Spherical coordinates surface integral

[daudaudaudau]

Hi.

I have this integral
[tex]
\int_0^{2\pi}\int_0^\pi \mathbf A\cdot\hat r d\theta d\phi
[/tex]
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] ?

Thanks.

 

[GPPaille]

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.