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Homework Help: Green's Theorem

  1. May 7, 2014 #1
    1. The problem statement, all variables and given/known data


    I know this might be trivial but,
    can you please tell me what I am missing?
    Here is my problem :

    2. Relevant equations
    Given F = (yi+x3j).dr
    Q = x3
    P = y
    => ∂Q/∂x = 3x2 and ∂P/dy = 1
    c (3x2 -1).dr = ∫2Pi010 (3r2cosθ2-1) rdrdθ
    2Pi010 (3r3cosθ2-r) drdθ
    Is it true so far ?

    Last edited: May 7, 2014
  2. jcsd
  3. May 7, 2014 #2

    What you've done is right. Go ahead with the problem.
  4. May 7, 2014 #3
    2Pi010 (3r3cosθ2-r) drdθ
    ->∫2Pi0(3/4cos2θ-1/2) dθ
    ->3/4∫2Pi0(cos2θ-1/2) dθ

    Is this now correct?
    If it is correct what about part b)
    it looks to me like 1/4 circle when it is plotted but here I have y = x^2 and no radius to start the same thing again and I am not sure about the 1st integral neither, becuase I think it cannot be from 0 to Pi/2 because it doesnt start from x= 0 but from y = 0
    Last edited: May 7, 2014
  5. May 7, 2014 #4
    Yup the answer is right.

    Stick to cartesian coordinates on this one. What are the limits on y (in terms of x) and x?
  6. May 7, 2014 #5
    Can we take that r = 1 because the line from (0,1) to (1,1) is the radius if we take it as 1/4 of a circle and this is the bottom right part?
    A bit confused by the limits though. I think x is from 0 to [itex]\sqrt{}y[/itex]
    and y is from 0 to 1?
  7. May 7, 2014 #6


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    Science Advisor

    No, there is no "quarter circle". To integrate over the region, take x going from 0 to 1 and, for each x, y going from [itex]x^2[/itex] to x.

    (You could do in the other order, taking y from 0 to 1 and, for each y, x from y to [itex]\sqrt{y}[/itex].)
  8. May 7, 2014 #7
    Isn't it supposed to be y from 0 to x2 and and x from 0 to 1 ?

    Alternatively, x from √y to 1 and y from 0 to 1.
  9. May 7, 2014 #8
    With these boundaries I found it to be 4/15 which is the answer indeed
    Thanks very much !
    Can I ask you one more thing, when if ever we can refer to a circle to solve these kind of questions, or I should look for x^2 + y^2 = r , in order to refer to a circle?
  10. May 7, 2014 #9
    Yes that's right. Only if your boundary is a circle can you refer to a circle. You must use the equation of the boundary given. Here, for example, the boundaries were y=x2, y=0 and x=1.
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