Question 27.18: Consider a solid object such as that shown in Figure 27.17. ......
Consider a solid object such as that shown in Figure 27.17. The density, ρ, of this object may vary from point to point.
So, at any point P with coordinates (x, y, z), the density is a function of position, that is ρ = ρ(x, y, z). Since density is a scalar, this is an example of a scalar field, like those discussed in Section 7.4. Suppose we select a very small piece of this object having volume δ V and located at P(x, y, z). Recall from elementary physics that
\text { density }=\frac{\text { mass }}{\text { volume }}Then the mass of this small piece, δm, is given by
\delta m=\rho \delta VIf we wish to calculate the total mass, M, of the object we must sum all such contributions from the entire volume. This is found by integrating throughout the volume. We write this as
\text { total mass, } M=\int_V \rho \mathrm{d} VThis is an example of a volume integral, so called because the integration is performed throughout the volume. It will usually take the form of a triple integral such as those discussed in Section 27.6.2. Technically, there are three integral signs, but for brevity these have been replaced by the single \int_V where it is to be understood that the integral is to be performed over a volume. In any specific problem care must be taken to ensure that the entire volume is included when the integration is performed.
For example, consider the case of a solid cube with sides of length 1 unit. Let one corner be positioned at the origin and let the edges coincide with the positive x, y and z axes. Suppose the density of the cube varies from point to point, and is given by ρ(x, y, z) = x + y + Z. Then the integral which gives the mass of the cube is \int_V(x+y+z) \mathrm{d} V, where the volume V is the region occupied by the cube. This integral has been evaluated in Example 27.16 and found to be \frac{3}{2}, representing the mass of the cube.