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Verify Stokes' Theorm

  1. May 3, 2013 #1
    1. The problem statement, all variables and given/known data
    Verify Stokes' Theorem

    F(x,y,z) = (xz,-y,x2y) and S is the surface bounded by the planes x = 0, y = 0, z = 0, and 2x + y + 2z = 8, excluding the part contained in the xz-plane.


    2. Relevant equations
    Stokes' Theorem:
    ∫∫A∇xF dA = ∫dAF dR


    3. The attempt at a solution
    Did the right hand side, got my answer of 4.

    The left hand side is where I have trouble.
    Curl F = {x2, x-2xy, 0}

    How do I define dA? It's three separate planes. ...???
    What are my bounds?

    Thank you for your help
     
  2. jcsd
  3. May 3, 2013 #2

    LCKurtz

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    You do three separate integrals. In the xy plane dA = dydx, in the yz plane it is dydz and you integrate over the corresponding triangles. Do you know how to calculate dA for the slanted plane, like you would do for any other surface?
     
  4. May 6, 2013 #3
    Thanks for the help!

    Okay....here's where I got:

    You said I need three integrals. Here they are:

    1.) xy plane, z = 0
    ∫∫(x2, x-2xy, 0) dot (0,0,1)dxdy, where (0,0,1) is perpendicular vector to the plane
    = 0

    2.) yz plane, x = 0
    0408-2z (x2, x-2xy, 0) dot (1,0,0)dydz, where (1,0,0) is perpendicular vector. Are these bounds correct?
    x = 0, so the integral is zero. But were the bounds correct?

    3.) plane 2x + y + 2z = 8
    0408-2x (x2, x-2xy, 0) dot (2,1,2) dydx, where (2,1,2) is perpendicular vector to the plane. I know this is wrong because it does not account for z. How should I do it?
     
  5. May 6, 2013 #4

    HallsofIvy

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    Actually, the boundary consists of four separate planes: x= 0, y= 0, z= 0, and 2x + y + 2z = 8.

    On x= 0, the yz-plane, the outward pointing normal is (-1, 0, 0) and dS= dydz.

    On y= 0, the xz-plane, the outward pointing normal is (0, -1, 0) and dS= dxdz.

    On z= 0, the xy-plane, the outward pointing normal is (0, 0, -1) and dS= dxdy.

    On 2x + y + 2z = 8 we can write y= 8- 2x- 2z so the "position vector" is [tex]\vec{r}= x\vec{i}+ (8- 2x- 2z)\vec{j}+ z\vec{k}[/tex]. [tex]\vec{r}_x= \vec{i}- 2\vec{j}[/tex] and [tex]\vec{r}_z= -2\vec{j}+ \vec{k}[/tex]. The cross product and so the outward pointing normal is [tex]2\vec{i}+\vec{j}+ 2\vec{k}[/tex] and dS= dxdz.
     
  6. May 6, 2013 #5

    LCKurtz

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    I have a nit-pick on both 1 and 2. Actually it's more than a nit-pick; it is important. Stokes theorem refers to an oriented surface. You have not given the orientation of the surface. If you assume it is oriented outward from the (almost) enclosed volume, your normals are both incorrect. Do you see why? Luckily, the answer is 0 anyway, but that's just luck. And yes, the limits on 2 are correct but you didn't show the limits on 1.

    You use the formula for dS for a surface. For z = f(x,y), what is that formula?
     
  7. May 6, 2013 #6

    LCKurtz

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    No, it is an open surface. And dS isn't dxdz on the plane.
     
    Last edited: May 6, 2013
  8. May 6, 2013 #7
    Yes, I see the error - my normal vectors are indeed perpendicular, but they point inward. Thanks for noticing that!

    And as for the third integral....

    z = (1/2)(8-2x-y).....when z = 0, ---> 2x + y = 8
    Thus, 0 <= y <= 8 - 2x
    And when y= 0 and z = 0 ---> x = 4
    Hence, 0 <= x <= 4

    Does that imply that my original integral was indeed correct?
     
  9. May 6, 2013 #8

    LCKurtz

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    dS does not equal dxdy when you are on the slanted plane.
     
  10. May 6, 2013 #9
    So would it be dxdydz??

    I am wary to do that because I am under the impression that dxdydz does not represent 2-dimensional surfaces that are planes.

    But....I presume then that the integral would be a TRIPLE integral, with the bounds of z: 0 <= z <= (1/2)(8-2x-y), and the other bounds for x and y would remain as I stated.

    I did have a strong suspicious that setting z = 0 was erroneous....
     
  11. May 6, 2013 #10

    LCKurtz

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    You can solve the plane for z and get the form z = f(x,y). Look in your book to find the formula for dS when z = f(x,y).
     
  12. May 6, 2013 #11
    I tried looking in my book but I wasn't really sure where to look...I looked at a couple example problems of Stokes' Theorem, but none of them involved planes...

    If dS is not dxdy when z = f(x,y) and it's not dxdydz....then what could it possibly be??
     
  13. May 6, 2013 #12

    LCKurtz

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    Look here then:
    http://tutorial.math.lamar.edu/Classes/CalcIII/SurfaceIntegrals.aspx
     
  14. May 6, 2013 #13
    In his example, f is a scalar function. In my example, it is a vector function.

    From that, I presume that I use the perpendicular vector and not its magnitude. I noticed that after he multiplies by the magnitude of the perpendicular vector, he made dA = dy dz, and he had x = g(y,z).

    This all leads me to think that this is the correct integral:

    0408-2x( (x/2)(8-2x-y) , -y , x2y ) dot (2,1,2) dy dx

    But you said that dy dx is not correct??
     
  15. May 6, 2013 #14

    LCKurtz

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    Well that integrand certainly isn't correct because it is supposed to be ##\nabla \times \vec F## in there. Perhaps you are getting confused between ##dA##, ##dS## and ##d\vec S##. Stokes theorem says$$
    \iint_S\nabla \times \vec F\cdot d\vec S =\int_C \vec F\cdot d\vec R$$where ##\vec R## is a parameterization of the bounding curve.

    Let's parameterize the plane as ##x=x,\, y = y,\, z = \frac{8-2x-y} 2## so$$
    \vec r(x,y) = \langle x, y, \frac {8-2x-y} 2\rangle$$In this setting ##dA = dydx##.
    The scalar element of surface area is defined as ##dS = |\vec r_x\times\vec r_y|dydx##. The vector area element is defined as$$
    d\vec S =\hat n dS =\frac{\pm\vec r_x\times\vec r_y}{|\vec r_x\times\vec r_y|}
    |\vec r_x\times\vec r_y|dydx =\pm\vec r_x\times\vec r_ydydx$$where ##\hat n## is a unit normal to the surface in the proper direction given by the orientation (choose the ##\pm## accordingly).

    So calculate ##d\vec S = \pm\vec r_x\times\vec r_ydydx## with the correct sign, dot ##\nabla\times \vec F## into it and use that as your integrand. By the way I think you should be getting 32/3 on both sides for your answer.
     
    Last edited: May 6, 2013
  16. May 6, 2013 #15
    Of course - it should be the curl, I suppose I was a little exhausted when I did that.

    And yes, I've actually reworked the right hand side, and got a result of 32/3.

    And, rather than compute the cross product, could I not deduce the normal vector from the equation of the plane? 2x + 1y + 2z = 8 ----> (2,1,2)/3 -----> positive for outward pointing normal.

    EDIT: And divide by 3 to make it a unit vector
     
  17. May 6, 2013 #16

    LCKurtz

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    Yes. But then you must remember that ##dS \ne dydx##. It is ##|r_x\times \vec r_y| dydx## so you still need the partials.
     
  18. May 6, 2013 #17
    Ahhhhhhhhhhhh, finally, I got the answers to match.

    Thank you so much for your help! You really cleared everything up. Thanks!
     
  19. Nov 12, 2015 #18
    I totally confused with this question. can explain me with the sketched diagram please?
     
  20. Nov 13, 2015 #19

    LCKurtz

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    This is a two year old thread. Please start a new thread with your question and use the homework template.
     
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