When do triple integrals involve a fourth dimension?

Of course, heuristically, a single integral gives area under a curve, and a double integral of a function gives the volume under the integrand and above a two-dimensional domain. Now, I understand that a triple integral of the number 1 gives the volume of the three-dimensional shape described by the limits of integration, but my professor told us that triple integrals are just integrals over "a 3D domain."

I suppose my confusion is this: does the value represented by a triple integral depend on the specific context of the problem, or are there different types of triple integrals that correspond to different meanings?

• If you continue with mathematics and enter into studying non-Euclidean geometry or objects of various and sundry strange topologies, I guess the concerns you mention might become important. And even at the level where you are now, there might be sometimes where hypothesizing a "fictional" fourth dimension makes some calculations more convenient somehow. But, no, at the level where you are right now, a triple integral is simply a triple integral, no further context needed. If you know the function and the shape of the volume, that is all there is to be said. – bob.sacamento Dec 10 '18 at 16:03
• Possible duplicate of What does a triple integral represent? – BigbearZzz Dec 10 '18 at 16:04
• A single integral can give the area under a curve, but often that's not what the integral really means. For example, a single integral can give the height of a ball after it has fallen for some time with an accelerating downward speed. We can make a graph of the speed over time and use "area under the curve" to help find the ball's height, but the height is not actually area under a curve and we don't really need to think about any areas in order to compute the height. – David K Dec 10 '18 at 22:02

In some instances, one can use a triple integral to measure the volume of a $$3D$$ region, but triple integrals can also be used to find 'volume' between the graph of a $$4D$$ function and a $$3D$$ region.
Given a $$3D$$ region $$E$$, the volume of $$E$$, which we'll denote as $$V(E)$$, is given by $$V(E)=\iiint\limits_{E}\mathrm dx\mathrm dy\mathrm dz$$ But if you have a function $$f(x,y,z)$$, then you know that it's graph is going to be $$4D$$. But we can still find the 'volume' of $$f$$ over $$E$$: $$V(f,E)=\iiint\limits_{E}f(x,y,z)\mathrm dx\mathrm dy\mathrm dz$$