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Homework Help: Divergence Theorem

  1. Aug 31, 2010 #1
    1. The problem statement, all variables and given/known data
    Check the divergence theorem using the function:

    [tex] \mathbf{v} = y^2\mathbf{\hat{x}} + (2xy + z^2) \mathbf{\hat{y}} + (2yz)\mathbf{\hat{z}} [/tex]

    2. Relevant equations

    [tex] \int_\script{v} (\mathbf{\nabla . v }) d\tau = \oint_\script{S} \mathbf{v} . d\mathbf{a} [/tex]

    3. The attempt at a solution

    taking the dot product it becomes

    [tex] \frac{\partial}{\partial x} y^2 \mathbf{\hat{x}} + \frac{\partial}{\partial y} ( 2xy + z^2) \mathbf{\hat{y}} + \frac{\partial}{\partial z} (2yz)\mathbf{\hat{z}} [/tex]

    so by only differentiating the certain parts:

    i get y^2 + 2x + z^2 + 2y,
    however the dot product of del and my vector v, should've been 2(x+y)
    how come I've got y^2 and z^2?

    does [tex] \frac{\partial}{\partial x} y^2 [/tex] not equal y^2???
  2. jcsd
  3. Aug 31, 2010 #2


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    Your divergence should not actually include any vectors. It's just a number. So you should have
    \frac{\partial}{\partial x} y^2 + \frac{\partial}{\partial y} ( 2xy + z^2) + \frac{\partial}{\partial z} (2yz)

    Once you take the dot product

    Then it looks like you differentiated your unit vectors as though they were variables

    [tex]\frac{\partial}{\partial x} y^2[/tex] is 0, not y2
  4. Aug 31, 2010 #3


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    this isn't correct, you applied a dot product, it should be
    [tex] \frac{\partial}{\partial x} y^2 + \frac{\partial}{\partial y} ( 2xy + z^2) + \frac{\partial}{\partial z} (2yz)[/tex]

    to see where you went wrong, what is
    [tex] \frac{\partial}{\partial y} z^2 [/tex]
  5. Aug 31, 2010 #4
    Is it 0? is that because z^2 is a constant since it's being differentiated with respect to y?
    and d/dx of a constant is 0?


    now knowing that [tex]\frac{\partial}{\partial y} z^2 = 0[/tex]
    I think I can solve the rest,
    \int_\script{v} (\mathbf{\nabla . v }) d\tau = \oint_\script{S} \mathbf{v} . d\mathbf{a}
    \frac{\partial}{\partial x} y^2 + \frac{\partial}{\partial y} ( 2xy + z^2) + \frac{\partial}{\partial z} (2yz)
    [tex] \oint_{\script{V}} 2(x+y) d\tau [/tex]

    taking the constant out and making tau dxdydz = [tex] 2 \int_0^1 \int_0^1 \int_0^1 (x+y) dxdydz [/tex]

    separating the three integrals out, (i think I can do that?)

    dx part: [tex] \int_0^1 (x+y)dx = \left[ \frac{x^2}{2}+y \right]_0^1 = (\frac{1}{2} + y)[/tex]

    dy part, using the result of the dx part:
    [tex] \int_0^1 (\frac{1}{2} + y) dy = \left[ \frac{1}{2} + \frac{y^2}{2} \right]_0^1 = (\frac{1}{2} + \frac{1}{2})[/tex]

    dz part using the result of the dy part:
    [tex] \int_0^1 (1)dz = \left[ 1*z\right]_0^1 = 1 [/tex]

    bringing in the constant 2, of what I brought outside the integral,

    [tex] \int_{\script{V}} \mathbf{\nabla . v} d\tau = 2[/tex]

    so that means the integral of a derivative over the vector is equal to 2,
    and to confirm the divergence theorem,

    \int_\script{v} (\mathbf{\nabla . v }) d\tau = \oint_\script{S} \mathbf{v} . d\mathbf{a}

    I also need to integrate over a closed surface for the cube?

    that means i need to solve for 6 integrals,
  6. Aug 31, 2010 #5


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    yes that is the definition of a partial derivative, the other variables are held constant
  7. Aug 31, 2010 #6


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    It does indeed require you to compute 6 double integrals over the surface of the unit cube.
  8. Aug 31, 2010 #7


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    not sure why you would separate it, anyway though you get the right answer (luck), the integral isn't correct. y is treated as a constant, what is teh integral of a constant?
  9. Aug 31, 2010 #8

    oh right, I should've separated and differentiated each bit individually,
    so \int x dx, |_0^1
    then added \int y dy |_0^1 to it
    and then multiplied by int dz,
  10. Aug 31, 2010 #9


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    no i meant
    [tex] \int_0^1 dx y = yx|_0^1 = y(1-0) = y[/tex]
  11. Aug 31, 2010 #10


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    just treat it as
    [tex] 2 \int_0^1 \left(\int_0^1 \left(\int_0^1 (x+y) dx\right)dy\right)dz [/tex]
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