Physical Meaning of Volume (or surface) Integrals with limits So when we do triple integrals of a 3D object, and apply the own object as the limit, we will get the volume of the object (from slicing the object and add them together)
But if we apply a different limit, what does the end result actually mean?
 A: So you said that
$$\iiint_R\,dV$$
is the volume of region $R$, which is correct.
But say we do
$$\iiint_R \mu(x,y,z)\, dV$$
where $\mu$ is the density function of an object (i.e, the density of the function at a point). Then integral evaluates to the mass of the object.
Another example is if you took the integral
$$\iiint_R (x,y,z) \, dx\,dy\,dz$$
which would give the "sum" of the $x,y,z$ coordinates of the region. Dividing this sum by the volume of the object gives the center of mass of the object, assuming constant density.
One final example:
$$\iiint_R (x,y,z)\mu(x,y,z) \, dx\,dy\,dz$$
which is the weighted sum of a non-homogenous object's $x,y,z$ coordinates. Dividing this quantity by the mass (the second integral) gives a formula to the center of mass of an object without constant density.
These formulae can all be adapted to other physical properties, for instance, the average temperature, by integrating the temperature and dividing by the volume, etc.
A: The integral of a function on a domain is the "sum" of its values. If you divide by the volume, you get an average.
In other words, the integral of a function is its average value over the domain times the volume of the domain (the volume is given by the integral of $1$).
E.g.
$$\int_0^1\int_0^2\int_0^3 x\,dx\,dy\,dz=\left.\left.\left.\frac{x^2}2\right|_0^3\right|_0^2\right|_0^1=9=\frac9{3\cdot2\cdot1}\int_0^1\int_0^2\int_0^3 \,dx\,dy\,dz.$$
The average value of $x$ over a box $3\times2\times1$ is $\dfrac32$.
