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Limits of integration for regions between polar curves

  1. Apr 8, 2012 #1
    Alright. I completely confused about determining the area between regions of polar curves. However, I do feel that I have a solid grasp in finding areas for single functions. For a given function in polar form, I know that I find the limits of integration by setting the function equal to zero and solving for those theta values. This is the area "swept out". When presented with two functions, I understand that I must set them equal to find points of intersection. I do not know how to apply these points of intersection. For two given curves, do I take the difference of the integrals from those two shared points?

    Here is an example:
    The region inside the rose r = 4sin2θ and r = 2.

    Equation = 1/2 ∫α→β (f(θ)^2 - g(θ)^2)dθ

    I set these two functions equal.

    sin2θ = 1/2

    2θ = pi/6 , 5pi/6...

    θ = pi/12 , 5pi/12...

    When I look at the graph of this function, I see that I want the areas inside the circle r = 2 that belong to the other function. I should calculate the area of one section and multiply it by 4 to find the total area. I do not see how I can compute the partial area for this by using those two points of intersection. There seems to be one point I am missing or something and I cannot seem to figure out how to find it. Am I missing a theta value for which sin = 1/2 ?Even if I had two points, I do not understand how the I would take difference of those two areas to find this. If I took the difference in area of r = 4sin2θ and r = 2 , I would have the area outside of the circle. If I did the opposite, I would have negative area.

    Thanks for reading this and helping.
  2. jcsd
  3. Apr 8, 2012 #2


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    Hey velouria131 and welcome to the forums.

    For this kind of problem you do the same kind of thing for enclosed areas with functions integrated to dx: you find the x-values that enclose the areas and they become your limits.

    In this case, what you do is instead of finding the x-values you want to find your theta values which you have already solved.

    The caveat though is that you need to find when the rose petal function crosses the circle function r = 2 (i.e. goes inside of the circle). When this happens you have to ignore this region. You will probably have to draw a graph to know when this happens because unlike the case when you are doing something with respect to dx you are going to have multiple regions to integrate over (in the dx case you only get one interval to integrate over, but in this case you will multiple).

    Have you had experience at graphing things in something like Maple or MATLAB (or the free version Octave)?
  4. Apr 8, 2012 #3
    Thanks for the reply and welcoming! I still have a few questions about this nonetheless. I do not understand what you mean be "ignore the region". In the particular example I provided, I could not find the limits of integration because setting the two equations equal resolved two points. Only one of these points is a point on any given leaf. If I did find these two points of intersection, how would I set up the integral? For the area of one leaf portion within the circle r = 2, would I just integrate r = 2 over that region where the points of intersection are the limits of integration? Is the point I am missing the point where zero = 4sin2θ? Also, I have not used MATLAB. Thanks again!
  5. Apr 8, 2012 #4


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    If you found the region of integration where the formula of the leaf was given in terms of r(theta) (the radius is a function of theta given by the function r(theta) = r in terms of theta) and the leaf function was 'outside' r = 2 (outside of the circle with r = 2) then your area would be calculated by:

    Integral from [θ1 to θ2] (r(θ) - 2)dr = Area enclosed in this region.

    When I mean ignore the region, I am talking about the parts of your polar curve where the leaf or petal curve (whatever you want to call it) falls below the radius of 2. In other words when |r(θ)| < 2 because in these regions if you want areas only when |r(θ)| < 2 you ignore them but if you want to consider these you will have to reverse the integration from (r(θ) - 2)dr to (2 - (r(θ))dr because if you don't you will get negative answers which will screw up completely your final answer.
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