Suppose that
\(D\subset\mathbb R^3\) is a closed bounded set, and
\(f\colon D\to\mathbb R\) a continuous function. We now subdivide
\(D\) into small rectangular boxes, either approaching
\(D\) from the in- or the outside, similarly as shown in
Figure 5.7 for plane domains. Suppose the partition has
\(n\) boxes. Number them in some order from
\(1\) to
\(n\text{,}\) and denote the
\(i\)-th box by
\(R_i\text{.}\) These boxes then have all the form
\begin{equation*}
R_i =[x_i,x_i+\Delta x_i]\times[y_i,y_i+\Delta y_i] \times[z_i,z_i+\Delta z_i]\text{,}
\end{equation*}
and their volume is \(\Delta x_i\Delta y_i\Delta z_i\text{.}\) We then choose an arbitrary point \((x_i^*,y_i^*,z_i^*)\in R_i\) and consider the sum
\begin{equation*}
\sum_{i=1}^nf(x_i^*,y_i^*,z_i^*)\Delta x_i\Delta y_i\Delta z_i,
\end{equation*}
which is the counterpart of the Riemann sum
(5.1). If
\(D\) is `reasonably' smooth and
\(f\) is continuous then one can show that the above sum, also called a Riemann sum, has a limit as
\(\max_i\{\Delta x_i,\Delta y_i,\Delta z_i\}\to 0\text{.}\) Moreover, the limit is independent of the choice of
\((x_i^*,y_i^*,z_i^*)\in R_i\text{.}\) The limit is denoted by
\begin{equation*}
\iiint_D f(x,y,z)\,dx\,dy\,dz,
\end{equation*}
or in more modern notation simply by
\begin{equation*}
\int_D f(\vect x)\,d\vect x,
\end{equation*}
and called the (triple) integral of \(f\) over \(D\).
In the next section we see how to evaluate triple integrals over certain classes of domains.
