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Homework Help: Vector Surface Integrals

  1. Sep 5, 2009 #1
    1. The problem statement, all variables and given/known data

    I am given a vector field associated with a square area of a certain side, lets call this side dx, centered on the x-axis at a certain point, say x = x1. The sides of this cross sectional square are parallel to the axis of y and z. I have to compute the flux of this vector field, or rather, the vector surface integral.

    Now I understand how to set up the equation, substitute the relevant variables, and such. What I'm not totally certain of is how to determine the upper and lower limits over the z and y paths.

    Also, what does the unit vector, n, do?

    2. Relevant equations

    V n dS, where dS = dydz, and V and n are vectors.

    3. The attempt at a solution

    As I said, I have already set up the (double) integral according to the above formula, but I need to determine the limits of the definite integral.
  2. jcsd
  3. Sep 5, 2009 #2
    I think they are usually provided. Could you post one question where you think they are not provided?
  4. Sep 6, 2009 #3
    I am given a vector field that permeates a square area element of a side (of certain value), centered on the x-axis (at a given point on x). The square's sides are parallel to the y and z axes. The unit vector n is equal to the unit vector x, which is equal to the cross product of unit vectors y and z.

    That is all I have and I am told to calculate the vector surface integral.

    I cannot think of a way to determine the limits. But other than that, the expression is pretty straightforward.
  5. Sep 6, 2009 #4


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    Please "post one question where you think they are not provide". Don't just repeat what you said before, post the actual question!
  6. Sep 6, 2009 #5
    That is the actual question. Only thing I can add is that the square area has a side of 1, centered on the x-axis at x = 1 (don't know how much this helps). It looks like a flux problem.
  7. Sep 6, 2009 #6


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    If you are "given" a square, you must have been given the length of one of its sides, call it w. If you know at it as some point on the x-axis, say x = a, then you know the square must have the parametric form < a, x, y > where x and y vary over the square. Now you just have to use what you are given by the placement of the square. Does x go from 0 to w or from -w/2 to w/2 or what? Ditto for y. Draw a picture.

    You haven't given us the vector field. If the vector field is constant it might not matter where the square is located, but usually it would matter.
  8. Sep 6, 2009 #7
    The vector field is V = xyz (in the xth direction).

    If the side of the square is 1 and its centered around the x-axis with the sides parallel to the y and z axes, then wouldn't the limits be (0, 1/2) for y and z, respectively?
    Last edited: Sep 6, 2009
  9. Sep 6, 2009 #8
    I collected that
    1) <1,0,0> is the normal
    2) surface area is a square with length of 1 and center of x0
    3) the vector field is V = xyz

    Draw a square with center at x0, so are the limits for y and z from -1/2 to 1/2 or 0 to 1/2?

    P.S. you should provide all the information in one post. Please note that other people cannot read your mind.
  10. Sep 6, 2009 #9
    I am not sure about the limits. I am trying to think of any information that would be indicative of what the limits should be.

    Also, I apologize if I have not been clear.
  11. Sep 6, 2009 #10


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    With regard to 1) ..so dS = <1, 0, 0> dy dz, right?

    With regard to 2) The [bold]surface[\bold] is a square. Its area is a number. And when you say "center at x0" do you mean (x0, 0, 0)?

    With regard to 3) V = xyz is not a vector field; it is an ordinary scalar function.
    I guess you mean V = <xyz, 0 , 0 >

    As for whether the limits on y and z are -1/2 to 1/2 or 0 to 1/2, which choice would describe a square of side 1 centered at (x0, 0, 0)?

    In my reply I said:

  12. Sep 6, 2009 #11
    Everything you have stated sounds legitimate to me. The only confusion I have is deciding between (0,w) or (-w/2, w/2). I am sure I am missing something very trivial here which ought to resolve this.
  13. Sep 6, 2009 #12


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    If your square is centered at (x0, 0, 0) and is parallel to the yz plane, as far as the yz plane is concerned, it is centered at (0,0). Draw a picture of a square centered at (0,0) whose sides have length w. That should answer your question.
  14. Sep 6, 2009 #13
    Thank you!
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