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Take the convention of spherical coordinates for mathematics defined here. It's convention that $\phi$ should range at most from $[0,\pi]$ and $\theta$ from $[0,2\pi]$. Suppose I wanted to integrate over the upper unit hemisphere. The conventional way of doing so would be to use the bounds $\theta \in [0, 2\pi], \phi \in [0, \pi/2]$, which yields for the function $\sin\phi$,

$$\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}\sin\phi\ d\theta d\phi = 2\pi$$

Suppose I am a rebel and want to integrate over the same region in a different way. My understanding of the range convention on $\phi$, $[0,\pi]$, is to avoid accidentally counting the same points on a surface twice. I'll attempt to avoid this mistake by only using the range of $\theta$ to be $[-\pi/2, \pi/2]$, and I want $\phi \in [-\pi/2, \pi/2]$.

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin\phi\ d\theta d\phi = 0$$

Given that the integral over these different bounds (which I think should still yield the upper unit hemisphere) is a different answer, I'm led to believe there must be a better reason for the $\phi$ constraint.

Why do these integrals yield different answers?

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The area element on the unit sphere is $|\sin\phi| d\theta d\phi$, not $\sin\phi \, d\theta d\phi$. But you don't have to worry about the absolute value signs if you only use $\phi \in [0,\pi]$.

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  • $\begingroup$ What does "the area element" mean? $\endgroup$ Jun 3, 2021 at 17:41
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    $\begingroup$ Oh, sorry. Since you wrote down the integral for the area correctly in the first case, I assumed you were familiar with the terminology. The area element is the thing that you integrate to get the area. Intuitively speaking, it's the infinitesimal area that you cover on the sphere if the angle coordinates vary in the infinitesimal rectangle $[\phi,\phi+d\phi] \times [\theta,\theta+d\theta]$. And it can't be negative, obviously, which is why you need the absolute value if $\sin \phi < 0$. $\endgroup$ Jun 3, 2021 at 18:53

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