11.7: Triple Integrals

Triple Riemann Sums and Triple Integrals

Through the application of a mass density distribution over a three-dimensional solid, Preview Activity $11.7.1$ suggests that the generalization from double Riemann sums of functions of two variables to triple Riemann sums of functions of three variables is natural. In the same way, so is the generalization from double integrals to triple integrals. By simply adding a $z$-coordinate to our earlier work, we can define both a triple Riemann sum and the corresponding triple integral.

Definition $11.7.1$

Let $f = f(x,y,z)$ be a continuous function on a box $B = [a,b] \times [c,d] \times [r,s]\text{.}$ The triple Riemann sum of $f$ over $B$ is created as follows.

• Partition the interval $[a,b]$ into $m$ subintervals of equal length $\Delta x = \frac{b-a}{m}\text{.}$ Let $x_0\text{,}$ $x_1\text{,}$ $\ldots\text{,}$ $x_m$ be the endpoints of these subintervals, where $a = x_0\lt x_1\lt x_2 \lt \cdots \lt x_m = b\text{.}$ Do likewise with the interval $[c,d]$ using $n$ subintervals of equal length $\Delta y = \frac{d-c}{n}$ to generate $c = y_0\lt y_1\lt y_2 \lt \cdots \lt y_n = d\text{,}$ and with the interval $[r,s]$ using $\ell$ subintervals of equal length $\Delta z = \frac{s-r}{\ell}$ to have $r = z_0\lt z_1\lt z_2 \lt \cdots \lt z_l = s\text{.}$
• Let $B_{ijk}$ be the sub-box of $B$ with opposite vertices $(x_{i-1},y_{j-1},z_{k-1})$ and $(x_i, y_j, z_k)$ for $i$ between $1$ and $m\text{,}$ $j$ between $1$ and $n\text{,}$ and $k$ between 1 and $\ell\text{.}$ The volume of each $B_{ijk}$ is $\Delta V = \Delta x \cdot \Delta y \cdot \Delta z\text{.}$
• Let $(x_{ijk}^*, y_{ijk}^*, z_{ijk}^*)$ be a point in box $B_{ijk}$ for each $i\text{,}$ $j\text{,}$ and $k\text{.}$ The resulting triple Riemann sum for $f$ on $B$ is
  $$\sum_{i=1}^m \sum_{j=1}^n \sum_{k=1}^{\ell} f(x_{ijk}^*, y_{ijk}^*, z_{ijk}^*) \cdot \Delta V.$$

If $f(x,y,z)$ represents the mass density of the box $B\text{,}$ then, as we saw in Preview Activity $11.7.1$, the triple Riemann sum approximates the total mass of the box $B\text{.}$ In order to find the exact mass of the box, we need to let the number of sub-boxes increase without bound (in other words, let $m\text{,}$ $n\text{,}$ and $\ell$ go to infinity); in this case, the finite sum of the mass approximations becomes the actual mass of the solid $B\text{.}$ More generally, we have the following definition of the triple integral.

Definition $11.7.2$

With following notation defined as in a triple Riemann sum, the triple integral of $f$ over $B$ is

$$\iiint_B f(x,y,z) \, dV = \lim_{m,n,\ell \to \infty} \sum_{i=1}^m \sum_{j=1}^n \sum_{k=1}^{\ell} f(x_{ijk}^*, y_{ijk}^*, z_{ijk}^*) \cdot \Delta V.$$

As we noted earlier, if $f(x, y, z)$ represents the density of the solid $B$ at each point $(x, y, z)\text{,}$ then

is the mass of $B\text{.}$ Even more importantly, for any continuous function $f$ over the solid $B\text{,}$ we can use a triple integral to determine the average value of $f$ over $B\text{,}$ $f_{\operatorname{AVG}(B)}\text{.}$ We note this generalization of our work with functions of two variables along with several others in the following important boxed information. Note that each of these quantities may actually be considered over a general domain $S$ in $\mathbb{R}^3\text{,}$ not simply a box, $B\text{.}$

• The triple integral
  $$\displaystyle V(S) = \iiint_S 1 \, dV$$

  represents the volume of the solid $S$.

• The average value of the function $f = f(x,y,x)$ over a solid domain $S$ is given by
  $$f_{\operatorname{AVG}(S)} = \displaystyle \left(\frac{1}{V(S)} \right) \iiint_S f(x,y,z) \, dV,$$

  where $V(S)$ is the volume of the solid $S\text{.}$

• The center of mass of the solid $S$ with density $\delta = \delta(x,y,z)$ is $(\overline{x}, \overline{y}, \overline{z})\text{,}$ where
  $$\begin{align*} \overline{x} & = \frac{\iiint_S x \ \delta(x,y,z) \, dV}{M},\\[4pt] \overline{y} & = \frac{\iiint_S y \ \delta(x,y,z) \, dV}{M}, \\[4pt] \overline{z} & = \frac{\iiint_S z \ \delta(x,y,z) \, dV}{M}, \end{align*}$$

  and $M = \displaystyle \iiint_S \delta(x,y,z) \, dV$ is the mass of the solid $S\text{.}$

In the Cartesian coordinate system, the volume element $dV$ is $dz \, dy \, dx\text{,}$ and, as a consequence, a triple integral of a function $f$ over a box $B = [a,b] \times [c,d] \times [r,s]$ in Cartesian coordinates can be evaluated as an iterated integral of the form

If we want to evaluate a triple integral as an iterated integral over a solid $S$ that is not a box, then we need to describe the solid in terms of variable limits.

Activity $11.7.2$

a. Set up and evaluate the triple integral of $f(x,y,z) = x-y+2z$ over the box $B = [-2,3] \times [1,4] \times [0,2]\text{.}$

b. Let $S$ be the solid cone bounded by $z = \sqrt{x^2+y^2}$ and $z=3\text{.}$ A picture of $S$ is shown at right in Figure $11.7.2$. Our goal in what follows is to set up an iterated integral of the form

$$\int_{x=?}^{x=?} \int_{y=?}^{y=?} \int_{z=?}^{z=?} \delta(x,y,z) \, dz \, dy \, dx$$

to represent the mass of $S$ in the setting where $\delta(x,y,z)$ tells us the density of $S$ at the point $(x,y,z)\text{.}$ Our particular task is to find the limits on each of the three integrals.

 

i. If we think about slicing up the solid, we can consider slicing the domain of the solid's projection onto the $xy$-plane (just as we would slice a two-dimensional region in $\mathbb{R}^2$), and then slice in the $z$-direction as well. The projection of the solid onto the $xy$-plane is shown at left in Figure $11.7.4$. If we decide to first slice the domain of the solid's projection perpendicular to the $x$-axis, over what range of constant $x$-values would we have to slice?

ii. If we continue with slicing the domain, what are the limits on $y$ on a typical slice? How do these depend on $x\text{?}$ What, therefore, are the limits on the middle integral?

iii. Finally, now that we have thought about slicing up the two-dimensional domain that is the projection of the cone, what are the limits on $z$ in the innermost integral? Note that over any point $(x,y)$ in the plane, a vertical slice in the $z$ direction will involve a range of values from the cone itself to its flat top. In particular, observe that at least one of these limits is not constant but depends on $x$ and $y\text{.}$

iv. In conclusion, write an iterated integral of the form \ref{eq_11_7_TI_not_box} that represents the mass of the cone $S\text{.}$

Note well: When setting up iterated integrals, the limits on a given variable can be only in terms of the remaining variables. In addition, there are multiple different ways we can choose to set up such an integral. For example, two possibilities for iterated integrals that represent a triple integral $\iiint_S f(x,y,z) \, dV$ over a solid $S$ are

• $\displaystyle \int_a^b \int_{g_1(x)}^{g_2(x)} \int_{h_1(x,y)}^{h_2(x,y)} f(x,y,z) \, dz \, dy \, dx$
• $\displaystyle \int_r^s \int_{p_1(z)}^{p_2(z)} \int_{q_1(x,z)}^{q_2(x,z)} f(x,y,z) \, dy \, dx \, dz$

where $g_1\text{,}$ $g_2\text{,}$ $h_1\text{,}$ $h_2\text{,}$ $p_1\text{,}$ $p_2\text{,}$ $q_1\text{,}$ and $q_2$ are functions of the indicated variables. There are four other options beyond the two stated here, since the variables $x\text{,}$ $y\text{,}$ and $z$ can (theoretically) be arranged in any order. Of course, in many circumstances, an insightful choice of variable order will make it easier to set up an iterated integral, just as was the case when we worked with double integrals.

Example $11.7.1$

Find the mass of the tetrahedron in the first octant bounded by the coordinate planes and the plane $x + 2 y + 3 z = 6$ if the density at point $(x,y,z)$ is given by $\delta(x, y, z) = x + y + z\text{.}$ A picture of the solid tetrahedron is shown at left in Figure $11.7.3$.

 

We find the mass, $M\text{,}$ of the tetrahedron by the triple integral

$$M = \iiint_S \delta(x,y,z) \, dV,$$

where $S$ is the solid tetrahedron described above. In this example, we choose to integrate with respect to $z$ first for the innermost integral. The top of the tetrahedron is given by the equation

$$x + 2 y + 3 z = 6;$$

solving for $z$ then yields

$$z = \frac{1}{3}(6 - x - 2y).$$

The bottom of the tetrahedron is the $xy$-plane, so the limits on $z$ in the iterated integral will be $0 \leq z \leq \frac{1}{3}(6-x-2y)\text{.}$

To find the bounds on $x$ and $y$ we project the tetrahedron onto the $xy$-plane; this corresponds to setting $z = 0$ in the equation $z = \frac{1}{3}(6 - x - 2y)\text{.}$ The resulting relation between $x$ and $y$ is

$$x + 2 y = 6.$$

The right image in Figure $11.7.3$ shows the projection of the tetrahedron onto the $xy$-plane.

If we choose to integrate with respect to $y$ for the middle integral in the iterated integral, then the lower limit on $y$ is the $x$-axis and the upper limit is the hypotenuse of the triangle. Note that the hypotenuse joins the points $(6,0)$ and $(0,3)$ and so has equation $y = 3 - \frac{1}{2}x\text{.}$ Thus, the bounds on $y$ are $0 \leq y \leq 3 - \frac{1}{2}x\text{.}$ Finally, the $x$ values run from 0 to 6, so the iterated integral that gives the mass of the tetrahedron is

$$M = \int_{0}^{6} \int_{0}^{3-(1/2)x} \int_{0}^{(1/3)(6-x-2y)} x+y+z \, dz \, dy \, dx.$$

Evaluating the triple integral gives us

$$\begin{align*} M & = \int_{0}^{6} \int_{0}^{3-(1/2)x} \int_{0}^{(1/3)(6-x-2y)} x+y+z \, dz \, dy \, dx\\[4pt] & = \int_{0}^{6} \int_{0}^{3-(1/2)x} \left[xz+yz+\frac{z}{2}\right]\biggm|_{0}^{(1/3)(6-x-2y)} \, dy \, dx\\[4pt] & = \int_{0}^{6} \int_{0}^{3-(1/2)x} \frac{4}{3}x - \frac{5}{18}x^2 - \frac{}{9}xy + \frac{2}{3}y - \frac{4}{9}y^2 + 2 \, dy \, dx\\[4pt] & = \int_{0}^{6} \left[\frac{4}{3}xy - \frac{5}{18}x^2y - \frac{7}{18}xy^2 + \frac{1}{3}y^2 - \frac{4}{27}y^3 + 2y \right]\biggm|_{0}^{3-(1/2)x} \, dx\\[4pt] & = \int_{0}^{6} 5 + \frac{1}{2}x - \frac{7}{12}x^2 + \frac{13}{216}x^3 \, dx\\[4pt] & = \left[5x + \frac{1}{4}x^2 - \frac{7}{36}x^3 + \frac{13}{864}x^4 \right] \biggm|_{0}^{6}\\[4pt] & = \frac{33}{2}. \end{align*}$$

Setting up limits on iterated integrals can require considerable geometric intuition. It is important to not only create carefully labeled figures, but also to think about how we wish to slice the solid. Further, note that when we say “we will integrate first with respect to $x\text{,}$” by “first” we are referring to the innermost integral in the iterated integral. The next activity explores several different ways we might set up the integral in the preceding example.

Activity $11.7.3$

There are several other ways we could have set up the integral to give the mass of the tetrahedron in Example $11.7.1$.

1. How many different iterated integrals could be set up that are equal to the integral in Equation \ref{eq_11_7_Tetrahedron_mass}?
2. Set up an iterated integral, integrating first with respect to $z\text{,}$ then $x\text{,}$ then $y$ that is equivalent to the integral in Equation \ref{eq_11_7_Tetrahedron_mass}. Before you write down the integral, think about Figure $11.7.6$, and draw an appropriate two-dimensional image of an important projection.
3. Set up an iterated integral, integrating first with respect to $y\text{,}$ then $z\text{,}$ then $x$ that is equivalent to the integral in Equation \ref{eq_11_7_Tetrahedron_mass}. As in (b), think carefully about the geometry first.
4. Set up an iterated integral, integrating first with respect to $x\text{,}$ then $y\text{,}$ then $z$ that is equivalent to the integral in Equation \ref{eq_11_7_Tetrahedron_mass}.

Now that we have begun to understand how to set up iterated triple integrals, we can apply them to determine important quantities, such as those found in the next activity.

Activity $11.7.4$

A solid $S$ is bounded below by the square $z=0\text{,}$ $-1 \leq x \leq 1\text{,}$ $-1 \leq y \leq 1$ and above by the surface $z = 2-x^2-y^2\text{.}$ A picture of the solid is shown in Figure $11.7.4$.

1. First, set up an iterated double integral to find the volume of the solid $S$ as a double integral of a solid under a surface. Then set up an iterated triple integral that gives the volume of the solid $S\text{.}$ You do not need to evaluate either integral. Compare the two approaches.
2. Set up (but do not evaluate) iterated integral expressions that will tell us the center of mass of $S\text{,}$ if the density at point $(x,y,z)$ is $\delta(x,y,z)=x^2+1\text{.}$
3. Set up (but do not evaluate) an iterated integral to find the average density on $S$ using the density function from part (b).
4. Use technology appropriately to evaluate the iterated integrals you determined in (a), (b), and (c); does the location you determined for the center of mass make sense?

Summary

• Let $f = f(x,y,z)$ be a continuous function on a box $B = [a,b] \times [c,d] \times [r,s]\text{.}$ The triple integral of $f$ over $B$ is defined as
  $$\iiint_B f(x,y,z) \, dV = \lim_{\Delta V \to 0} \sum_{i=1}^m \sum_{j=1}^n \sum_{k=1}^l f(x_{ijk}^*, y_{ijk}^*, z_{ijk}^*) \cdot \Delta V,$$

  where the triple Riemann sum is defined in the usual way. The definition of the triple integral naturally extends to non-rectangular solid regions $S\text{.}$

• The triple integral $\iiint_S f(x,y,z) \, dV$ can tell us
  • the volume of the solid $S$ if $f(x,y,z) = 1\text{,}$
  • the mass of the solid $S$ if $f$ represents the density of $S$ at the point $(x,y,z)\text{.}$

  Moreover,

  $$f_{\operatorname{AVG}(S)} = \displaystyle \frac{1}{V(S)} \iiint_S f(x,y,z) \, dV,$$

  is the average value of $f$ over $S\text{.}$