Skip to main content

Section 1.5 Triple integrals

The definition of a double integral of a function of two variables extends easily to that of a triple integral of a function of three variables, and indeed more generally to integrals of functions of \(n\) variables for any \(n\geq 1\text{.}\) In this course we will stop at \(n=3\text{.}\) As you will see the double and triple integral serve as the foundation for all further types of integrals we treat in this course (e.g., line, surface and volume integrals).

The definition of the triple integral is exactly analogous to that of the double integral, only with more subindices! As such we will give a more compressed treatment.

Definition 1.5.1. Triple integral over a box.

Let \(f\) be a function of three variables. Given a box

\begin{equation*} B=[a,b]\times[c,d]\times [e,f]=\{(x,y,z)\in\R^3\colon a\leq x\leq b, c\leq y\leq d, e\leq z\leq f\} \end{equation*}

lying within the domain of \(f\text{,}\) a partition \(P\) of \(B\) into \(n\) subboxes \(B_k\) of volume \(\Delta V_k=\Delta x_k\Delta y_k\Delta z_k\text{,}\) and a selection of sample points \((x_k,y_k,z_k)\in B_k\) from each subbox, the associated Riemann sum is defined as

\begin{equation*} S=\sum_{k=1}^n f(x_k,y_k,z_k)\, \Delta V_k\text{.} \end{equation*}

The norm of \(P\text{,}\) denoted \(\norm{P}\) is the largest value of any width, depth, or height of any subbox of \(P\text{.}\)

The function \(f\) is integrable over \(B\) if there is a number \(I\in \R\) such that

\begin{equation} I=\lim_{\norm{P}\to 0}\sum_{k=1}^nf(x_k,y_k,z_k)\Delta V_k\tag{1.5.1} \end{equation}

for any sequence of partitions \(P\) of \(B\) with norm tending to 0. When this is the case we call \(I\) the integral of \(f\) over \(B\text{,}\) denoted

\begin{equation*} I=\iiint\limits_{B}f(x,y,z)\, dV\text{.} \end{equation*}

Interpretation 1.5.2. Riemann sums as approximations.

As with double integrals, it is best to consider any given Riemann sum

\begin{equation*} S=\sum_{k=1}^n f(x_k,y_k,z_k)\, \Delta V_k \end{equation*}

of \(f\) over a box \(B\subseteq \R^3\) as an approximation of the integral \(\iiint\limits_{B}f(x,y,z)\, dV\text{.}\) As the partitions \(P\) become finer and finer (\(\norm{P}\rightarrow 0\)), these approximations get better and better. In the limit we get the integral precisely.

Definition 1.5.3. Triple integral over bounded region.

Let \(f\) be a function of three variables defined over the bounded region \(D\subseteq\R^3\text{.}\) Let \(B\) be any box containing \(D\) and let \(f^*\) be the function defined on the box \(B\) as follows:

\begin{equation*} f^*(x,y,z)=\begin{cases}f(x,y,z)\amp \text{if } (x,y,z)\in D\\ 0\amp \text{if } (x,y,z)\notin D \end{cases}\text{.} \end{equation*}

The function \(f\) is integrable over \(D\) if the function \(f^*\) is integrable over the box \(B\text{,}\) and in this case we define

\begin{equation*} \iiint\limits_Df(x,y,z)\, dV=\iiint\limits_{B}f^*(x,y,z)\, dV\text{.} \end{equation*}

Interpretation 1.5.4. Interpreting integrals.

The key to understanding the meaning of a double (\(n=2\)) or triple (\(n=3\)) integral in a given context lies in going back to the Riemann sums

\begin{align} n=2:\amp \sum_{k=1}^n f(x_k,y_k)\Delta A_k \amp n=3:\amp \sum_{k=1}^n f(x_k,y_k,z_k)\Delta V_k \text{.}\tag{1.5.2} \end{align}

If we can understand what the Riemann sums compute, then we understand what the integral computes, as the former are simply approximations of the latter. The following list of common interpretation schema illustrates this principle.

  1. Physical: density functions.

    Assume \(f\) is a denisty function of some quantity \(Q\text{:}\) e.g. \(f\) computes the mass per unit area (\(n=2\)) or unit volume (\(n=3\)) at a given point in the region; or \(f\) computes the charge per unit area (\(n=2\)) or unit volume (\(n=3\)) at a given input in the region. In this case the Riemann sums can be understood as approximations of the total quanty \(Q\) over the region; and thus the integral computes the total quantity \(Q\) for the region.

  2. Geometric: integrating 1.

    Assume we are integrating the constant function \(f(x,y)=1\) (\(n=2\)) or \(f(x,y,z)=1\) (\(n=3\)). Here the Riemann sums (1.5.2) can be understood as approximations of the area (\(n=2\)) or volume (\(n=3\)) of the region. We conclude

    \begin{align*} \iint\limits_{\mathcal{R}}1\, dA\amp=\operatorname{area} \mathcal{R} \amp \iiint\limits_D1\, dV=\operatorname{vol} D \end{align*}
  3. Geometric: integrating nonnegative \(f(x,y)\).

    If \(f(x,y)\geq 0\) on the region \(\mathcal{R}\text{,}\) we can interpret it as the height of the graph of \(f\) over the point \((x,y)\text{.}\) Riemann sums of \(f(x,y)\) can then be thought of approximations of the volume of the solid region \(\mathcal{S}\) lying above \(\mathcal{R}\) and below the graph of \(f\text{;}\) the integral \(\iint\limits_Rf(x,y)\, dA\) is thus understood as the precise volume of this region.

Definition 1.5.5. Volume of solid region.

A bounded region \(D\subseteq \R^3\) is measurable if the constant function \(f(x,y,z)=1\) is integrable over \(D\text{.}\) In this case we define the volume of \(D\) as

\begin{equation} \operatorname{vol}D=\iiint\limits_D 1 \, dV\text{.}\tag{1.5.3} \end{equation}

Definition 1.5.6. Elementary solid region.

An elementary solid region is a subset \(D\subseteq \R^3\) of one of the following forms.

  • Type A.

    The projection of \(D\) onto the \(xy\)-plane is a planar region \(\mathcal{R}\subseteq \R^2\) of type 1 or 2, and we have

    \begin{equation*} D=\{(x,y,z)\colon (x,y)\in \mathcal{R}, p_1(x,y)\leq z\leq p_2(x,y)\}\text{,} \end{equation*}

    where \(p_1, p_2\) are continuous on \(\mathcal{R}\text{.}\)

  • Type B.

    The projection of \(D\) onto the \(yz\)-plane is a planar region \(\mathcal{R}\subseteq \R^2\) of type 1 or 2, and we have

    \begin{equation*} D=\{(x,y,z)\colon (y,z)\in \mathcal{R}, q_1(y,z)\leq x\leq q_2(y,z)\}\text{,} \end{equation*}

    where \(q_1, q_2\) are continuous on \(\mathcal{R}\text{.}\)

  • Type C.

    The projection of \(D\) onto the \(xz\)-plane is a planar region \(\mathcal{R}\subseteq \R^2\) of type 1 or 2, and we have

    \begin{equation*} D=\{(x,y,z)\colon (x,z)\in \mathcal{R}, r_1(x,z)\leq y\leq r_2(x,z)\}\text{,} \end{equation*}

    where \(r_1, r_2\) are continuous on \(\mathcal{R}\text{.}\)

Observe that there are in fact 6 distinct types of elementary regions, depending on (a) which coordinate plane we project onto (3 choices), and (b) whether the projection \(\mathcal{R}\) is of type 1 or type 2. At the risk of going too far in our taxonomy, we could label these different types as A1, A2, B1, B2, C1, etc..

Remark 1.5.8.

Observe that equations (1.5.4)–(1.5.6) each describe the triple integral as an iterated integral, just as in our earlier Fubini theorems. Only now the inner one is a single integral, and the outer one is a double integral! Since the double integral is computed over an elementart region \(\mathcal{R}\text{,}\) we can finish the computation by using Theorem 1.3.6. Furthermore, since \(\mathcal{R}\) is either of type 1 or type 2, we can expand the equations (1.5.4)–(1.5.6) into six integral identities involving only single integrals. Theorem 1.5.9 does precisely this for your convenience.

Observe that a box \(B=[a,b]\times [c,d]\times [e,f]\) is itself an elementary region of \(\R^3\) of a particularly simple type: all of the defining boundary functions are constant. Furthermore, we see that \(B\) is elementary of any the six types. Fubini's theorem over boxes then follows as a direct corollary of Theorem 1.5.9.

Example 1.5.11. Triple integral over a box.

Compute the integral of \(f(x,y,z)=xz\sin(\pi x^2yz)\) over the box \(B=[0,1]\times [0,1]\times [0,1]\text{.}\)

Solution.

Making full use of Fubini's theorem we use the sequence of integration \(dy\, dx\, dz\)We have

\begin{align*} \iiint\limits_B f(x,y,z)\, dV \amp=\int_0^1\int_0^1\int_0^1xz\sin(\pi x^2yz)\, dy\, dx\, dz\\ \amp= \int_0^1\int_0^1-\frac{1}{\pi}xz(\cos(\pi xz) \end{align*}

Example 1.5.12. Volume of tetrahedron.

Let \(D\) be the tetrahedron with vertices \((0,0,0), (0,1,0), (1,1,0), (0,1,1) \text{.}\)

  1. Set up three separate integrals, of type A, B, and C, that compute the volume of \(D\text{.}\)

  2. Compute the volume of \(D\) using one of the integrals above.

Solution.

Type A.

The projection of \(D\) onto the \(xy\)-plane is just its bottom face:

\begin{equation*} \mathcal{R}=\{(x,y,0)\colon 0\leq x\leq 1, x\leq y\leq 1\}\text{.} \end{equation*}

The top part of the tetrahedron lies in the plane containing \((0,1,1),(1,1,0),(0,0,0)\text{,}\) which has defining equation \(x-y+z=0\text{,}\) or \(z=y-x\text{.}\) Thus we can describe \(D\) as a type-A region as

\begin{equation*} D=\{(x,y,z)\colon 0\leq x\leq 1, x\leq y\leq 1, 0\leq z\leq y-x\text{,} \end{equation*}

and the corresponding volume iterated integral is

\begin{equation*} \operatorname{vol} D=\int_0^1\int_x^{1}\int_0^{y-x}1\, dz\, dy\, dx\text{.} \end{equation*}

Type B.

The projection of \(D\) onto the \(yz\)-plane is just its left vertical face:

\begin{equation*} \mathcal{R}=\{(0,y,z)\colon 0\leq y\leq 1, 0\leq z\leq y\}\text{.} \end{equation*}

Starting with a point \((0,y,z)\) in this projection and moving in the positive \(x\)-direction, we see that we lie within \(D\) until hitting its oblique plane at \(x=y-z\text{.}\) Thus we can describe \(D\) as a type-B region as

\begin{equation*} D=\{(x,y,z)\colon 0\leq y\leq 1, 0\leq z\leq x, 0\leq x\leq y-z\text{,} \end{equation*}

and the corresponding volume iterated integral is

\begin{equation*} \operatorname{vol} D=\int_0^1\int_0^{y}\int_0^{y-z}1\, dx\, dz\, dy\text{.} \end{equation*}

Type C.

I claim the projection of \(D\) onto the \(xz\)-plane is

\begin{equation*} \mathcal{R}=\{(x,0,z)\colon 0\leq x\leq 1, 0\leq z\leq 1-x\}\text{.} \end{equation*}

Indeed, the back face of \(D\) is the triangular region

\begin{equation*} \{(x,1,z)\colon 0\leq x\leq 1, 0\leq z\leq 1-x\}\text{,} \end{equation*}

which projects onto \(\mathcal{R}\text{.}\) It is also clear that this is the “widest” section of \(D\) in terms of casting a shadow onto the \(xz\)-plane, and hence that the projection is no bigger than \(\mathcal{R}\text{.}\) Next, given a point \((x,0,z)\in \mathcal{R}\text{,}\) as we move in the \(y\) direction we do not enter the tetrahedron until we hit the oblique plane at \(y=x+z\text{,}\) and we stay within the tetrahedron until we hit its back wall at \(y=1\text{.}\) Thus we can describe \(D\) as a type-C region as

\begin{equation*} D=\{(x,y,z)\colon 0\leq x\leq 1, 0\leq z\leq 1-x,x+z\leq y\leq 1\}\text{,} \end{equation*}

and the corresponding volume integral is

\begin{equation*} \operatorname{vol} D=\int_0^1\int_0^{1-x}\int_{x+z}^1 1\, dy\, dz\, dx\text{.} \end{equation*}
All these integrals evaluate to \(1/6\text{,}\) the volume of \(D\text{.}\)

Example 1.5.13. Volume of slice of parabolic cylinder.

Let \(D\subseteq\R^3\) be the region bounded by the parabolic cylinder \(x=y^2\text{,}\) the plane \(x+z=1\text{,}\) and the \(xy\)-plane.

  1. Set up two integrals, of type A and B, that compute the volume of \(D\text{.}\)

  2. Compute the volume of \(D\) using one of the integrals above.

Solution.

Evaluate the Sage cell below to see a graph of the three surfaces delimiting the region \(D\text{.}\)

Type A.

The given description of \(D\) is basically already a type-A one:

\begin{equation*} D=\{(x,y,z)\colon -1\leq y\leq 1, y^2\leq x\leq 1, 0\leq z\leq 1-x\}\text{.} \end{equation*}

Thus

\begin{align*} \operatorname{vol} D \amp =\iiint\limits_{D} 1\, dV \\ \amp= \int_{-1}^1\int_{y^2}^1\int_0^{1-x}1\, dz\, dx\, dy \\ \amp = \int_{-1}^1\int_{y^2}^11-x\, dx\, dy\\ \amp = \frac{1}{2}\int_{-1}^1(1-y^2)^2\, dy\\ \amp = \int_0^1 1-2y^2+y^4\, dy \amp (\text{even function})\\ \amp = 1-2/3+1/5=8/15\text{.} \end{align*}

Type B.

The type-B description of \(D\) is somewhat trickier, as we must first identify the projection of \(D\) onto the \(yz\)-plane. Since all points \((x,y,z)\in D\) satisfy \(y^2\leq x\) and \(x\leq 1-z\text{,}\) we conclude that \(y^2\leq 1-z\) for all \((x,y,z)\in D\text{,}\) and hence the projection onto the \(yz\)-plane lies within the set

\begin{equation*} \mathcal{R}=\{(0,y,z)\colon -1\leq y\leq 1, y^2\leq 1-z\}=\{(0,y,z)\colon -1\leq y\leq 1, 0\leq z\leq 1-y^2\}\text{.} \end{equation*}

To see that the projection is in fact all of \(\mathcal{R}\text{,}\) notice that for any \(z\leq 1-y^2\) the point \((y^2,y,z)\) lies above the the curve \(x=y^2\) and below the plane \(x+z=1\text{,}\) and hence is an element of \(D\text{.}\) Thus any point \((0,y,z)\in \mathcal{R}\) is the projection of the point \((y^2,y,z)\in D\text{.}\) Finally, for any point \((0,y,z)\in \mathcal{R}\text{,}\) as we move in the \(x\)-direction we enter \(D\) at the cylinder \(x=y^2\) and exit \(D\) at the plane \(x=1-z\text{.}\) This yields the type-B description

\begin{equation*} D=\{(x,y,z)\colon -1\leq y\leq 1,0\leq z\leq 1-y^2, y^2\leq x\leq 1\}\text{.} \end{equation*}

Verify for yourself that the volume computation

\begin{equation*} \operatorname{vol} D=\int_{-1}^1\int_0^{1-y^2}\int_{y^2}^11\, dx\, dz\, dy \end{equation*}

yields \(8/15\text{,}\) just as above.

Example 1.5.14. Region from iterated integral.

Sketch the region \(D\subseteq \R^3\) whose volume is computed by the iterated integral

\begin{equation*} \int_0^1\, \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\, \int_2^{3-y^2-z^2}1\, dx\, dz\, dy\text{.} \end{equation*}
Solution.

The iterated integral corresponds to the type-B description of the region

\begin{equation*} D=\{(x,y,z)\colon 0\leq y\leq 1, -\sqrt{1-y^2}\leq z\leq \sqrt{1-y^2}, 2\leq x\leq 3-y^2-z^2\}\text{.} \end{equation*}

The projection of \(D\) onto the \(yz\)-plane is the right half of the unit disc

\begin{equation*} y^2+z^2\leq 1, x=0\text{.} \end{equation*}

The equation \(x=3-y^2-z^2\) defines a paraboloid that extends in the negative \(x\)-direction from the vertex \((3,0,0)\text{.}\) Its intersection with the plane \(x=2\) is the circle

\begin{equation*} 1=y^2+z^2\text{.} \end{equation*}

We conclude \(D\) consists of all points \((x,y,z)\) satisfying

\begin{align*} 2\amp \leq x\leq 3-y^2-z^2 \\ 0 \amp\leq y \text{.} \end{align*}

Evaluate the Sage cell below for a graph.

Not surprisingly Theorem 1.3.9 and Definition 1.4.4 extend easily to the case of triple integrals. We make these generalizations official below.

Definition 1.5.16. Average value over solid region.

Assume \(f\) is integrable over the measurable solid region \(D\subseteq \R^3\text{.}\) The average value of \(f\) over \(D\text{,}\) denoted \(\operatorname{avg}_D(f)\text{,}\) is defined as

\begin{equation} \operatorname{avg}_D(f)=\frac{1}{\operatorname{vol} D}\iiint\limits_Df(x,y,z)\, dV\text{.}\tag{1.5.13} \end{equation}

Although I want you to practice sketching three-dimensional regions by hand, it is also helpful to use Sage as assistance. The cell below uses Sage's implicit_plot3d routine, which creates a graph of a surface defined by an equation. Sage also has a plot3d which graphs functions \(z=f(x,y)\text{,}\) which is also useful. (See Sage example 1.1.1 for an example using plot3d.)

Sage allows us to easily integrate over elementary regions via iterated integrals. As usual, the hard part is identifying the correct iterated integral to compute; Sage can then do the rest. Consider Example 1.5.13, where we needed to compute \(\iiint\limits_D1\, dV\text{.}\) The hard part was coming up with a suitable description of \(D\text{:}\) e.g.,

\begin{equation*} D=\{(x,y,z)\colon -1\leq y\leq 1, y^2\leq x\leq 1, 0\leq 0\leq 1-x\}\text{.} \end{equation*}

Once this step is completed, the relevant triply iterated integral

\begin{equation*} \int_{-1}^1\int_{y^2}^1\int_0^{1-x}1\, dz\, dx\, dy \end{equation*}

is easy enough to compute.