# interpreting triple integrals

The limits of integration on a double integral are justified because I'm essentially finding the area underneath two functions. Namely, first I integrate (say, with respect to dx) and I get an area function in terms of 'y'; then computing the next integral (with respect to dy) will give me the area of another function, which in this case means a volume.

But I don't see how I can justify the limits of integration for a triple integral. Specifically I'm doing problems including the divergence theorem. How do I interpret the triple integral?

I don't like thinking of the limits of integration as just "its from here to here, simply because those are the dimensions of the region you're integrating over". I think theres a more complete interpretation of those limits.

EDIT: I think my question basically boils down to how is the triple integral related to the fundamental theorem of calculus? I'm finding increments of what?

Take for instance ∫∫∫ p(x,y,z)dxdydz where p is the density function. How do you interpret that integral (like what does each individual integral represent, and why does it have to be an iterated integral)? And if it were the case that instead of p(x,y,z) I had Div(E) what would each integral represent?

I focused on the limits of integration, because I think these have a lot to do with the fundamental theorem of calculus.

I'm sorry if my question is unclear. Thanks.

• You could make the question more clear with a specific example. Commented Aug 2, 2014 at 16:28
• I added an "edit:" part I hope it makes more sense now :(.
– DLV
Commented Aug 2, 2014 at 16:36

If $\rho$ is the density function, and $\Sigma$ is a solid on space, then $$\iiint_\Sigma \rho~\mathrm{d}V$$ represents the mass of $\Sigma$. The divergence theorem states that, if $\mathbf{F}$ is a vector field, then: $$\iint_{\partial \Sigma} \mathbf{F}~\mathrm{d}\boldsymbol{\sigma} = \iiint_\Sigma \mathrm{div} \ \mathbf{F}~\mathrm{d}V$$ Think of $\mathbf{F}$ as the motion of a fluid, let's say, water, inside $\Sigma$, and $\partial \Sigma$ as permeable membrane. The left-hand side is the flux of water that goes through $\partial \Sigma$ and the right-hand side would be the resulting volume of water that leaves/enters $\Sigma$. Notice that $\mathrm{div} \ \mathbf{F} \equiv 0$ doesn't mean that there is no fluid entering or leaving the surface, it just means that the quantity that enters is equal to the quantity that leaves.

A "triple integral" arises when you have some function $f:\>{\mathbb R}^3\to{\mathbb R}$ and a body $B\subset{\mathbb R}^3$, and you want to compute the "total impact of $f$ over $B$". This "total impact" is denoted by $\int_B f({\bf x})\ {\rm d}({\bf x})$ and has the following properties:

It is linear in $f$, additive with respect to $B$, and $\int_B 1\ {\rm d}({\bf x})={\rm vol}(B)$.

From this idea it follows that $\int_B f({\bf x})\ {\rm d}({\bf x})$ is the limit of sums of the following kind: $$\int_B f({\bf x})\ {\rm d}({\bf x})=\lim_\ldots \sum_{k=1}^N f(\xi_k){\rm vol}(B_k)\ ,$$ where $B=\bigcup_{k=1}^N B_k$ is a partition of $B$ into small essentially disjoint subdomains $B_k$, and the $\xi_k$ are sampling points chosen arbitrarily in $B_k$ $\>(1\leq k\leq N)$.

The above is a geometrically intuitive description of what one has in mind, and coincides for in the one-dimensional case with the definition of the Riemann integral. Now for the one-dimensional case we have the miraculous FTC, and we want to bring that into bear for double or triple integrals as well. That's where Fubini's theorem comes in, which provides a reduction process that converts a triple integral into a nested sequence of three "simple" integrals involving only one variable. But doing this properly for a complicated body $B$ involves dealing carefully with lower and upper limits of the intermediate integrals.

It means that you are summing the contributions from all the space. With your example:

$$$$\int\int\int p(x,y,z)\ dx\ dy\ dz$$$$

you can imagine for a scalar problem that p(x,y,z) represents a spatial distribution, and you want to sum the equivalent elements in a volume. In electromagnetism, you have for example the expression of the potential vector

$$$$A(r) = \int J(r) \frac{exp(-jkr)}{r} dr$$$$

where your triple integrals are replaced by the equivalent polar coordinates, and where you take a look at the contribution of field sources depending on their positions of the space.

• Does this make sense: As I integrate I can see that my units go kg/m3 -> kg/m2 -> kg/m -> kg, this means that each time I integrated I "got the job done" for each coordinate. ?
– DLV
Commented Aug 2, 2014 at 16:55
• yes, for this case, it is equivalent to the contributions of some mass over one, two, or three dimensions. Since the integral is summing, but also normalizing by the integration step ( dx, dz ... ) it is equivalent to a mass density Commented Aug 2, 2014 at 17:03