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Difficult Surface Integrals

  1. May 23, 2004 #1
    I expected Stokes theorem to make my life easier but these problems are even harder than the normal ones ive been doing.


    Use Stokes' Theorem to evaluate [tex]\int\int_ScurlFdS[/tex]

    where F(x, y, z) = < x^2*y^3*z, sin(xyz) ,xyz >

    S: Part of cone [tex]y^2 = x^2 + z^2[/tex] that lies between the planes y = 0 and y = 3 oriented in the direction of the positive y - axis.

    not sure about this one... I parameterized the curve C as r(t) = < 3cost, 3, -3sint > to get a curve with positive orientation induced by S having normal vector in pos-y direction...a circle in plane y = 3,
    answer: 2187*pi/4

    Use Stokes' Theorem to evaluate [tex]\int\int_ScurlFdS[/tex]

    where [tex]F(x, y, z)=< e^{xy}*cos(z) ,x^2*z, xy >[/tex]

    S: Hemisphere [tex]x = sqrt{1-y^2-z^2}[/tex] oriented in the direction of the positive x - axis.

    curve C : r(t) = <0, cost, sint> ... a circle again in plane x=0, hmmm, wonder if I'm messing these up...?
    answer: 0 (this seemed way too easy...everything was zero!)
     
    Last edited: May 23, 2004
  2. jcsd
  3. May 23, 2004 #2

    arildno

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    I assume you're talking about this verson of Stokes' theorem:
    [tex]\iint_{S}\nabla\times\vec{F}\cdot{d}\vec{S}=\oint_{C}\vec{F}\cdot{d}\vec{r}[/tex]

    In that case, your last calculation is correct; even if it is indecently easy.. :smile:
     
  4. May 23, 2004 #3
    Even for the first one? The answer just seemed completely off considering its size.
     
  5. May 23, 2004 #4

    arildno

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    At a glance, the first one ought to be zero as well..
     
  6. May 23, 2004 #5

    arildno

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    Sorry, first glance wrong..
     
    Last edited: May 23, 2004
  7. May 23, 2004 #6

    arildno

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    If 3^{7}=2187, then the first should be correct as well.
     
  8. May 23, 2004 #7
    thank you :)
     
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