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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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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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