Spherical coordinates surface integral

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SUMMARY

The discussion focuses on evaluating the surface integral of a constant vector \(\mathbf A\) dotted with the position unit vector \(\hat r\) in spherical coordinates. The integral is expressed as \(\int_0^{2\pi}\int_0^\pi \mathbf A\cdot\hat r d\theta d\phi\). The conclusion reached is that the integral evaluates to zero due to the symmetry of the sphere, as every vector cancels out with its opposite when integrated over the full surface.

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daudaudaudau
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Hi.

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

Thanks.
 
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I guess that the answer is 0:

\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}

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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