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Surface and line integrals

  1. Nov 18, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider a vector A = (2x-y)i + (yz^2)j + (y^2z)k. S is a flat surface area of a rectangle bounded by the lines x = +-1 and y = +-2 and C is its rectangular boundary in the x-y plane. Determine the line integral ∫A.dr and its surface integral ∫(∇xA).n dS


    2. Relevant equations



    3. The attempt at a solution
    First I found ∇xA, which ended up simplifying to 1k
    so ∫k.n dS
    at this part I'm confused as to how to simplify it further.


    For the line integral,
    ∫A.dr
    =∫(2x-y)dx + ∫(yz^2)dy + ∫(y^2z)dz
    this is as far as i was able to understand, I'm not quite sure how to break up the integrals or what intervals to use. But for this line integral, I still tried
    =∫(2x-y)dx of side #1 + ∫(2x-y)dx of side #2 + .... of side #3 + .... of side #4
    the 2nd and 4th sides I got were 0 since the change of dx was 0
    =∫(2x+2)dx (from -1 to 1) - ∫(2x -2)dx (from -1 to 1)
    =8
    not sure if this is a correct procedure at all, but I got an answer out of it. My problem with these questions is coming up with the integrals to solve, I'm sort of lost in how to determine them.
     
  2. jcsd
  3. Nov 18, 2013 #2

    HallsofIvy

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    It's not clear what you mean by "side 1" but I think you mean the line y= 2. Then dr= dx, x going from -1 to 1. 2x- y= 2x- 2 so you integrate that from -1 to 1: [itex]\int_{-1}^1 (2x- 2)dx[/itex]. On "side 2", x= 1, and "side 4", x= -1, x does not change so dx= 0 and there is no integer with respect to x. On "side 3", y= -2, the integral is [itex]\int_{1}^{-1} (2x+ 2)dx[/itex] (Notice the order of the limits of integration. If we go from x= -1 to 1 on y= 2, we are going "counterclockwise" around the boundary and must go from x= 1 to x= -1 on the other side to continue counterclockwise.).

    The thing that may be bothering you is that the vector function given is has three components and three variables, x, y, and z. But the path of integration is in the xy-plane. z= 0 there so the y and z components of the integrand are 0.

    You also ask about the surface integral, [itex]\int (\nabla\times A)\cdot \vec{n}dS[/itex]. Again, the surface is the xy-plane so "[itex]\vec{n}[/itex]" is the unit vector in the z direction, [itex]\vec{k}[/itex]. That means that after finding [itex]\nabla A[/itex] you only need to integrate the z-component. Of course, dS= dxdy and the double integral is for x from -1 to 1 and y from -2 to 2.
     
  4. Nov 18, 2013 #3
    Wow, that was very clear. Thanks for the amazing explanation. I definitely have a better understanding of the surface integrals now, and it looks like I was doing my line integrals right after all.
     
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