Finding Area Of Polar Function

In summary: 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.
  • #1
Lancelot59
646
1
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
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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  • #2
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
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
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
Mark44 said:
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
Lancelot59 said:
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
I'm confused here. Let me start over with the sketching.
 

FAQ: Finding Area Of Polar Function

1. How do you find the area of a polar function?

To find the area of a polar function, you can use the formula A = 1/2 ∫(r^2) dθ, where r is the radius and θ is the angle. This formula is derived from the concept of finding the area under a curve using integration.

2. What is the difference between finding area in rectangular coordinates and polar coordinates?

In rectangular coordinates, the area is calculated by multiplying the length and width of a rectangle. In polar coordinates, the area is calculated by integrating the function r^2 with respect to θ, which takes into account the distance from the origin and the angle.

3. Can you use the same method to find the area of all polar functions?

Yes, the formula A = 1/2 ∫(r^2) dθ can be used to find the area of any polar function, as long as the function is continuous and bounded.

4. How can you tell if a polar function is bounded?

A polar function is bounded if the maximum and minimum values of r are finite. This can be determined by graphing the function or by setting r equal to 0 and solving for θ.

5. Are there any special cases to consider when finding the area of a polar function?

Yes, there are a few special cases to consider. If the polar function has a loop or self-intersecting curve, the area must be calculated separately for each loop. Additionally, if the polar function has negative values, the absolute value of the function must be used in the integration.

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