An area of the sphere in spherical coordinates is:

$$A=\int_0^{2\pi}\int_0^\pi d\Omega=\int_0^{2\pi}\int_0^\pi \sin \theta d\theta d\phi$$

Where $\Omega$ is a solid angle. Integral gives old good $4\pi$ as required.

Now, lets say that I want to weight every solid angle element with a unit vector:

$$\vec{r}=(\cos \theta \cos\phi, \cos \theta \sin \phi, \sin \theta)$$

This unit vector points to an area of a solid angle. For every vector $\vec{v}_0$ there would exist $-\vec{v}_0$ canceling each others contribution to the integral, therefore I would expect that due to spherical symmetry the following integral vanishes, however it does not. Result is:

$$\int_0^{2\pi}\int_0^\pi \vec{r} d\Omega = (0, 0, \pi^2) \neq \vec{0}$$

Where is my intuition failing?

  • $\begingroup$ "...of the sphere" ...of radius $\;1\;$ , of course. $\endgroup$ – DonAntonio Jan 16 '17 at 12:33
  • 1
    $\begingroup$ Note that $\forall \theta \in [0,\pi] \,$, $\, \sin \theta \ge 0 \, $ so your $\vec{r}$ double counts the upper hemisphere. The correct radial vector should read $$\mathbf{e}_{r}=(\color{red}{\sin \theta} \cos \phi, \, \color{red}{\sin \theta} \sin \phi, \, \color{red}{\cos \theta})$$ $\endgroup$ – Ng Chung Tak Jan 16 '17 at 13:23

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