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Area enclosed by one petal of the rose ?

  • Thread starter hamburgler
  • Start date
22
0
I would appreciate any help on this problem because I am completely confused. :confused:

1. The problem statement...

What is the area enclosed by one petal of the rose r=6cos3[tex]\theta[/tex]?
(w/o a calc)


2. The sorry attempt...

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

and the integral is from [tex]\pi[/tex] to [tex]\pi[/tex]/3

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

So then [tex]\int[/tex](6cos3[tex]\theta[/tex])^2d[tex]\theta[/tex]

and then I get stuck at the integral of the function...
 
Last edited:

Answers and Replies

334
0
You can use the so-called "half angle" formulas to find the integral: [itex]cos(\frac{\theta}{2})= \sqrt{\frac{cos(\theta)+1}{2}}[/itex]. 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 [itex]\theta > 0[/itex] where [itex]cos(3\theta)=0[/itex]
 
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