1. Definition of the Triple Integral
You have seen this construction twice before. For the single integral you sliced an interval into sub-intervals, sampled, summed, and took a limit. For the double integral you sliced a flat region into small rectangles (cells), sampled, summed, and took a limit. The triple integral repeats the pattern with one more dimension: we partition a solid into small three-dimensional cells (tiny boxes), evaluate the function at a sample point inside each cell, sum, and take the limit as the cells shrink to zero. Nothing structurally new — just one dimension more.
Formally, let $f(x,y,z)$ be continuous on a box $B = [a,b]\times[c,d]\times[p,q]$. Partition each interval into sub-intervals of widths $\Delta x_i$, $\Delta y_j$, $\Delta z_k$, forming sub-boxes of volume $\Delta V = \Delta x_i\,\Delta y_j\,\Delta z_k$. Choose a sample point $(x_{ijk}^*,\,y_{ijk}^*,\,z_{ijk}^*)$ in each sub-box. Then:
$$\iiint_B f(x,y,z)\,dV = \lim_{m,n,\ell\to\infty} \sum_{i=1}^{m}\sum_{j=1}^{n}\sum_{k=1}^{\ell} f(x_{ijk}^*,\,y_{ijk}^*,\,z_{ijk}^*)\,\Delta V$$The definition extends naturally to non-rectangular solid regions $S$ by enclosing $S$ in a box and setting $f=0$ outside $S$.
- Mass from spatial density. If $f(x,y,z) = \rho(x,y,z)$ is a mass density (mass per unit volume), then $\iiint_S \rho\,dV$ gives the total mass of the solid $S$. This is the primary physical reading.
- Volume of the solid. Set $f(x,y,z) = 1$. Then $\iiint_S 1\,dV = \text{Volume}(S)$. The constant function $1$ is the workhorse of this topic — it turns the triple integral into a volume calculator.
- Why not "volume under a surface"? The double integral $\iint_R f(x,y)\,dA$ can be read as the volume under the surface $z=f(x,y)$. For the triple integral, the analogous reading would require a fourth spatial dimension — which we honestly decline. There is no "volume under a hyper-surface" picture available to us.