Chapter 17

Triple Integrals

17.1Triple Integrals as Iterated Integrals

Triple integrals follow a very similar general theory to double integrals: starting with a function $𝑓 (𝑥, 𝑦, 𝑧)$ on $ℝ 3$ , we define the integral over a region $𝐸$ by breaking that region into small cubical volumes of size $Δ 𝑉$ and building a 3-dimensional Riemann sum. Taking the limit gives the triple integral, or

∭ 𝐸 𝑓 𝑑 𝑉 .

Before chopping the region into pieces, it is worth remembering what is being accumulated. A scalar field attaches a number to every point in a three-dimensional region.

Figure 17.1 Before anything is chopped up, here is the thing being added: a number attached to every point of a room. Warm where the haze is red, cool where it is blue, and clear where there is nothing---the field passing through zero rather than stopping. Read it as a temperature, or as the density of something the room is full of. A triple integral answers the obvious question about a picture like this: how much is there altogether?

Definition 17.1 (Triple Integral). If the Riemann sums approach a limit as the volumes of the pieces become arbitrarily small, we define the triple integral of $𝑓$ over $𝐸$ to be

∭ 𝐸 𝑓 (𝑥, 𝑦, 𝑧) 𝑑 𝑉 = l i m m a x (Δ 𝑉) \to 0 \sum 𝑖 \sum 𝑗 \sum 𝑘 𝑓 (𝑥 𝑖, 𝑦 𝑗, 𝑧 𝑘) Δ 𝑉 .

The next animation is this definition on a coordinate box. The first approximation uses only one cell; dividing each direction makes the Riemann sum settle toward the integral.

Figure 17.2 Start with one cell---the whole room, and a single reading of

𝑓

taken at its middle. That one number times the volume of the room is a terrible estimate, and it is already a Riemann sum. Now cut each side in two, and again: every cell takes its own reading, the block colours begin to reproduce the haze from the figure above, and the total settles down. Nothing here is new except the number of directions there are to cut in.

Taking $𝑓 = 1$ counts the volume of the solid, taking $𝑓 = 𝜌$ accumulates a density to find mass, and dividing by the volume of the solid gives the average value:

V o l u m e (𝐸) = ∭ 𝐸 1 𝑑 𝑉, 𝑀 = ∭ 𝐸 𝜌 𝑑 𝑉, 𝑓 a v g = 1 V o l u m e (𝐸) ∭ 𝐸 𝑓 𝑑 𝑉 .

The definition works for regions that are not boxes as well. The pieces that belong to the solid are kept; those outside it are discarded.

Figure 17.3 The same subdivision, over solids that are not boxes. Now most cells of the grid are simply not in the object, and the ones that are make a staircase along its boundary---which shrinks as the cells do. Switch to

𝑓 = 1

and the sum counts volume instead; that is the only difference between "how much stuff is in here" and "how big is this". What the figure cannot do is tell you which cells to keep in symbols, and that is the rest of this chapter.

To evaluate such an expression, we need to break the integral into slices, and evaluate them one at a time. Such slicing relies on understanding the volume element in three dimensions, which is the volume of an infinitesimal box. Since a box's volume is given by length times width times height, we can express $𝑑 𝑉$ as a product of three infinitesimal lengths:

𝑑 𝑉 = 𝑑 𝑥 𝑑 𝑦 𝑑 𝑧 .

This lets us separate the triple integral into three consecutive integrals. There are six total Cartesian orders:

𝑑 𝑥 𝑑 𝑦 𝑑 𝑧, 𝑑 𝑥 𝑑 𝑧 𝑑 𝑦, 𝑑 𝑦 𝑑 𝑥 𝑑 𝑧,

𝑑 𝑦 𝑑 𝑧 𝑑 𝑥, 𝑑 𝑧 𝑑 𝑥 𝑑 𝑦, 𝑑 𝑧 𝑑 𝑦 𝑑 𝑥 .

Their equivalence does not follow merely from rearranging the differential symbols; it follows from the theorem that allows a multiple integral to be evaluated as an iterated integral.

Theorem 17.2 (Iterated Integration in Three Dimensions). Suppose $𝑓$ is continuous on a bounded solid $𝐸$ , and that $𝐸$ can be described in the nested form

𝑎 \leq 𝑥 \leq 𝑏, 𝑔 1 (𝑥) \leq 𝑦 \leq 𝑔 2 (𝑥), ℎ 1 (𝑥, 𝑦) \leq 𝑧 \leq ℎ 2 (𝑥, 𝑦),

with continuous boundary functions. Then

∭ 𝐸 𝑓 𝑑 𝑉 = \int 𝑏 𝑎 \int 𝑔 2 (𝑥) 𝑔 1 (𝑥) \int ℎ 2 (𝑥, 𝑦) ℎ 1 (𝑥, 𝑦) 𝑓 (𝑥, 𝑦, 𝑧) 𝑑 𝑧 𝑑 𝑦 𝑑 𝑥 .

The analogous statement holds for any other order in which the solid has such a description.

When the domain $𝐸 \subset ℝ 3$ is a coordinate box, described as

𝐸 = {(𝑥, 𝑦, 𝑧) ∣ 𝑎 \leq 𝑥 \leq 𝑏, 𝑐 \leq 𝑦 \leq 𝑑, 𝑒 \leq 𝑧 \leq 𝑘},

the triple integral splits into an iterated integral with constant bounds:

∭ 𝐸 𝑓 𝑑 𝑉 = \int 𝑏 𝑎 \int 𝑑 𝑐 \int 𝑘 𝑒 𝑓 𝑑 𝑧 𝑑 𝑦 𝑑 𝑥 .

All six orders are available because every bound is constant. The geometry does not favor an order, although the integrand still might.

Example 17.3. Compute

∭ 𝐸 𝑥 𝑒 𝑦 𝑑 𝑉, 𝐸 = {(𝑥, 𝑦, 𝑧) ∣ 1 \leq 𝑥 \leq 2, 0 \leq 𝑦 \leq 1, 2 \leq 𝑧 \leq 5} .

First we choose an order: say we integrate $𝑑 𝑧$ first. These bounds go from $2$ to $5$ so the integral along our slice is

S l i c e = \int 52 𝑥 𝑒 𝑦 𝑑 𝑧 = 𝑥 𝑒 𝑦 𝑧 ∣ 52 = 𝑥 𝑒 𝑦 (5 - 2) = 3 𝑥 𝑒 𝑦 .

Now we just have to do the double integral of this over the rectangle $𝑅 = {1 \leq 𝑥 \leq 2, 0 \leq 𝑦 \leq 1}$ containing all of the slices:

∭ 𝐸 𝑥 𝑒 𝑦 𝑑 𝑉 = \iint 𝑅 3 𝑥 𝑒 𝑦 𝑑 𝐴 .

Again we choose a direction to slice: starting with $𝑑 𝑦$ , we decompose this into an iterated integral as

\iint 𝑅 3 𝑥 𝑒 𝑦 𝑑 𝐴 = \int 21 \int 10 3 𝑥 𝑒 𝑦 𝑑 𝑦 𝑑 𝑥

= \int 21 3 𝑥 (\int 10 𝑒 𝑦 𝑑 𝑦) 𝑑 𝑥 = \int 21 3 𝑥 (𝑒 𝑦 ∣ 10) 𝑑 𝑥

= \int 21 3 𝑥 (𝑒 - 1) 𝑑 𝑥 = 32 𝑥 2 (𝑒 - 1) ∣ 21 = 92 (𝑒 - 1) .

17.2Slicing Solids and Describing Their Projections

Because we are going to evaluate a triple integral as three iterated integrals, we can reuse a lot of what we learned about double integrals along the process. In particular, we can think of the process of computing a triple integral as first choosing one of the directions to integrate, and then treating the two remaining directions as a double integral over a region parameterizing all the slices.

In short,

a t r i p l e - i n t e g r a l s e t u p i s a l i n e - s l i c i n g p r o b l e m f o l l o w e d b y a d o u b l e - i n t e g r a l s e t u p .

The slicing direction is a real choice. Different directions produce different line slices and different shadows, and one direction may describe the solid much more simply than another.

Figure 17.4 First decide which way to slice. The solid is drawn as the family of lines that sweeps it out, and there are three families to choose from; switching between them fills the same solid with lines running a different way and casts its shadow onto a different coordinate plane. For the tetrahedron every face is a plane and all three cost the same, which is what "six orders are available" looks like. For the paraboloid they are not equal at all: slicing downward gives clean bounds, while slicing sideways puts a square root in the innermost limit. Choosing the direction is the new skill here; everything after it is a double integral over the shadow you just chose.

For example, suppose we choose line slices parallel to the $𝑧$ -axis. A solid that each vertical line enters once and leaves once can be written

𝐸 = {(𝑥, 𝑦, 𝑧) ∣ (𝑥, 𝑦) \in 𝑅, 𝑏 (𝑥, 𝑦) \leq 𝑧 \leq 𝑡 (𝑥, 𝑦)} .

The bottom and top surfaces determine the integral along one line:

S l i c e (𝑥, 𝑦) = \int 𝑡 (𝑥, 𝑦) 𝑏 (𝑥, 𝑦) 𝑓 (𝑥, 𝑦, 𝑧) 𝑑 𝑧 .

The region $𝑅$ is the projection of $𝐸$ onto the $𝑥 𝑦$ -plane: the shadow the solid casts when viewed in the $𝑧$ -direction. Adding up the line slices over that shadow gives

∭ 𝐸 𝑓 𝑑 𝑉 = \iint 𝑅 (\int 𝑡 (𝑥, 𝑦) 𝑏 (𝑥, 𝑦) 𝑓 (𝑥, 𝑦, 𝑧) 𝑑 𝑧) 𝑑 𝐴 .

This gives a consistent setup routine:

Choose a direction for the line slices.
Find where each line enters and leaves the solid.
Find the projection of the solid on the perpendicular coordinate plane.
Describe that projected region using the methods from Double Integrals.
Assemble the iterated integral from the inside outward.

The next demonstration shows these layers together: one highlighted line in the solid, the projection underneath it, and the ordinary double-integral slices that describe that projection.

Figure 17.5 Now the setup, drawn the way you would draw it. On the left is the solid with the direction of integration running through it as a family of lines, and the face they stand on shaded: the shadow. One strip of those lines is picked out in gold---that strip is the innermost integral, and it rises from one surface to the other. On the right the shadow is drawn again face-on, as the plane region it is, with the gold slice showing where that strip sits in it. Choosing where to slice that region is the double-integral problem from the last chapter, and the whole triple integral is those two decisions stacked. Work through the menu: the projections run rectangle, triangle, disc, and the vertical bounds run constant, variable top, variable top and bottom.

This allows us to think about triple integrals as not a new thing, but just adding one more direction to a process we already understand well.

Example 17.4. For example, consider the following domain $𝐸$ :

𝐸 = {(𝑥, 𝑦, 𝑧) ∣ 0 \leq 𝑥 \leq 2, 0 \leq 𝑦 \leq 3, 0 \leq 𝑧 \leq 𝑥 + 𝑦} .

Here the $𝑧$ bound is different depending on which point $(𝑥, 𝑦)$ you are at, so we do the $𝑧$ -integral first:

S l i c e (𝑥, 𝑦) = \int 𝑥 + 𝑦 0 𝑓 𝑑 𝑧 .

Since both the $𝑥$ and $𝑦$ bounds are constants, the remaining region in the $𝑥 𝑦$ -plane is a rectangle, and we know how to integrate over these:

∭ 𝐸 𝑓 𝑑 𝑉 = \iint 𝑅 S l i c e 𝑑 𝐴 = \int 20 \int 30 S l i c e 𝑑 𝑦 𝑑 𝑥 .

Putting this all together gives a representation of the triple integral as an iterated integral:

∭ 𝐸 𝑓 𝑑 𝑉 = \int 20 \int 30 \int 𝑥 + 𝑦 0 𝑓 𝑑 𝑧 𝑑 𝑦 𝑑 𝑥 .

We then evaluate this triple integral as three one dimensional integrals from Calculus I.

The bounds also tell us how the pieces are nested. For a fixed order, the inner bounds may depend on both variables that remain, the middle bounds may depend only on the outer variable, and the outer bounds are constants. This is a check on one representation of the solid, not a claim that the solid has only one possible order.

17.2.1Reading an Integral Backward

We should also be able to begin with an iterated integral and reconstruct its geometry. Read from the inside outward: first the line slices, then their projection.

Example 17.5. Describe the solid represented by the integral and evaluate it:

\int 20 \int 2 - 𝑥 0 \int 2 𝑥 + 𝑦 1 𝑑 𝑧 𝑑 𝑦 𝑑 𝑥 .

The inner bounds say that a vertical line enters the solid at the plane $𝑧 = 𝑥 + 𝑦$ and leaves at the plane $𝑧 = 2$ . The remaining bounds describe the triangular projection

𝑅 = {(𝑥, 𝑦) ∣ 0 \leq 𝑥 \leq 2, 0 \leq 𝑦 \leq 2 - 𝑥}

in the $𝑥 𝑦$ -plane. Thus the solid lies above that triangle, above $𝑧 = 𝑥 + 𝑦$ , and below $𝑧 = 2$ . Its volume is

\int 20 \int 2 - 𝑥 0 \int 2 𝑥 + 𝑦 1 𝑑 𝑧 𝑑 𝑦 𝑑 𝑥 = \int 20 \int 2 - 𝑥 0 (2 - 𝑥 - 𝑦) 𝑑 𝑦 𝑑 𝑥 = 12 \int 20 (2 - 𝑥) 2 𝑑 𝑥 = 43 .

17.2.2Discovering the Projection

Sometimes we have to do some work to solve for the bounds given equations describing the region $𝐸$ . Below is one such example.

Example 17.6. Write the integral of $𝑓$ over the region $𝐸$ as an iterated integral, where $𝐸$ is bounded by the $𝑥 𝑦$ -plane and the surface $𝑧 = 1 - 𝑥 2 - 𝑦 2$ .

Since the $𝑥 𝑦$ -plane is given by $𝑧 = 0$ , the surfaces we are given directly describe the $𝑧$ bounds, which suggests we begin with the $𝑑 𝑧$ integral:

S l i c e (𝑥, 𝑦) = \int 1 - 𝑥 2 - 𝑦 2 0 𝑓 𝑑 𝑧 .

But now we are left to discover the remaining bounds for ourselves! How do we do this? The region $𝑅$ over which we need to add up the $𝑧$ slices can be described as the points $(𝑥, 𝑦)$ where the $𝑧$ bounds define an interval of integration. The boundary of this region is where the $𝑧$ bounds collapse to become equal! That is, $𝑅$ is cut out by what we get from equating the bounds

𝑧 = 1 - 𝑥 2 - 𝑦 2 a n d 𝑧 = 0 .

This implies $𝑥 2 + 𝑦 2 = 1$ , so the region $𝑅$ is bounded by the unit circle. We are now down to a double integral where we must again choose an order of slicing. If the bounds we had were given in $𝑥 =$ or $𝑦 =$ form such a choice might be obvious, but here there is no natural one, so we just need to choose and solve for the correct bounds.

Choosing to do $𝑦$ next, we solve to get $𝑦 = \pm \sqrt 1 - 𝑥 2$ as the upper and lower bounds of the integral. The interval for $𝑥$ integration is similarly given by the points on the line where these $𝑦$ bounds define an interval: its boundary is where they collapse to be equal! Setting $\sqrt 1 - 𝑥 2 = - \sqrt 1 - 𝑥 2$ , we see the only solution is where this quantity is zero, so $𝑥 = \pm 1$ . This gives the iterated integral

∭ 𝐸 𝑓 𝑑 𝑉 = \int 1 - 1 \int \sqrt 1 - 𝑥 2 - \sqrt 1 - 𝑥 2 \int 1 - 𝑥 2 - 𝑦 2 0 𝑓 𝑑 𝑧 𝑑 𝑦 𝑑 𝑥 .

This last example looks rather intimidating with the square roots in the bounds, as this means the final integral we have to do will likely involve $\sqrt 1 - 𝑥 2$ after we plug these in. Such integrals often require trigonometric substitution to evaluate, and while this is often possible, we will learn in the next chapter a way to simplify such integrals with a change of coordinates.

Before accepting any setup, make three quick checks:

The inner bounds lie on boundary surfaces of the solid.
The remaining four bounds describe the complete projection.
The lower bound stays below the upper bound throughout that projection.

Finally, a line in the chosen direction may intersect the solid in more than one separated interval. In that case the solid may need to be subdivided, just as a planar region sometimes had to be subdivided when setting up a double integral.