How to calculate mass of closed Universe?

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How does one integrate the mass density over a closed Universe (a 3-sphere?) to obtain the total mass of that Universe?

Is this the correct integral?

[tex] M = R(t)^3 \rho\int_0^1 4 \pi r^2 \frac{dr}{\sqrt{1-r^2}}[/tex]

where [itex]R(t)[/itex] is the radius of the Universe at cosmological time [itex]t[/itex].

By making the substitution [itex]r=\sin \chi[/itex] one finds that the above integral gives:

[tex] M = \pi^2 R(t)^3 \rho.[/tex]

According to wikipedia the hyperarea of a 3-sphere is [itex]2\pi^2 R^3[/itex] so I'm out by a factor of two.
 
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The former applies the cube to the brackets, the latter applies the cube to the function.
##R(t)^3=Rt^3##
##R^3(t)=\big [ R(t)\big ]^3##

compare: ##\cos(kx)^3## with ##\cos^3(kx)##
Since it is also valid to write ##\cos kx## the first version does not make it clear what is intended.
Is it the cosine of the cube of a product or the cube of the cosine of a product? OP only had one variable in the brackets, which gives notice to the reader.

... its really just a notation foible.
I don't think anyone would get confused in such a small equation but in big long multiline equations it gets easier to be misread.
 
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Simon Bridge said:
The former applies the cube to the brackets, the latter applies the cube to the function.
##R(t)^3=Rt^3##
##R^3(t)=\big [ R(t)\big ]^3##

compare: ##\cos(kx)^3## with ##\cos^3(kx)##
Since it is also valid to write ##\cos kx## the first version does not make it clear what is intended.
Is it the cosine of the cube of a product or the cube of the cosine of a product? OP only had one variable in the brackets, which gives notice to the reader.

... its really just a notation foible.
I don't think anyone would get confused in such a small equation but in big long multiline equations it gets easier to be misread.

I understand, thanks!