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$r=2+\cos \theta$

Short 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-by-step solution

Given information

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

Evaluate

We know that arc length can be given as

$$\begin{gathered} L={\int }_a^b\sqrt{{(r\text{'})}^2+{\left(r\right)}^2}dr \\ Substitutingthevaluesweget \\ L={\int }_0^{2\pi }\sqrt{{(-\sin \theta )}^2+{(2+\cos \theta )}^2}d\theta \\ Wefirstsimplifythegivenexpression \\ L={\int }_0^{2\pi }\sqrt{{\sin }^2\theta +4+4\cos \theta +{\cos }^2\theta } \\ L={\int }_0^{2\pi }\sqrt{5+4\cos \theta } \\ U\sin gCAScalculator,Weevaluatetheintegral \\ L=13.36units \end{gathered}$$