Finding Area Of Polar Function

  • Thread starter Lancelot59
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  • #1
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Well this problem started off simply enough. I was given this function:
[tex]r=2cos(3\theta)[/tex]
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:
[tex]\frac{1}{2}\int_{0}^{2\pi}{(2cos(3\theta))^{2} d\theta}[/tex]

Got this:[tex]2[\theta+sin(6\theta)][/tex] 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

  • #2
13
0
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.
 
  • #3
634
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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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  • #4
35,013
6,765
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.
 
  • #5
634
1
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.
 
  • #6
35,013
6,765
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.
 
  • #7
634
1
I'm confused here. Let me start over with the sketching.
 

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