From my understanding if the integrand is 1, then it gives you the volume of the region defined by the bounds. But what does the value of a triple integral represent if the integrand is a function for a surface in space?

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    $\begingroup$ The volume of the 3 dimensional object the integrand defines. The regions you can create using only the integrals bounds are quite limits. $\endgroup$ Commented Apr 18, 2016 at 0:51
  • $\begingroup$ @TonyS.F. A double integral with a 1 in the integrand gives you area. When you have a function for a surface in space in the integrand of a double integral, it multiplies the area by the height of that surface, giving you a volume. If we have a triple integral with an integrand of 1 however, we have a volume. If we have a function for a surface in the integrand, than the volume is being multiplied by it.. and this is where my logic breaks. It doesn't seem to make sense to think of this as volume the same way a triple integral with an integrand of 1 is volume. $\endgroup$ Commented Apr 18, 2016 at 0:57
  • $\begingroup$ You can think of them as being weighted volumes? $\endgroup$ Commented Apr 18, 2016 at 1:01
  • $\begingroup$ @TonyS.F. I'm not sure what you mean by that. $\endgroup$ Commented Apr 18, 2016 at 1:08
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    $\begingroup$ You are still adding up "blocks" of volume but each block is made of a different material with a different density, which is precribed by the integrand. $\endgroup$ Commented Apr 18, 2016 at 1:10

6 Answers 6


You can think of the integrand as the "density" of the region and the value of the integral as the "mass" of the object.

For example, $$ \int_0^1\int_0^1\int_0^1 1 \, \text{d}x \, \text{d}y \,\text{d}z $$ can represent the volume of the unit cube within the region $0\le x\le 1$, $0\le y\le 1$ and $0\le z\le 1$.

For $$ \int_0^1\int_0^1\int_0^1 (x^2+y^2+z^2) \, \text{d}x \, \text{d}y \, \text{d}z\ , $$ you can think about it as the mass of the same cube where its density is given by the function $f(x,y,z)=x^2+y^2+z^2$. This means that the cube is light near the origin and is getting heavier as you move away from it.

In general, $f$ can be negative so you must consider the signed-mass which means that the mass can be negative somewhere...

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    $\begingroup$ This is a good demonstration of what I meant by weighted volumes. $\endgroup$ Commented Apr 18, 2016 at 1:11
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    $\begingroup$ For a more realistic example than signed mass, you can think of $f(x,y,z)$ as the density of electrical charge (which can be positive as well as negative). Then the integral gives the total charge of the body that you integrate over. $\endgroup$ Commented Apr 18, 2016 at 6:10
  • $\begingroup$ @HansLundmark That's pretty neat, much more realistic that my "negative mass" haha. Anyway, I think using the term "mass" might convey more intuition than electric charges for beginner. $\endgroup$
    – BigbearZzz
    Commented Apr 18, 2016 at 8:05
  • $\begingroup$ For which values of $x$, $y$ and $z$ will $f$ be negative? $\endgroup$
    – nluigi
    Commented Apr 18, 2016 at 20:50
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    $\begingroup$ @nluigi What I meant is that for a different choice of the function $f$, it can sometimes be negative. For example we can let $f=x+y-z$. $\endgroup$
    – BigbearZzz
    Commented Apr 19, 2016 at 1:08

When $1$ is your integrand, you have these geometric interpretations:

$\quad\int dx$ is to length

$\quad\iint dA$ is to area

$\quad\iiint dV$ is to volume

When your integrand is some function, then you've likely heard:

$\quad\int f(x)\ dx$ is the signed area underneath the curve.

But wait, you say, why is it that $\int dx$ is a length, but $\int f(x)dx$ is an area? First, it's all an interpretation. One could say that $\int dx$ is the area under the curve $f(x)=1$. The difference is that if $f(x)\ne1$, then we're assigning some non-trivial value to every point in the evaluated space. Note that if $f(x)$ and $x$ have physical units $[f(x)]$ and $[x]$ respectively, then that "area" has units $[f(x)]\times[x]$. For example, integrating a velocity function over some period of time $\int_{t_0}^{t_f} v(t)\ dt$ would give you a change in displacement.

So this is a sort of density. At every point in the interval, you're summing how much of the integrand is present. You're finding how much "stuff" is associated with the interval of integration. With that in mind, let's revise our previous statement:

$\quad\int f(x)\ dx$ is the signed amount of stuff associated with the interval of integration.

It follows that

$\quad\iint f(x,y)\ dxdy$ is the signed amount of stuff associated with the region of integration.

This could certainly be a volume, but we shouldn't limit ourselves. Integrating a surface charge density over a region would give you the total amount of charge, wouldn't it? That's not a volume. What we're really doing is assigning a weighting factor to every point in the region and summing all of these contributions via integration.

$\quad\iiint f(x,y,z)\ dxdydz$ is the signed amount of stuff associated with the volume of integration.

If your integrand was a mass density $\rho(x,y,z)$, then integrating over the volume would give the total mass.

In terms of understanding what an integral actually does, what I presented is mostly a heuristic, but I hope it's a useful one. It certainly rings true in physics. If there's one thing to note, it's that if you're integrating quantities with physical dimensions, then the units of your result will have units equal to that of the integrand times those of the differential elements.

That is, if $I = \int f d\tau$, then $[I] = [f][d\tau]$.

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    $\begingroup$ Indeed more generally, in manifold theory, the objects one integrates are called "densities" (see en.m.wikipedia.org/wiki/Density_on_a_manifold). I don't know the origin of this term, but I suspect it was chosen to convey this natural interpretation of the objects one integrates (with respect to "volume," at least). $\endgroup$ Commented Apr 18, 2016 at 1:42

I warmly endorse the "weighted volume" interpretation found in other answers. But since IgnorantCuriosity asked about parallels with e.g. using a double integral to represent the volume underneath a surface in 3 dimensions, here's how that goes: a triple integral represents the volume underneath a hypersurface in four dimensions.


  • $\int f(x)\,dx$ is the area under the curve $y=f(x)$ in two dimensions $(x,y)$.
  • $\int\int f(x,y)\,dx\,dy$ is the volume under the surface $z=f(x,y)$ in three dimensions $(x,y,z)$.
  • $\int\int\int f(x,y,z)\,dx\,dy\,dz$ is the hypervolume under the hypersurface $t=f(x,y,z)$ in four dimensions $(x,y,z,t)$. ("Under" means $t<f(x,y,z)$.)

This depends on what the function is inside the integrand. If the function is a density function, the integral would give us the total mass of the object. It could also be the volume of some 4 dimensional object as well.


Mass ... $$ \iiint_E w(x,y,z)\;dx\,dy\,dz $$ is the total mass of a region $E$ in space, where $w$ is the "density" (mass per unit volume) that may vary from one point to another.

Similarly, to illustrate the integration of a signed function (that has potentially both positive and negative values), you can think of computing the total charge in a region of space, where $w$ is the "charge density" (charge per unit volume).

In your calculus course, you probably have some problems where you integrate over a volume of water, where the pressure varies with depth.


I too favour the weighted volume analogy, but thinking fundamentally it is a continuous summation of a volume. It does not have to be a volume, even if we conceptually model it as such, it could be input voltages, currents, magnetic fields over time. Hypersurface indeed.

Integrating my acceleration gives my velocity. Integrating my velocity indicates the distance I have travelled, etc. These values are not necessarily scalar and accumulate continuously. Fundamental calculus.


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