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Partial integration of the function f = 1

  1. Jan 18, 2012 #1
    Given a function z(x,y), is the following expression valid:

    ∫(1)∂z = z

    Does this even make sense? I came up with this in my differential equations class, but I'm not sure if it actually means anything.
     
  2. jcsd
  3. Jan 19, 2012 #2

    HallsofIvy

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    Correct except that you do not have the "constant of integration". Which, because the partial derivative with respect to z treats other variables as constants, may be a function of those other variables. That is, if your variables are x, y, and z, the "anti-derivative" of 1, with respect to z, is [itex]z+ f(x,y)[/itex] where f can be any function of two variables.
     
  4. Jan 19, 2012 #3
    I think that makes sense. Thanks!
     
  5. Jan 20, 2012 #4

    HallsofIvy

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    Note that if you have the three partial differential equations
    [tex]\frac{\partial f}{\partial x}= 1[/tex]
    [tex]\frac{\partial f}{\partial y}= 1[/tex]
    [tex]\frac{\partial f}{\partial z}= 1[/tex]

    The last equation gives, as I said, f(x,y,z)= z+ g(x,y). Differentiating that with respect to y,
    [tex]\frac{\partial f}{\partial y}= \frac{\partial g}{\partial y}= 1[/tex]
    so that g(x,y)= y+ h(x). Differentiating that with respect to x,
    [tex]\frac{\partial g}{\partial x}= \frac{dh}{dx}= 1[/tex]
    which gives h(x)= x+ C. Putting all of those together, f(x,y,z)= z+ g(x,y)= z+ y+ h(x)a= z+ y+ x+ C.
     
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