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Homework Help: What the hell am i doing? lol

  1. Apr 12, 2005 #1
    find the flux of the vector
    [tex]\vec{A} = \frac{6ka^2y}{\sqrt{x^2+y^2+a^2}} \hat{e_x}[/tex]
    [tex]\frac{3ka^2z}{\sqrt{y^2+z^2+4a^2}}\hat{e_y}[/tex]

    [tex]\frac{2ka^2x}{\sqrt{x^2+z^2+9a^2}}\hat{e_z}[/tex]

    a) intergrating over the surface of the box, of
    0 <= x <= 2a
    0 <= y <= 3a
    0 <= z <= a
    b) divergence thrm over the volume of the box


    a)...I setup 6 flux intergrals and sum them for the total flux is the first step..
    each of these intergrals is setup like (going directly to the dot product result)...
    [tex]\oint \vec{A} \bullet d\vec{a}[/tex]

    [tex]\int_0^a \int_0^{3a} A_x dydz [/tex] however when i setup the opposite side of the box...
    [tex]\int_0^a \int_0^{3a} -A_x dy dz [/tex] because of the [tex]d\vec{a} = dydz(-\vec{\hat{e_x}})[/tex]

    which means they just cancel which i do for all of them...and get 0 flux which is not what it should be nothing loops back on itself, and the divergence gives me something totally diffrent so whats up?
     
    Last edited: Apr 12, 2005
  2. jcsd
  3. Apr 12, 2005 #2

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    The function has a different value at opposite sides of the box.
     
  4. Apr 12, 2005 #3
    soooo..? [tex]\int_0^a \int_0^{3a} A_x dydz[/tex]
    [tex]6ka^3\int_0^{3a} \frac{y}{\sqrt{x^2+y^2+a^2}} dy[/tex]
    [tex]6ka^3(\sqrt{10a^2+x^2}-\sqrt{a^2+x^2})[/tex]


    [tex]\int_0^a \int_0^{3a} -A_x dydz[/tex]
    [tex]-\int_0^a \int_0^{3a} A_x dydz[/tex] how is this not opposite what I just did?

    do i have to setup the first intergral like...
    [tex]x=2a[/tex] at this side of the surface
    [tex]6ka^3\int_0^{3a} \frac{y}{\sqrt{(2a)^2+y^2+a^2}} dy[/tex]
     
    Last edited: Apr 12, 2005
  5. Apr 12, 2005 #4

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    You're at two different values of x. The final answer should only be in terms of a.
     
  6. Apr 12, 2005 #5
    check my edit is that what u mean?
     
  7. Apr 12, 2005 #6

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    yes, when your integrating at x=2a, you need to plug in 2a for x.
     
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