15.5: Triple Integrals

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:

$$\int_{x_0}^{x_1}\int_{y_0}^{y_1}\int_{z_0}^{z_1} dz\,dy\,dx. \nonumber$$

If the limits are constant, we are simply computing the volume of a rectangular box.

Example $\PageIndex{1}$

We use an integral to compute the volume of the box with opposite corners at $(0,0,0)$ and $(1,2,3)$.

Solution

$$\int_0^1\int_0^2\int_0^3 dz\,dy\,dx=\int_0^1\int_0^2\left.z\right|_0^3 \,dy\,dx =\int_0^1\int_0^2 3\,dy\,dx =\int_0^1 \left.3y\right|_0^2 \,dx =\int_0^1 6\,dx = 6. \nonumber$$

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

Example $\PageIndex{2}$

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

Solution

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 $\PageIndex{1}$.

Figure $\PageIndex{1}$: A tetrahedron (AP).

Now the volume is

$$\eqalign{ \int_0^2\int_{3x/2}^3\int_0^{5x/2} dz\,dy\,dx &=\int_0^2\int_{3x/2}^3\left.z\right|_0^{5x/2} \,dy\,dx\cr &=\int_0^2\int_{3x/2}^3 {5x\over2}\,dy\,dx\cr &=\int_0^2 \left.{5x\over2}y\right|_{3x/2}^3 \,dx\cr &=\int_0^2 {15x\over2}-{15x^2\over4}\,dx\cr &=\left. {15x^2\over4}-{15x^3\over12}\right|_0^2\cr &=15-10=5.\cr } \nonumber$$

Pretty much just the way we did for two dimensions we can use triple integration to compute mass, center of mass, and various average quantities.

Example $\PageIndex{3}$

Suppose the temperature at a point is given by $T=xyz$. Find the average temperature in the cube with opposite corners at $(0,0,0)$ and $(2,2,2)$.

Solution

In two dimensions we add up the temperature at "each'' point and divide by the area; here we add up the temperatures and divide by the volume, $8$:

$$\eqalign{ {1\over8}\int_{0}^2\int_{0}^2\int_{0}^2 xyz\,dz\,dy\,dx &={1\over8}\int_{0}^2\int_{0}^2\left.{xyz^2\over2}\right|_0^2\,dy\,dx ={1\over16}\int_{0}^2\int_{0}^2 xy\,dy\,dx\cr &={1\over4}\int_{0}^2\left.{xy^2\over2}\right|_0^2\,dx ={1\over8}\int_{0}^2 4x\,dx ={1\over2}\left.{x^2\over2}\right|_0^2 =1.\cr } \nonumber$$

Example $\PageIndex{4}$

Suppose the density of an object is given by $xz$, and the object occupies the tetrahedron with corners $(0,0,0)$, $(0,1,0)$, $(1,1,0)$, and $(0,1,1)$. Find the mass and center of mass of the object.

Solution

As usual, the mass is the integral of density over the region:

$$\eqalign{ M&=\int_{0}^1\int_{x}^1\int_{0}^{y-x} xz\,dz\,dy\,dx =\int_{0}^1\int_{x}^1 {x(y-x)^2\over2}\,dy\,dx ={1\over2}\int_{0}^1 {x(1-x)^3\over3}\,dx\cr &={1\over6}\int_{0}^1 x-3x^2+3x^3-x^4\,dx ={1\over120}.\cr } \nonumber$$

We compute moments as before, except now there is a third moment\index{moment}:

$$\eqalign{ M_{xy} &= \int_{0}^1\int_{x}^1\int_{0}^{y-x} xz^2\,dz\,dy\,dx ={1\over360},\cr M_{xz} &= \int_{0}^1\int_{x}^1\int_{0}^{y-x} xyz\,dz\,dy\,dx ={1\over144},\cr M_{yz} &= \int_{0}^1\int_{x}^1\int_{0}^{y-x} x^2z\,dz\,dy\,dx ={1\over360}.\cr } \nonumber$$

Finally, the coordinates of the center of mass are $\bar x=M_{yz}/M=1/3$, $\bar y=M_{xz}/M=5/6$, and $\bar z=M_{xy}/M=1/3$.

Contributors

• David Guichard (Whitman College)

• Integrated by Justin Marshall.