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Showing that a three-dimensional vector field is conservative

  1. Apr 23, 2006 #1
    Alright, so the field is [tex]\mathbf{F} = (z^2 + 2xy,x^2,2xz)[/tex]
    it's a gradient only when [tex]f_x = z^2 + 2xy[/tex], [tex]f_y = x^2[/tex] and [tex]f_z = 2xz[/tex]
    integrate the first equation with respect to x to get [tex]f(x,y,z) = \int z^2 +2xy\,dx = xz^2 + x^2y + g(y,z) [/tex]
    now, [tex]f_z(x,y,z) = g_z(y,x)[/tex] which is 2xz
    integrate that with respect to z, [tex]g(y,z) = \int 2xz\,dz = xz^2 + h(y)[/tex]
    plug that into our previous expression [tex]f(x,y,z) = xz^2 + x^2y + g(y,z) = 2xz^2 + x^2y + h(y)[/tex]
    derive that with respect to y for [tex]f_y(x,y,z) = x^2 + h'(y) = x^2[/tex]
    so [tex]h'(y) = 0[/tex] and [tex]h(y) = c[/tex] and we can set [tex]c = 0[/tex] and now the potential function is [tex]f(x,y,z) = 2xz^2 + x^2y[/tex] which is wrong. It should be [tex]f(x,y,z) = xz^2 + x^2y[/tex]. Help me please. Sorry if my LaTeX is wonky.
     
    Last edited: Apr 23, 2006
  2. jcsd
  3. Apr 23, 2006 #2

    nrqed

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    no...the second term is x^2y
     
  4. Apr 23, 2006 #3

    nrqed

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    No, your derivative is wrong. should be z^2 + h'
     
  5. Apr 23, 2006 #4
    Hold on, I'm trying to fix what I typed, as it doesn't match my paper here... alright, fixing the first thing fixed the second thing, so now this reflects how I did the problem.
     
    Last edited: Apr 23, 2006
  6. Apr 23, 2006 #5

    nrqed

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    no. Look at your f(x,y,z). f_z is 2xz + g_z
     
  7. Apr 23, 2006 #6
    So then what do you do from there? f_z = 2xz + g_z and we know that f_z = 2xz so that g_z = 0 and g = h(y), is that right? That would give the right answer.
     
  8. Apr 23, 2006 #7

    nrqed

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    yes..
    You are done. the derivative with respect to y of f must give x^2. But now you know that f= xz^2 + x^2 y + h(y) so this shows that h is a constant.
    You are done.
     
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