The Main Idea 

Triple integrals extend the concept of double integrals into three dimensions. Just as a double integral sums up values over a flat region, a triple integral sums up values throughout a three-dimensional rectangular box. Think of it as filling a room with tiny cubes and adding up the function’s value at each cube—then shrinking those cubes infinitely small.

The Setup

A rectangular box [latex]B[/latex] in 3D space is defined by intervals for [latex]x[/latex], [latex]y[/latex], and [latex]z[/latex]:

[latex]B = {(x,y,z) \mid a \leq x \leq b, c \leq y \leq d, e \leq z \leq f}[/latex].

We slice this box into smaller subboxes, each with volume [latex]\Delta V = \Delta x \Delta y \Delta z[/latex]. The triple integral is the limit of the Riemann sum as these subboxes get infinitely small:

[latex]{\displaystyle\lim_{{l,m,n}{\rightarrow}{\infty}}} \ {\displaystyle\sum_{i=1}^{l}}{\displaystyle\sum_{j=1}^{m}}{\displaystyle\sum_{k=1}^{n}} \ {f}{({{x}^{*}_{ijk}},{{y}^{*}_{ijk}},{{z}^{*}_{ijk}})}{{\Delta}{x}}{{\Delta}{y}}{{\Delta}{z}}[/latex]

Computing Triple Integrals

Fubini’s theorem is your computational workhorse here. For continuous functions over rectangular boxes, you can evaluate the triple integral as an iterated integral in any order:

[latex]\underset{B}{\iiint} f(x,y,z) , dV = \int_{e}^{f} \int_{c}^{d} \int_{a}^{b} f(x,y,z) , dx , dy , dz[/latex]

There are six possible orderings (integrate [latex]x[/latex], [latex]y[/latex], [latex]z[/latex] in any sequence), and for rectangular boxes, all produce the same result.

Problem-Solving Strategy

Work from the inside out—integrate with respect to the innermost variable first, treating the other variables as constants. Then move outward, evaluating at the bounds after each integration step. For rectangular regions, the order doesn’t affect difficulty, so choose whichever feels most natural.

Key Condition

The function must be continuous on [latex]B[/latex] for the triple integral to exist, though continuity everywhere except possibly on the boundary is sufficient.

The Main Idea 

Moving beyond rectangular boxes, triple integrals can handle more complex three-dimensional regions by using projections and bounding surfaces. The key is to visualize the solid region as sitting between two surfaces, with its footprint (projection) on one of the coordinate planes.

The Core Approach

For a general region [latex]E[/latex], identify which coordinate plane works best for projection. If projecting onto the [latex]xy[/latex]-plane, describe the region as:

[latex]E = {(x,y,z) \mid (x,y) \in D, u_1(x,y) \leq z \leq u_2(x,y)}[/latex]

Here, [latex]D[/latex] is the shadow of [latex]E[/latex] on the [latex]xy[/latex]-plane, and [latex]z[/latex] varies between two surfaces [latex]z = u_1(x,y)[/latex] (bottom) and [latex]z = u_2(x,y)[/latex] (top). The triple integral becomes:

[latex]\underset{E}{\iiint} f(x,y,z) , dV = \underset{D}{\iint} \left[\int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z) , dz\right] dA[/latex]

You can project onto any coordinate plane—just adjust which variable is bounded by surfaces and which two form the projection region.

Setting Up the Integral

The projection region [latex]D[/latex] can be Type I or Type II. For Type I regions in the [latex]xy[/latex]-plane, [latex]x[/latex] has constant bounds and [latex]y[/latex] depends on [latex]x[/latex]. For Type II regions, [latex]y[/latex] has constant bounds and [latex]x[/latex] depends on [latex]y[/latex]. Choose the type that simplifies your bounds.

Problem-Solving Strategy

1. Sketch the region and identify natural bounding surfaces
2. Determine which projection plane makes the bounds clearest
3. Write out the region description carefully, noting which variables have constant bounds versus function bounds
4. Integrate from innermost to outermost, working through one variable at a time

Finding Volume

To find the volume of a solid region [latex]E[/latex], evaluate [latex]\underset{E}{\iiint} 1 , dV[/latex]. The integrand is simply 1, so you’re adding up all the tiny volume elements in the region.

The Main Idea 

For triple integrals over general regions, the order of integration can dramatically affect how difficult the computation becomes. Unlike rectangular boxes where order doesn’t matter much, choosing the right order for bounded regions can mean the difference between a manageable integral and an impossible one.

Why Change the Order?

Sometimes the given order leads to complicated integrands or difficult bounds. By changing the order, you might simplify the function you’re integrating or get cleaner limits. Occasionally, switching to polar coordinates in one of the coordinate planes can also make the calculation much easier.

How to Change the Order

The key tool is visualization. Sketch the region [latex]E[/latex] and all three of its projections onto the [latex]xy[/latex]-, [latex]xz[/latex]-, and [latex]yz[/latex]-planes. Each projection gives you a different way to describe the same region. From the original integral’s bounds, determine what the region [latex]E[/latex] looks like. Then rewrite those bounds to reflect your new desired order.

Problem-Solving Strategy

1. Sketch the region [latex]E[/latex] in three dimensions if possible
2. Draw all three projections onto the coordinate planes
3. Write out the region description in set notation from the original bounds
4. Determine which projection plane makes the most sense for your new order
5. Rewrite the bounds based on that projection, ensuring each variable’s limits match the geometry
6. If you see circular or radial symmetry in a projection, consider using polar coordinates in that plane

When to Consider Polar Coordinates

If a projection onto the [latex]xz[/latex]– or [latex]yz[/latex]-plane shows circular symmetry (like [latex]x^2 + z^2 \leq r^2[/latex]), use polar substitution in that plane just as you would in the [latex]xy[/latex]-plane.

The Main Idea 

Just as you can find the average value of a function over a region in the plane, you can find the average value of a function throughout a three-dimensional solid. The idea is intuitive: integrate the function over the entire solid, then divide by the solid’s volume to get the “typical” value.

The Formula

For a function [latex]f(x,y,z)[/latex] that’s integrable over a solid region [latex]E[/latex] with positive volume [latex]V(E)[/latex], the average value is:

[latex]f_{\text{ave}} = \frac{1}{V(E)} \underset{E}{\iiint} f(x,y,z) \, dV[/latex]

This formula has two parts: the numerator is the triple integral of the function over [latex]E[/latex], and the denominator is the volume of [latex]E[/latex], which you find by evaluating [latex]V(E) = \underset{E}{\iiint} 1 \, dV[/latex].

Problem-Solving Strategy

1. Set up the region [latex]E[/latex] with appropriate bounds based on the geometry
2. Evaluate the triple integral of the function: [latex]\underset{E}{\iiint} f(x,y,z) \, dV[/latex]
3. Evaluate the volume integral: [latex]V(E) = \underset{E}{\iiint} 1 \, dV[/latex]
4. Divide the function integral by the volume to get the average value

Why This Makes Sense

Think of it like finding the average temperature in a room. You’re summing up all the temperature values at every point in the space, then dividing by the total volume to get a single representative temperature. The triple integral accumulates all the values, and dividing by volume normalizes it to a meaningful average.