exercise 31-36 find a definite integral that represents the length of the specified polar curve, and then use graphing calculator or computer algebra system to approximate the value of integral 

The limacon $r=2+\cos \theta$

Step-by-Step Solution

Answer

The integral can be given as ${\int }_0^{2\pi }\sqrt{{(-\sin \theta )}^2+{(2+\cos \theta )}^2}d\theta$ and length can be given as 13.36 units.

Step 1: Given information

We are given a limacon $r=2+\cos \theta$

Step 2: Evaluate

We know that arc length can be given as 

$$\begin{gathered} L={\int }_a^b\sqrt{{(r\text{'})}^2+{\left(r\right)}^2}dr \\ Substituting the values we get \\ L={\int }_0^{2\pi }\sqrt{{(-\sin \theta )}^2+{(2+\cos \theta )}^2}d\theta \\ We first simplify the given expression \\ L={\int }_0^{2\pi }\sqrt{{\sin }^2\theta +4+4\cos \theta +{\cos }^2\theta } \\ L={\int }_0^{2\pi }\sqrt{5+4\cos \theta } \\ U\sin g CAS calculator, We evaluate the integral \\ L=13.36units \end{gathered}$$