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Polar Double Integral Between 2 Regions

  1. Mar 27, 2016 #1
    1. The problem statement, all variables and given/known data

    Find the area in the first quadrant that is inside the circle ##r=100sin(\theta)## and outside the leminscate ##r^2=200cos(2\theta)##.

    I have graphed the region as I interpreted it below. The area I am trying to find is the non-shaded, white region.

    graph.jpg

    2. Relevant equations

    ##\int_{\alpha}^{\beta} \int_{r_1(\theta)}^{r_2(\theta)} f(r, \theta)rdrd\theta##

    3. The attempt at a solution

    1. Determining the limits of the integrals:

    Since the area we are solving for is in the first quadrant, and spans the whole first quadrant...

    ## \alpha = 0## and ##\beta = \frac {\pi} {2}##

    As the leminscate is the first polar region we interact...

    ##r_1(\theta) = \sqrt{200cos(2\theta)}## and ##r_2(\theta) = 100sin(\theta)##

    2. Setting up & Evaluating the Integral:

    ##\int_{0}^{\frac {\pi} {2}} \int_{\sqrt{200cos(2\theta)}}^{100sin(\theta)} r dr d\theta =##

    ##= \int_{0}^{\frac {\pi} {2}} [ \frac 1 2 r^2 |_{\sqrt{200cos(2\theta)}}^{100sin(\theta)}]d\theta##

    ##= \int_{0}^{\frac {\pi} {2}} [5000sin(\theta) - 100cos(2\theta)] d\theta##

    ##= 5000 \int_{0}^{\frac {\pi} {2}} sin(\theta)d\theta - 100 \int_{0}^{\frac {\pi} {2}} cos(2\theta)d\theta ##

    ##= 5000 [-cos(\theta) |_{0}^{\frac {\pi} {2}}] - 100 [\frac {sin(2\theta)} {2} |_{0}^{\frac {\pi} {2}}] ##

    ##= 5000 [0+1] - 100 [0-0]##

    ##\int_{0}^{\frac {\pi} {2}} \int_{\sqrt{200cos(2\theta)}}^{100sin(\theta)} r dr d\theta = 5000##

    Apparently, this is an incorrect answer... I think I am having more of a conceptual problem with the regions than anything else.
     
  2. jcsd
  3. Mar 27, 2016 #2

    andrewkirk

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    The inside of the square root gives negative answers for part of your ##\theta## range. What part?
    So split your outer integral into two. One for the part where the inside of the square root is positive, and another where it isn't.
    What are the limits of ##\theta## for that split?

    What should your lower bound for the inner integral be in your second inner integral (Hint: look at the diagram)?
     
  4. Mar 27, 2016 #3
    I don't understand, where, and why is my square root giving negative values? And how am I splitting this up? I don't see where I went wrong.
     
  5. Mar 27, 2016 #4

    Nathanael

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    Look at the leminscate ##r=\pm \sqrt{200\cos(2\theta)}## ... What happens when cos(2θ) becomes negative? When does this occur?

    We are interested in the portion of the leminscate in the first quadrant; does that correspond with the the θ interval [0, π/2] ?
     
  6. Mar 27, 2016 #5

    Ray Vickson

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    Look again at the diagram. For some values of ##\theta \in (0, \pi/2)## the ##r##-integration runs from some ##r_1(\theta)## to ##r_2(\theta)##, but for other values of ##\theta## it runs from ##r=0## to ##r = r_2(\theta)##.
     
  7. Mar 29, 2016 #6
    I apologize for the late response, but I'm sorry I still don't understand what you mean by when it goes negative? I've tried conceptualizing it.

    The solution for this problem was also posted by my teacher, but I am still trying to grasp it.

    upload_2016-3-29_19-16-59.png

    In the first integral, I can understand where the ##\frac {\pi} {4} ## is coming from, as it's the upper bound of the leminscate, but I plotted the image, and the ##\frac {\pi} {6}## seems to correspond to nothing on the region, it seems arbitrary.

    I also don't understand why the second integral changes its bounds to (pi/6, pi/2) all of a sudden... I'm just not understanding these bounds at all.
     

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  8. Mar 29, 2016 #7

    SteamKing

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    Well, you are interested in only one part of the lemniscate which is located inside the white circle.

    If you were to plot the equation for the lemniscate, starting at θ = 0, you would have a point located at approximately (14.14,0) on the positive x-axis. However, this point is located outside the region enclosed by the white circle.

    What you want to do is calculate the angle at which the lemniscate and the circle intersect after θ = 0 and before θ = π/4; in other words, solve ##100 sin (θ) = \sqrt{200 ⋅ cos (2θ)}## for θ. This value of θ becomes the lower bound of your integration.
     
  9. Mar 29, 2016 #8
    Okay, thank you, it completely slipped my mind that the theta still traversed the leminscate even after pi/4. My only issue now is solving this equation, which although it is algebra, I'm not have a very good time with it.
     
  10. Mar 29, 2016 #9

    SteamKing

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    To make things a little easier, there's a trig identity which you can use on cos (2θ) to convert it to an expression in sin (θ).
     
  11. Mar 29, 2016 #10
    ##100sin(\theta) = \sqrt{200cos(2\theta)}##
    ##50sin^2(\theta) - cos(2\theta) = 0##
    ##50sin^2(\theta) - (1 - 2sin^2(\theta))=0##
    ##50sin^2(\theta) - 1 + 2sin^2(\theta))=0##
    ##52sin^2(\theta) - 1 =0##
    ##sin(\theta) = \frac {1} {\sqrt{52}} ##
    ##\theta = sin^-1(\frac {1} {\sqrt{52}}) ##

    Not really sure what to make of this. How do I get an exact value for the bounds?
     
  12. Mar 29, 2016 #11

    SteamKing

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    You can't, because there isn't one, except as shown in your calculations.

    You can check your solution by substituting this angle into the equations for the lemniscate and the circle and obtaining the coordinates of the point of intersection.
     
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