# Finding Area Of Polar Function

Well this problem started off simply enough. I was given this function:
$$r=2cos(3\theta)$$
And I had to find the area bound by it. I sketched it out from zero to 2pi and got this:
[PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP58019c987a3dd6i59cg00000i1ghg0cgb4b0a00?MSPStoreType=image/gif&s=6&w=364&h=302 [Broken]
So I did that correctly. I then proceeded to integrate the function like so:
$$\frac{1}{2}\int_{0}^{2\pi}{(2cos(3\theta))^{2} d\theta}$$

Got this:$$2[\theta+sin(6\theta)]$$ from 0 to 2pi. My final answer was 2pi...but the correct answer is pi. Where did I go wrong. From looking at my notes I followed the right steps. I think I chose the limits incorrectly, however they make sense to me because the function runs from 0 to 2pi.

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## Answers and Replies

You just forgot a factor of 1/2 outside the integral. The integral for polar area is I = 1/2 integral of r^2 dtheta.

Actually there was. I forgot to put it in here. The one half became a two after I moved the square in and pulled out the resulting constant of 4. I fed the same definite integral into WolframAlpha and it gave me the same thing (I had to move the 1/2 inside).[URL]http://www2.wolframalpha.com/Calculate/MSP/MSP181619c98e1e6ehhfe63000049e3g64375g56ae7?MSPStoreType=image/gif&s=16&w=269&h=37[/URL]

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Mark44
Mentor
The polar curve traces out all three petals as theta ranges from 0 to pi. You integrated from 0 to 2pi, so you went completely around twice, thus got twice the area. Change the limits of integration and you'll get the right value.

The polar curve traces out all three petals as theta ranges from 0 to pi. You integrated from 0 to 2pi, so you went completely around twice, thus got twice the area. Change the limits of integration and you'll get the right value.

Then why did I have to go from zero to two pi to get back to where I started from while I was sketching it? I'll go back and look at that. Thanks for the help.

Mark44
Mentor
Then why did I have to go from zero to two pi to get back to where I started from while I was sketching it? I'll go back and look at that. Thanks for the help.
Going from 0 to pi takes you around once; 0 to 2pi takes you all the way around twice.

IIRC, the figure is called a three-petaled rose. Here are some of the points on the rose.
(2, 0) - Start
(0, pi/6)
(-2, pi/3)
(0, pi/2)
(2, 2pi/3)
(0, pi)
(-2, pi) - back to starting point.

I'm confused here. Let me start over with the sketching.