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

one petal of the polar rose$r=\cos 3\theta$

Short answer

The integral can be given as$L=2{\int }_0^{\frac{\pi }{6}}\sqrt{{\left(\cos 3\theta \right)}^2+9{(-\sin 3\theta )}^2}d\theta$ and the arc length is 2.22 units

Step-by-step solution

Given information

We are given a petal of the polar rose$r=\cos 3\theta$

Evaluate

We know that arc length of one petal of polar rose

can be given as

$L={\int }_0^{\frac{\pi }{2n}}\sqrt{{\left(\cos n\theta \right)}^2+{\left(n^2\sin n\theta \right)}^2}d\theta$

Substituting n=3 in the above equation we get

$L=2{\int }_0^{\frac{\pi }{6}}\sqrt{{\cos }^{2}3\theta +9{\sin }^{2}3\theta }d\theta L$

Now on using CAS calculator we get

$$\begin{gathered} L=2(1.1) \\ L=2.2unit \end{gathered}$$