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Homework Help: Green's Theorem (integration problem, not concept)

  1. Nov 19, 2006 #1
    So a previous problem said to show that the area of some simple closed curve C was:

    [tex]A= \frac{1}{2} \oint_{C} (xdy - ydx)[/tex]

    Simple enough. My problem says to find the area of the curve

    [tex]x^{2/3} + y^{2/3} = 4[/tex]

    using that formula.

    So off to polar coordinate land I go. r = 4, x = cos(t), dx = -sin(t), y = sin(t), dy = cos(t), right?

    So after plugging everything in and taking out the r, I get:

    [tex]A = 2 \oint_{0}^{2 \pi} (cos(t)^{5/3} + sin(t)^{5/3})dt[/tex]


    How do I integrate that? They aren't squared. And I can't find any trig formula to simplify that or any integration formula. Do I have to venture into infinite series?
    Last edited: Nov 19, 2006
  2. jcsd
  3. Nov 19, 2006 #2


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    I don't think you're parametrizing your curve correctly--one way to check is to substitute a point from your parametrized expression back into the original expression. If you let (x(t), y(t)) = (sin^3 t, cos^3 t), how might you tweak that?
  4. Nov 19, 2006 #3
    Now I'm completely lost. I've been doing this assignment and another lab since yesterday morning, so my mind is fried.

    I think I understand that r =/= 4 (although I am having a brain fart and can't figure out what it should be), but I still don't understand your last sentence.
  5. Nov 20, 2006 #4


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    You are trying to find a parametric representation of the curve [tex]x^{2/3} + y^{2/3} = 4[/tex]. That means you want a function c(t) = (x(t), y(t)) such that for all t, [tex]x(t)^{2/3} + y(t)^{2/3} = 4[/tex], and also you want every point (x, y) on the curve to be equal to (x(t), y(t)) for some t. Are you with me? I'm saying you could try choosing c(t) = (cos^3(t), sin^3(t)) and see how you can alter it so that it's right.
    Last edited: Nov 20, 2006
  6. Nov 21, 2006 #5
    Yeah, I ended up doing that but didn't know how to integrate sin^2(t)cos^2(t). I really hate that. I understand the concept but I don't know how to integrate. =/ It was like 1 - cos(4t) or something, and integrating that is a snap, but my trig sucks. :(
  7. Nov 21, 2006 #6
    If you wanted to simplify the curve eqn you could cube the whole equation to get x^2 + y^2 = 64.
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