It will come as no surprise that we can also do triple integrals—integrals over a three-dimensional region. The simplest application allows us to compute volumes in an alternate way.

To approximate a volume in three dimensions, we can divide the three-dimensional region into small rectangular boxes, each $\Delta x\times\Delta y\times\Delta z$ with volume $\Delta x\Delta y\Delta z$. Then we add them all up and take the limit, to get an integral: $$\int_{x_0}^{x_1}\int_{y_0}^{y_1}\int_{z_0}^{z_1} dz\,dy\,dx.$$ If the limits are constant, we are simply computing the volume of a rectangular box.

Example 15.5.1 We use an integral to compute the volume of the box with opposite corners at $(0,0,0)$ and $(1,2,3)$. $$\int_0^1\int_0^2\int_0^3 dz\,dy\,dx=\int_0^1\int_0^2\left.z\right|_0^3 \,dy\,dx =\int_0^1\int_0^2 3\,dy\,dx =\int_0^1 \left.3y\right|_0^2 \,dx =\int_0^1 6\,dx = 6.$$ $\square$

Of course, this is more interesting and useful when the limits are not constant.

Example 15.5.2 Find the volume of the tetrahedron with corners at $(0,0,0)$, $(0,3,0)$, $(2,3,0)$, and $(2,3,5)$.

The whole problem comes down to correctly describing the region by inequalities: $0\le x\le 2$, $3x/2\le y\le 3$, $0\le z\le 5x/2$. The lower $y$ limit comes from the equation of the line $y=3x/2$ that forms one edge of the tetrahedron in the $x$-$y$ plane; the upper $z$ limit comes from the equation of the plane $z=5x/2$ that forms the "upper'' side of the tetrahedron; see figure 15.5.1. Now the volume is \eqalign{ \int_0^2\int_{3x/2}^3\int_0^{5x/2} dz\,dy\,dx &=\int_0^2\int_{3x/2}^3\left.z\right|_0^{5x/2} \,dy\,dx\cr &=\int_0^2\int_{3x/2}^3 {5x\over2}\,dy\,dx\cr &=\int_0^2 \left.{5x\over2}y\right|_{3x/2}^3 \,dx\cr &=\int_0^2 {15x\over2}-{15x^2\over4}\,dx\cr &=\left. {15x^2\over4}-{15x^3\over12}\right|_0^2\cr &=15-10=5.\cr } $\square$