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Greens theorem-maximises area

  1. May 5, 2012 #1
    1. The problem statement, all variables and given/known data
    find a simple closed curve with counter-clockwise orientation that maximises the value of [itex]\oint (cos x^2 + 4yx^2) dx + (4x-xy^2)dy[/itex]


    2. Relevant equations



    3. The attempt at a solution

    from greens theorem [itex]\oint Pdx+qdy = \int\int \frac{\partial q}{\partial x} - \frac{\partial p}{\partial y} dA[/itex]

    so[itex]\int\int_{D} 4-y^2-4x^2 dxdy[/itex]

    the integrand is positive everywher inside the elipse,zero on the ellipse and negative outside. so to maximise the integral C should be the oriented boundary of the ellipse oriented in the counter-clockwise direction.

    x=2cosr y=sinr dx=rdr

    sooo i think the maximum value is [itex]\int^{2\pi}_{0} \int^{2}_{0} 4 - sin^2 r - 16 cos ^2 r drd\theta[/itex]

    which i the simplify.
    anyone know if i'm on the right lines here...
     
  2. jcsd
  3. May 5, 2012 #2

    tiny-tim

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    hi gtfitzpatrick! :smile:
    yeees (except you've missed out an "r", and the limits are wrong) …

    but the question only asks for the curve :wink:
     
  4. May 5, 2012 #3
    Hi tiny-tim,
    thanks for that, i see the missed r thanks. which of my limits are wrong, should my limit be 0 to 4?
     
  5. May 5, 2012 #4

    tiny-tim

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    it's an ellipse

    shouldn't r depend on θ ? :wink:
     
  6. May 5, 2012 #5
    ah yes i see, i had set my limits as if it was a circle. but as you say it is an ellipse. I know the question doesnt ask for it but, i'll have to think about it! cant see straight off what to set my limits to..
     
  7. Jul 23, 2012 #6
    just reading over this again. So the question only asks for the curve so [itex]\int\int_{D} 4-y^2-4x^2 dxdy[/itex] so the curve C = to the ellipse [itex]4-y^2-4x^2 [/itex] this is the curve that maximises the area right but how to i explain it better than that?
     
  8. Jul 23, 2012 #7

    tiny-tim

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    if you want to find the region D over which ∫∫D f(x,y) dxdy is maximised,

    then if you choose D to be the region in which f(x,y) ≥ 0, then clearly either enlarging or dikminishing D will decrease the integral :wink:
     
  9. Jul 23, 2012 #8

    LCKurtz

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    But that isn't the equation of the ellipse because there is no = sign.
     
  10. Aug 20, 2012 #9
    getting back to this...
    So my equation for an ellipse is[itex]\frac{x^2}{1^2} + \frac{y^2}{2^2} = 1[/itex] so parametizing this I get x=cos[itex]\theta[/itex] and y=2sin[itex]\theta[/itex] and from green the area is [itex]\frac{1}{2}\int^{2\pi}_{0} cos\theta(2cos\theta)-sin\theta(-2sin\theta d\theta[/itex] = [itex]\frac{1}{2}\int^{2\pi}_{0} 2(1) d\theta [/itex] = [itex]2\pi[/itex]
     
    Last edited: Aug 20, 2012
  11. Aug 20, 2012 #10

    tiny-tim

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    hi gtfitzpatrick! :smile:

    (ah! you corrected the "2" ! :biggrin:)

    to check: put z = 2x, and then halve the area of z2 + y2 = 4 :wink:
     
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