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Flux through a square and rectangle

  1. Aug 24, 2013 #1
    1. The problem statement, all variables and given/known data
    a. Calculate the flux of the vector v1 = (1, 3 5) through a 2×2 square in the x-z plane (i.e., y = 0).
    b. Calculate the flux of the vector v2=(z, y, -x) through this rectangle:0≤ x ≤3, 0≤ y ≤ 2, z = 0..

    3. The attempt at a solution
    I guess flux is suppose to be some kind of surface integral though I don't recall doing these in calculus before. That's not to say I didn't.. but I simply dont remember them. I've looked up the integral methods and this is what I got:
    A. [itex]\int \vec{v_1}da[/itex]
    z=0, da=dxdy[itex]\hat{z}[/itex]
    [itex]\int^2_0 5 dxdy[/itex] = 5(2)(2)=20

    B. [itex]\int \vec{v_3}da[/itex]
    again z=0, da=dxdy[itex]\hat{z}[/itex]
    [itex]\int -x dxdy[/itex]
    [itex]\int^3_0 -x dx[/itex]= -9/2
    [itex]\int^2_0 dy[/itex]=2

    First thing that strikes me is the negative in B. I don't think that I should have a negative flux if I'm understanding this right. Can anyone help me understand this?
  2. jcsd
  3. Aug 24, 2013 #2


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    hi bowlbase! :smile:
    it depends whether you regard the normal as being in the positive or negative z direction

    since the question doesn't say, it doesn't matter :wink:
    correct method, but
    i] you've misread the question … the normal is in the y direction!
    ii] please always write ∫∫ not ∫ for a double integral, and you're less likely to make mistake :wink:
    no, you seem to have evaluated ∫∫ xdxdy as a sum of two integrals

    you need to integrate wrt x first, then integrate that result wrt y
  4. Aug 24, 2013 #3
    A) its funny that I have done it correctly on paper but incorrectly here. I'm not paying enough attention I guess.

    B) no idea why I added those. I'm glad that I at least had the correct idea.

    thanks for the help!
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