So I've been wrestling with understanding exactly what line integral and surface integrals represent, in both scalar and vector form. Also standard double and triple integrals can solve for different things. Having searched the internet, this seems to be a common problem, and I struggle to find really good answers.

I've created a typology of integrals. I'd love if more people could correct my entries or add anything to them. When you study these in multivariable calculus, it all becomes somewhat overwhelming. So I hope this, when rendered comprehensive and correct, can help future students.

Please add corrections in comments and I'll edit the table. Thanks in advance for any insights that can be offered.enter image description here

  • $\begingroup$ "Double integral" = Volume. "Triple integral" = Volume. Huh?? What do you really expect from "answerers"?? $\endgroup$ Commented Dec 20, 2020 at 0:27
  • $\begingroup$ Yes, I think that's right. A triple integral when the integrand is 1 is volume. I was hoping people might offer suggestions about, esp. what line integrals and surface integrals represent. $\endgroup$
    – Abcderia
    Commented Dec 20, 2020 at 0:42
  • $\begingroup$ Um... a line integral is an integral along a line. A surface integral is an integral over a surface. What (really) don't you understand? $\endgroup$ Commented Dec 20, 2020 at 0:44
  • $\begingroup$ Yes, I can calculate those. But what exactly do they "represent." If, say, a single integral is area. What exactly is a surface vector over a scalar field represent? What does a line area scalar mean? What physical ideas do these represent? Having scoured the internet, this seems to be a common question, with little really helpful discussion. Thanks for your prompt look-see at this. I am grateful. $\endgroup$
    – Abcderia
    Commented Dec 20, 2020 at 0:47
  • $\begingroup$ Crossposted to physics.stackexchange.com/q/602017/2451 $\endgroup$
    – Qmechanic
    Commented Dec 21, 2020 at 11:19

1 Answer 1


I feel like this might almost be more of a physics than a math question, but I'll try to help by stating a few different things.

Generally, in terms of dimensional analysis, an $n$-fold integral over length-like coordinates will add $n$ length dimensions to the dimension of the integrand. This is why one calls triple integrals volume integrals and double integrals area integrals. But if the integrand itself is already of dimension $L^1$ (i.e., length), then a double integral of that integrand will have the dimensions of volume, $L^3$ (how one might calculate the volume of a cylinder, for example).

This dimensional analysis applies to integration with respect to non-length variables as well, but the interpretation may not always be as straightforward. Integration with respect to time $t$, for example, will typically be done on integrands of dimension $1/t$, though that is by no means a rigorous rule—it all depends on the mathematical or physical context. When calculating trajectories, for example, the integrand may be of dimension $L/t$ (velocity) and the integral with respect to $t$, leading to a length $L$ as a result. The key point, again, is that integration with respect to some variable increases the dimension of the result with respect to that variable by one.

This applies to line and surface integrals in the same way. General line and surface integrals are really just generalizations of 1D and 2D integrals to non-Euclidean (curved) spaces. It might be a good idea to look for pedagogical introductions into the basic concept behind integration on manifolds if you want to learn more about these generalizations.

  • $\begingroup$ Sure! The idea of adding the dimension of the variable to the dimension of the integrand can actually be seen pretty easily if you consider the construction of the Riemann integral as a sum $\sum_i f(x_i) \Delta x_i$: It is evident that this must have the dimensions of $f$ times the dimensions of $x$. One of the reasons one should never omit the $\mathrm{d}x$ in an integral! $\endgroup$ Commented Dec 20, 2020 at 1:17
  • $\begingroup$ Thanks, @MrArsGravis, for your generosity in sharing your understanding. $\endgroup$
    – Abcderia
    Commented Dec 20, 2020 at 2:03

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