# How can a single integral equal a triple integral?

Here is part of a discussion about the gravitational potential of a sphere:

Let $dx$ $dy$ $dz$ represent an infinitesimal volume containing matter of density $\rho$ and mass $dm$. Then the potential at distance $R$ from the element will be

$$dV = -\frac{G\rho}{R}dx\ dy\ dz.$$

We may express the total potential as an integral of the volume element over the spherical shell:

$$\int{dV} = V = -G\rho \int\!\!\!\! \int\!\!\!\! \int{\frac{dx\ dy\ dz}{R}} = -G\rho \int^{r+dr}_{r}\!\!\!\int^{\pi}_{\theta=0}\int^{2\pi}_{\phi=0}{\frac{r^2\sin{\theta} \ dr\ d\theta \ d\phi}{R}}$$

where the volume element has been expressed in polar coordinates.

I do not understand how the author integrated both sides and ended up with a single integral on the left and a triple integral on the right. Is it true that (where $v$ is volume): $$\int{dv} = v = \int\!\!\!\! \int\!\!\!\! \int{dx\ dy\ dz}$$ because I previously thought that $$\int\!\!\!\! \int\!\!\!\! \int{dv} = \int\!\!\!\! \int\!\!\!\! \int{dx\ dy\ dz}$$ i.e., that you must have the same number of nested integrals on either side of an equation.

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• This is simply a shorthand: $\int dV$ is "summing up all the volume elements", meaning the same thing as your triple integral. – WetSavannaAnimal Nov 1 '13 at 6:14
• So $\int{dV}$ means the same thing as $\iiint{dV}$? – Randy Randerson Nov 1 '13 at 6:19
• Yes, and as Chris White's answer says, it can be construed as more than only shorthand in the Lebesgue paradigm, where one can think of it either as a triple integral or, by organizing the whole of $\mathbb{R}^3$ at once into measurable sets, a single integral. A piece of relevant trivia: it is Fubini's theorem that shows these two ideas are equivalent (or, more precisely, when these two are equivalent, which they are in all practical cases). – WetSavannaAnimal Nov 1 '13 at 7:21

Many students start calculus by learning to integrate over intervals on the real line. Then they integrate functions over 2D and 3D domains, adding more integral signs to match the number of dimensions. I suppose this is nice notation in the typical cases where you can do your integral dimension-by-dimension, e.g. $$\iiint\limits_V f(x, y, z) \,\mathrm{d}V = \int_{x_0}^{x_1} \left(\int_{y_0(x)}^{y_1(x)} \left(\int_{z_0(x,y)}^{z_1(x,y)} f(x, y, z) \,\mathrm{d}z \right) \,\mathrm{d}y \right) \,\mathrm{d}x.$$ The three integrals on the left remind you that you can reduce this multidimensional problem to nested single-dimensional problems. If you are really lucky, the whole thing separates and you have $$\iiint\limits_V f(x) g(y) h(z) \,\mathrm{d}V = \left(\int_{x_0}^{x_1} f(x) \,\mathrm{d}x\right) \left(\int_{y_0}^{y_1} g(y) \,\mathrm{d}y\right) \left(\int_{z_0}^{z_1} h(z) \,\mathrm{d}z\right).$$
But that triple integral was just a single symbol. What did it mean? Basically two things: (1) sum up the contributions over some domain of some integrand to follow, and (2) this domain is a subset of $\mathbb{R}^3$. Now point (2) is already conveyed in the other symbol $\mathrm{d}V$, which can be written $\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$ or $r^2 \sin(\theta) \,\mathrm{d}r \,\mathrm{d}\theta \,\mathrm{d}\phi$ or any number of other ways. In fact, these expanded forms are even more informative, conveying some information about the interpretation of the variables. As a result, all we really need the integrals for is to convey point (1), and a single symbol does this just as well as three.
For this reason, it is just as correct to write $$\int\limits_V f \,\mathrm{d}V$$ as it is to write $$\iiint\limits_V f \,\mathrm{d}V.$$ They are perfectly equivalent.
Having this mindset is also more useful in more general/advanced branches of analysis, where integration is closely tied to measure theory, and one can put measures on $\mathbb{R}^3$ just as easily as on $\mathbb{R}$, so there isn't much point in distinguishing.