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Spherical coordinates surface integral

  1. Aug 7, 2009 #1

    I have this integral
    \int_0^{2\pi}\int_0^\pi \mathbf A\cdot\hat r d\theta d\phi
    where [itex]\hat r[/itex] is the position unit vector in spherical coordinates and [itex]\mathbf A[/itex] is a constant vector. Is it possible to evaluate this integral without calculating the dot product explicitly, i.e. without knowing that that [itex]\hat r=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta) [/itex] ?

  2. jcsd
  3. Aug 7, 2009 #2
    I guess that the answer is 0:

    [tex]\int_0^{2\pi}\int_0^{\pi}{A\cdot\widehat{r}d\theta d\phi} = A \cdot \int_0^{2\pi}\int_0^{\pi}{\widehat{r}d\theta d\phi}[/tex]

    But you are now integrating normal vectors over the sphere, which is perfectly symmetric. Every vector can be summed with its opposite to give 0.
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