Area enclosed by one petal of the rose ?

[hamburgler]

I would appreciate any help on this problem because I am completely confused.

1. The problem statement...

What is the area enclosed by one petal of the rose r=6cos3\theta?
(w/o a calc)2. The sorry attempt...

(1/2)\int(6cos3\theta)^2d\theta

and the integral is from \pi to \pi/3

I'm pretty sure that's how you set it up, but I saw other people doing it another way.

So then \int(6cos3\theta)^2d\theta

and then I get stuck at the integral of the function...

 

[jhicks]

You can use the so-called "half angle" formulas to find the integral: cos(\frac{\theta}{2})= \sqrt{\frac{cos(\theta)+1}{2}}. Just from what I figure the graph looks like, your integration limits are wrong also. The first petal will be bounded by the first two times for \theta > 0 where cos(3\theta)=0