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Verify Stokes's theorem with the given surface and vector field

  1. Jul 25, 2009 #1
    1. The problem statement, all variables and given/known data
    verify Stokes's theorem for the given surface and vector field.
    S is defined by x^2 + y^2 + z^2 = 4, z <= 4, oriented by downward normal;

    F = (2y-z, x + y^2 - z, 4y - 3x)

    2. Relevant equations

    double integral over S of the curl F ds = integral over S' of F ds.

    3. The attempt at a solution
    I calculated the curl F= del operator cross-product F = 5i - 2j - k, feel free to check if you wish.

    But in my vector calc book for a similar example, I noticed that they made an upward-pointing normal vector (as it was oriented upward) after del x F. I though del x F was a normal vector--am I wrong? The normal vector they came up with after the del x F (a separate calculation for my knowledge) was very different.

    Any hints from where to go from there?
  2. jcsd
  3. Jul 25, 2009 #2
    Check your curl again. Maple gave me a different answer by only a sign difference.
  4. Jul 26, 2009 #3
    Yeah, that's been confusing me lately. I took elem. linear algebra, so I know how to calculate a determinant, so this is kind of ridiculous. I'll post my curl...

    curl F = [d/dy(4y-3x)-d/dz(x+y^2-z)]i - [d/dx(4y-3x)-d/dz(2y-z)]j + [d/dx(x+y^2 -z)-d/dy(2y-z)]k. And from there I got = [4-(-1)]i -[-3-(-1)]j + [1-(2)]k = 5i+2j-k.
    If you mean a sign change, do you mean 5i-2j-k?
  5. Jul 26, 2009 #4
    Well what you just showed me is correct. I think you just wrote the wrong thing up there then. B/c it is 5i + 2j -k.
  6. Jul 26, 2009 #5
    Are the bounds for S correct? Somehow I see that as the entire sphere itself, which I thought meant you can't use Stokes' Theorem.
  7. Jul 26, 2009 #6
    Oh ok. Yeah, I probably didn't press the shift key. Do you know where I should go from here?
  8. Jul 26, 2009 #7
    Did you get my private message?
  9. Jul 26, 2009 #8
    just did. sry bounds are x^2 + y^2 + z^2 = 4 and z<= 0, just a typo.
  10. Jul 26, 2009 #9
    that makes a lot more sense now, lol. Okay so ignore the PM haha.

    So you want to take the flux integral instead of the line integral right? You could probably find it online, but here's the formula for a flux integral.

    [tex]\int_{S}\vec{F}\cdot d\vec{A}=\int_{R}\vec{F}(x,y,f(x,y))\cdot (-f_{x}\vec{i} - f_{y}\vec{j} + \vec{k})dxdy[/tex]
  11. Jul 26, 2009 #10
    I reread your question, and I thought I'd mention that given what you're given, verify Stokes' Theorem here probably means taking the line integral instead of the curl flux integral. It looks like it would be a lot easier than the flux integral.

    EDIT: Nevermind haha. That line integral would be pretty hairy.
  12. Jul 26, 2009 #11
    alright, understandable. I saw that equation earlier, but didn't think I needed it.
    In this case z = sqrt(4-x^2-y^2), but are x and y just as they are... for F(x,y,f(x,y))?
    or should I parametrize x,y and z?
  13. Jul 26, 2009 #12
    oh, ok. yeah, i've tried with the parametrization, and it doesn't seem that I can solve it by myself.
  14. Jul 26, 2009 #13
    They're just as they are. Plug in [tex]\sqrt(4-x^{2}-y^{2})[/tex] for z in the function. Do keep in mind that this is just one way to evaluate that flux integral. You *could* change to spherical coordinates and integrate. If you know how to parameterize that surface then you can always do that. Point is, see what each give you. Go back to the line integral and see if that's easier. Stokes' Theorem exists to simplify not complicate.
  15. Jul 26, 2009 #14
    Also, any other method you use to evaluate that flux integral would require a "different formula." I put that in quotes b/c all of the formulas are related to each other.
  16. Jul 26, 2009 #15
    Well, I plugged in everything and made the dot product...
    Here is what I came up with (thus a little stuck again):

    I was thinking of making u = 4-x^2-y^2, but that doesn't totally solve my problem.
  17. Jul 26, 2009 #16
    Well that's not right. Remember that you're taking the flux of [tex](\nabla \times \vec{F})\cdot d\vec{A}[/tex]. That means you should just have:

    [tex]\int\int (-5f_{x}-2f_{y}-1)dxdy[/tex]
  18. Jul 26, 2009 #17
    okay....but what about F(x,y,f(x,y))dot(-f_x,-f_y,k)?
  19. Jul 26, 2009 #18
    However, that's still one very disgusting integral. I think what they meant for you to do was to take the line integral. The line itself would be a closed circle where z=0. Hence, your curve should reduce to [tex]4=x^{2}+y^{2}[/tex].
  20. Jul 26, 2009 #19
    yeah, which is what i did originally. from there i thought about reparametrizing the function in cylindrical coordinates, where z = 0. x = 2cos(t), y = 2sin(t), 0<=t<=2pi. but you're saying that if I do the aformented -5f_x, -2f_y, -1 da, that will work as well? seems rather simple :)
  21. Jul 26, 2009 #20
    Looks can be deceiving, lol. But yeah that integral's fairly simple. I mean it's still not exactly a nice problem, but there are no tricks involved as far as the rest go. Just pure integration.

    I'm going to sleep, lol. I'll be up tomorrow if you need more help then.
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