Volume Integrals

A volume integral takes the form

$$\int\!\int\!\int_V f(x,y,z)\,dV,$$ (1.98)

where $V$ is some volume, and $dV=dx\,dy\,dz$ is a small volume element. The volume element is sometimes written $d^{\,3}{\bf r}$, or even $d\tau$.

As an example of a volume integral, let us evaluate the center of gravity of a solid pyramid. Suppose that the pyramid has a square base of side $a$, a height $a$, and is composed of material of uniform density. Let the centroid of the base lie at the origin, and let the apex lie at $(0,\,0,\,a)$. By symmetry, the center of mass lies on the line joining the centroid to the apex. In fact, the height of the center of mass is given by

$$\overline{z} = \left. \int\!\int\!\int z\,dV\right/ \int\!\int\!\int dV.$$ (1.99)

The bottom integral is just the volume of the pyramid, and can be written

$$\int\!\int\!\int dV = \int_0^a dz \int_{-(a-z)/2}^{(a-z)/2} dy\int_{-(a-z)/2}^{(a-z)/2} dx = \int_0^a (a-z)^2\,dz=\int_0^a (a^{\,2}-2\,a\,z+z^{\,2})\,dz$$

$$= \left[a^{\,2}\,z-a\,z^{\,2}+z^{\,3}/3\right]_0^a= \frac{1}{3}\,a^{\,3}.$$ (1.100)

Here, we have evaluated the $z$-integral last because the limits of the $x$- and $y$- integrals are $z$-dependent. The top integral takes the form

$$\int\!\int\!\int z\,dV = \int_0^a z\,dz \int_{-(a-z)/2}^{(a-z)/2} dy\int_{-(a-z)/2}^{(a-z)/2} dx = \int_0^a z\,(a-z)^2\,dz=\int_0^a (z\,a^{\,2}-2\,a\,z^{\,2}+z^{\,3})\,dz$$

$$= \left[a^{\,2}\,z^{\,2}/2-2\,a\,z^{\,3}/3+z^{\,4}/4\right]_0^a= \frac{1}{12}\,a^{\,4}.$$ (1.101)

Thus,

$$\bar{z} = \left.\frac{1}{12}\,a^{\,4}\right/\frac{1}{3}\,a^{\,3} = \frac{1}{4}\,a.$$ (1.102)

In other words, the center of mass of a pyramid lies one quarter of the way between the centroid of the base and the apex.