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Integral on end points

  1. Jan 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that the following line integral depends only on the end points of C by using a suitable theorem.

    Then evaluate without parameterising it.

    2. Relevant equations

    [itex]\int_{(0,0)}^{(3,2)} 2xe^y dx + x^2e^y dy[/itex]

    3. The attempt at a solution

    Is this the theorem...

    [itex] \displaystyle \int F(x,y) dr =\int_C \frac{\partial \phi}{\partial x} dx +\frac{\partial \phi}{\partial y} dy =\int_a^b (\frac{\partial \phi}{\partial x} \frac{dx}{dt} +\frac{\partial \phi}{\partial y} \frac{dy}{dt}) dt[/itex]...........?
     
  2. jcsd
  3. Jan 24, 2012 #2

    lanedance

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    do you know about conservative functions and vector fields ?
     
  4. Jan 24, 2012 #3
    Yes, this force vector field is conservative because g_x=f_y, hence is independent of path C..but I dont know what theorem to use?
     
  5. Jan 24, 2012 #4

    lanedance

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    how about the gradient theorem
     
  6. Jan 24, 2012 #5
    It must be simpler than I thought

    I calculate the potential function [itex] \phi (x,y)= x^2 e^y +C[/itex]. we know the gradient of the potential function is the force vector field given in the question. I integrated backwards to get the potential, s I guess thats the theorem..right?

    I evaluate the force field to be 9e^2....based on [itex]\displaystyle \int \nabla \phi \dot dr = \phi(x_1, y_1) - \phi (x_0, y_0)[/itex]...
     
  7. Jan 24, 2012 #6

    lanedance

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    looks good
     
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