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Upon integration,

$\int f(x) \implies $ area of curve.

$\iint f(x) \implies $ volume under the curve.

$\iiint f(x) \implies $ ? . We are expected to come up with something 4 dimensional?

I simply know.

$\iiint 1 \implies $ Volume.

$\iiint \rho(x) \implies $ gives mass.

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What's the question? –  Raskolnikov Nov 27 '12 at 10:53
Yes, it would be the hypervolume of a 4D object. This is a bit tough to imagine, but check out what James Stewart has to say in his Calculus text; in particular, look at his section on "Applications of Triple Integrals" (16.6 p.1031). –  Benjamin Dickman Nov 27 '12 at 11:13
The question is $\int\int\int f(x)=?$ As for James Stewart, He does not say anything more than what you have already said and what I already wrote. Thank you, though. –  user45099 Nov 27 '12 at 11:33
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up vote 1 down vote accepted

We know that a continuous function is integrable over a closed bounded domain. In fact if $f(x,y,z)=1$ on the domain $D$, then you certainly know that the triple integral gives the volume od $D$: $$V=\iiint_D dV$$ Now if $f$ be a positive function, then $$\iiint_D f(x,y,z) dV$$ can be interpreted as the hypervolume (i.e. the 4 dimensional volume) of the region in 4-space (which as @B.D noted is a bit hard to imagine)having the set $D$ as its 3-dimentional base and having its top on the hypersurface $u=f(x,y,z)$. Of course and indeed, this is not specially useful interpretation but, many more useful cases arise in application which you are aware of them.

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