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Section 6.1 Definition of Triple Integrals

Suppose that DR3 is a closed bounded set, and f:DR 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, and denote the i-th box by Ri. These boxes then have all the form
Ri=[xi,xi+Δxi]×[yi,yi+Δyi]×[zi,zi+Δzi],
and their volume is ΔxiΔyiΔzi. We then choose an arbitrary point (xi,yi,zi)Ri and consider the sum
i=1nf(xi,yi,zi)ΔxiΔyiΔzi,
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 maxi{Δxi,Δyi,Δzi}0. Moreover, the limit is independent of the choice of (xi,yi,zi)Ri. The limit is denoted by
Df(x,y,z)dxdydz,
or in more modern notation simply by
Df(x)dx,
and called the (triple) integral of f over D.

Remark 6.1.

If we choose f to be the constant function with value one, that is, f(x,y,z)=1, then the integral represents the volume of the domain:
(6.1)Volume of the domain D=D1dx.
Another physical interpretation is the following. Let D represent the region occupied by a solid made of some material. At every point (x,y,z) on that plate let ϱ(x,y,z) denote the mass density of the material. Then the
Total mass of the solid occupying D=Dϱ(x)dx.
In the next section we see how to evaluate triple integrals over certain classes of domains.