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Using Stokes' theorem

  1. Apr 7, 2009 #1
    1. The problem statement, all variables and given/known data
    Calculate
    [tex]\int \int _{S}\nabla \times \overline{F} \cdot \hat{N}dS[/tex]

    where [tex]\overline{F} = 3y\hat{i} - 2xz\hat{j} + (x^{2}-y^{2})\hat{k}[/tex] and S is a
    hemispherical surface x2 + y2 + z2 = a2, z ≥ 0 and [tex]\hat{N}[/tex] is a normal of the surface outwards. Can you use Stokes' theorem?

    2. Relevant equations
    I think I can use Stokes' theorem

    [tex]\int \int _{S}\nabla \times \overline{F} \cdot \hat{N}dS = \oint_{C} \overline{F} \cdot d\overline{r} [/tex]

    3. The attempt at a solution

    [tex] \oint_{C} \overline{F} \cdot d\overline{r} = \oint_{C} 3y\hat{i} - 2xz\hat{j} + (x^{2}-y^{2})\hat{k} \cdot (dx\hat{i} + dy\hat{j} + dz\hat{k}) [/tex]
    [tex] = \oint_{C} (3y dx - 2xz dy + (x^{2}-y^{2})dz) [/tex]

    I don't know how to continue!

    I should probably integrate closed line C that is the perimeter curve of the surface (circle radius a), but I have no further idea how to do that. I'm new to this subject and I simply don't understand what I could do/write .. :cry:

    Help is highly appreciated!
     
    Last edited: Apr 7, 2009
  2. jcsd
  3. Apr 7, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Odd, most people learn how to integrate on curves before getting to Stokes' theorem.

    The circle of radius a, in the xy-plane can be written in terms of the parameter [itex]\theta[/itex] as
    [itex]x= a cos(\theta)[/itex]
    [itex]y= a sin(\theta)[/itex]
    z= 0.

    [tex]dx= -a sin(\theta)d\theta[/tex]
    [tex]dy= a cos(\theta)d\theta[/tex]
    dz= 0
     
  4. Apr 7, 2009 #3
    :smile:
    Ok, I got a solution -3a2pi

    What I did I made
    [tex]\overline{r}(t) = a cos t + a sin t [/tex] where [tex]0 \leq t \leq 2 \pi[/tex]

    and then dot product of F(r) and r'(t)

    I integrated that from 0 to 2pi with respect to t.

    Can I somehow expect a negative result?
     
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