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When you integrate in spherical coordinates, the differential element isn't just $ d\theta d\phi $. No. It's $\sin\theta d\theta d\phi$, where $\theta$ is the inclination angle and $\phi$ is the azimuthal angle.

For example, attempting to integrate the unit sphere without the $\sin\theta$ term:

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

With the $\sin\theta$ term you get

$$ \int_0^{2\pi}\int_0^{\pi} \sin\theta d\theta d\phi = 4 \pi $$

But I'm puzzled why you need to multiply the differential solid angle by $\sin\theta$. It would seem it's because the chunks at the north/south pole are "worth less" than the chunks at the equator. That kind of makes sense because they will be closer together.

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    It comes from the Jacobian.2012-08-29
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    Geometrically/intuitively?2012-08-29
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    @bobobobo analytically, please, see my answer or this question [here](http://math.stackexchange.com/questions/17850/how-to-deduce-the-area-of-sphere-in-polar-coordinates). The square-form of the Jacobian must be 1 in the case of unit sphere if your first equation is correct -- is it? You need to calculate all differentials for the Jacobian as hinted by Sigur.2013-01-04

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