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Green's theorem: region inside 8 leaved petal

  1. Jul 31, 2009 #1
    1. The problem statement, all variables and given/known data
    Use Green's theorem to compute the area of one petal of the 8-leafed rose defined by r=9sin(4theta)
    It may be useful for recall that the area of a region D enclosed by a curve C can be expressed as A =(1/2)int xdy-ydx.

    2. Relevant equations
    A =(1/2)int xdy-ydx


    3. The attempt at a solution
    I graphed the region on my calculator, but I am not sure what to use for x and y. Would the best way be to change r into rectangular coordinates? If so what would my limit be? 0 to pi/4? perhaps?

    Thanks in advance!
     
  2. jcsd
  3. Aug 1, 2009 #2
    EDIT: I'm stupid, lol.
     
  4. Aug 1, 2009 #3
    Okay, so you're right with the 0 to pi/4. Now you just need to parameterize the curve. Think about finding the x and y coordinates of each of the r-vectors... Then you just integrate x(t)y'(t)dt and y(t)x'(t)dt. Hope that helps.
     
  5. Aug 1, 2009 #4
    Ok, so using the fact that x=rcostheta and y=rsintheta, would the parameter of x=9sin(4theta)cos(theta) and y=9sin(4theta)sin(theta) ?
     
  6. Aug 1, 2009 #5
    That's correct. Now you just have to figure out your limits...
     
  7. Aug 1, 2009 #6
    So limits are 0 to pi/4 correct? When I write out the integral, its huge. Is there any trick to combining the terms together neatly?
     
  8. Aug 1, 2009 #7
    Haha, yeah it's a pretty big integral. I guess this is why you wouldn't normally use Green's Theorem to find the area of a rose petal. You'd just use the polar formula...
     
  9. Aug 1, 2009 #8
    This should be your integral:

    [tex]\frac{1}{2}(\int_{0}^{\pi/4}[{9sin(4t)cos(t)[9sin(4t)cos(t)+36cos(4t)sin(t)]]dt}-\int_{0}^{\pi/4}[{9sin(4t)sin(t)[-9sin(4t)sin(t)+36cos(4t)cos(t)]]dt})[/tex]

    EDIT: Realistically, most people would use Maple, Mathematica, or a calculator to evaluate that monster. It doesn't look friendly even if there's something that can be done to shrink it down.
     
  10. Aug 1, 2009 #9
    Awesome got it! Thanks for all the help!!!
     
  11. Nov 17, 2009 #10
    FTW:

    This actually SIMPLIFIES to something much easier if you go through the distribution.
     
  12. Nov 17, 2009 #11
    Yeah I realized that after a few minutes but I didn't edit my post.

    Also, lol @ the response 3 months later... :-)
     
  13. Nov 18, 2009 #12
    yea i know but i just spent like an hour going through this and i felt like i needed to share it with future WEBWORK kids.
     
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