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Divergence theorem SUPER complex, maybe

  1. Jul 10, 2012 #1
    1. The problem statement, all variables and given/known data

    use divergence theorem to evaluate ∫s∫F dot n dA if

    F=[sinh yz, 0, y4] , S: r=[u,cosv,sinv], -4≤u≤4 , 0≤v≤pi



    3. The attempt at a solution

    Instructor surprised us with this one, I have no idea how to attempt. I know that ∫vdiv v dV=∫sn dot v dA, which is the divergence theorem, but I have no idea how to apply it.
     
  2. jcsd
  3. Jul 10, 2012 #2
    Do you know how to find [tex]\nabla \bullet F[/tex]
     
  4. Jul 10, 2012 #3
    In principle, yes. The gradient of a function.

    It should be (1/r)(∂/∂r)(rvr)+(1/r)(∂/∂θ)vθ+(∂/∂z)vz
     
  5. Jul 10, 2012 #4
    Gradient and divergence are different.
    One takes gradient of scalar field. The result is vector field.
    One takes divergence of vector field. The result is scalar field.

    You are given a vector function F in cartesian coordinates. The following link should help. http://en.wikipedia.org/wiki/Divergence#Application_in_Cartesian_coordinates

     
  6. Jul 10, 2012 #5

    HallsofIvy

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    Since your function as given in terms of x, y, and z, wouldn't it be better to use
    [tex]\nabla\cdot f\vec{i}+ g\vec{j}+ h\vec{k}= \frac{\partial f}{\partial x}+ \frac{\partial g}{\partial y}+ \frac{\partial h}{\partial z}[/tex]
    be better?

    Now, what is the divergence of this vector function, [itex]sinh(yz)\vec{i}+ y^4\vec{k}[/itex]?

    (Quite frankly, this problem is remarkably trivial!)
     
  7. Jul 10, 2012 #6
    OK!

    So (∂f/∂x)=(∂/∂x)sinh(yz)=0
    and (∂h/∂z)=(∂/∂z)y4=0

    so the divergence of F is zero! and anything dot zero is zero, right? so the answer is zero??
     
  8. Jul 10, 2012 #7
    Looks good.

    Whenever one sees a crazy closed surface integral or contour line integral, it would be wise to check if the vector function is divergence free or curl free respectively.

     
  9. Jul 10, 2012 #8
    Thank you so much! I took an incomplete due to family issues in this class and I'm trying to make it up this summer, but it's been such a mess trying to remember the 1st half of the class and do homework for the 2nd half with almost a year between. Thank you for your help!!!
     
  10. Jul 10, 2012 #9

    LCKurtz

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    Hold on a minute there everyone. The divergence theorem applies to a surface enclosing a volume, which this isn't. You can add three more obvious surfaces to your given surface ##S## to enclose volume. Call them ##S_1,\,S_2,\, S_3##. Now if your surface is oriented so your enclosed volume surface is directed outward (orientation wasn't given), you can apply the divergence theorem. So what you have so far is$$
    \iint_{S_1\cup S_2 \cup S_3 \cup S}\vec F \cdot d\vec S=\iiint_V \nabla\cdot \vec F\ dV
    = 0$$
    That just tells you the integral over ##S## is minus the sum of the surface integrals over the other 3 surfaces. You aren't finished until you check them.
     
  11. Jul 10, 2012 #10
    Okay, so I check them out by computing the dot product of F over each surface?
     
  12. Jul 10, 2012 #11

    LCKurtz

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    Yes. What you hope for in this type of problem is that the surface integral over the other three surfaces is easier to calculate than the original surface integral. I think you will find that is the case in this problem.
     
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