Do I have this right Trip integrals Just finished my proof of the volume of a cone using trip integrals. I think I noticed something. Wonder if I got it right.
The first integral defines the line/curve, the second defines the area under the line/curve (The displacement in 2D) and the third integral defines the volume under the surface.( The displacement in 3D).
Can I extend this to say the fourth integral defines the displacement of a volume in 4D? 
 A: Mathematically, there is a definition* (actually, more than one) of $n$-dimensional volume, such that when the definite quadruple integral exists, it really does calculate the (signed) 4-dimensional volume under the graph of the function.

*Basically, volume should satisfy:


*

*The $n$-volume of any subset of $\mathbb R^n$ is nonnegative (maybe $\infty$).

*The $n$-volume of the empty set is 0.

*A subset of a set with $n$-volume 0 also has $n$-volume 0.

*The $n$-volume of the set of points in $\mathbb R^n$ with each coordinate should be 1. (The length of [0,1] is 1, the area of the unit square is 1, the volume of the unit cube is 1, etc.).

*If you have a (countable) collection of disjoint sets in $\mathbb R^n$ then to get the $n$-volume of all of them together you just add the $n$-volumes of the pieces. (wikipedia picture, using $\mu$ for the area)

*If you move a set around (without rotating it or changing its shape) then it should have the same $n$-volume.
It turns out there's basically only one $n$-volume function that satisfies the above, and when, say, a Riemann multiple integral exists for some nonnegative function, it gives the $n$-volume under the curve.
A: I generally interpret triple integrals in the same way as you do, according to their geometric interpretations. (length,area and volume)
But I guess the 4th dimension and beyond really depends on the context. (especially in physics) It's a pity that our geometric intuition is limited to only 3 dimensions.
