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Vector area

  1. Aug 21, 2005 #1
    The vector area is given by the integral [itex]\vec a = \int_s d\vec a[/itex] for a surface s.

    I have three problems to prove based on this:

    1] Find the vector area of a hemispherical bowl of radius R.

    I solved a bit of this.
    The area element for a hemisphere is given by [tex]d\vec a = R^2\sin \theta d\theta d\phi \hat r[/tex].
    So,[tex]\vec a = R^2\int_{0}^{\frac{\pi}{2}}\sin \theta d\theta \int_{0}^{2\pi} d\phi \hat r[/tex]

    How am I supposed to integrate over the unit vector [itex]\hat r[/itex]?

    2] Show [tex]\vec a = 0[/tex] for any closed surface.

    3] Show [tex] \vec a[/tex] is the same for all surfaces sharing the same boundary.
     
  2. jcsd
  3. Aug 21, 2005 #2

    Tide

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    Hint:

    [tex]\hat r = \frac {x \hat i + y \hat j + z \hat k}{r}[/tex]
     
  4. Aug 21, 2005 #3
    Does this mean I have to convert all the factors into Cartesian coordinates, since I have used spherical polar in this case?
     
  5. Aug 21, 2005 #4

    Tide

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    No, but you should now convert x, y and z to their spherical forms.

    The basic problem is that the direction of the unit vector [itex]\hat r[/itex] varies with position. The Cartesian unit vectors, however, have fixed directions.
     
  6. Aug 28, 2005 #5
    I converted them and after making the necessary evalutions to the integral, I got:
    [tex]\vec a = \int_s d\vec a = \pi R^2\hat z[/tex]
    What does this equation mean?
    How is the vector area different from the scalar area particularly in this case?
    Why is [itex]\vec a = 0[/itex] for closed surfaces?
     
  7. Aug 28, 2005 #6

    HallsofIvy

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    Essentially what you are doing is adding the "differential of area" vector at each point. It should be easy to see from the symmetry of this problem that the resultant will be in the z direction. It should also be easy to see that for any closed surface, for each point there will be a point where the vectors are pointing in the opposite directions.
     
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