1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    StatusX

    User Avatar
    Homework Helper

    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

    StatusX

    User Avatar
    Homework Helper

    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

    StatusX

    User Avatar
    Homework Helper

    yes, when your integrating at x=2a, you need to plug in 2a for x.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: What the hell am i doing? lol
  1. What am I doing wrong? (Replies: 15)

  2. What Am I Doing Wrong? (Replies: 2)

  3. What am I suppose to do? (Replies: 20)

  4. What am I doing wrong? (Replies: 1)

Loading...