Triple Integrals — Volume, Mass & Evaluation Methods

Triple integrals add a function over a three-dimensional region. The main uses students see first are volume, when the integrand is $1$, and mass, when the integrand is a density function.

You usually write it as

$$\iiint_E f(x,y,z)\,dV,$$

where $E$ is the solid and $dV$ is a tiny volume element. To evaluate a triple integral, you usually rewrite it as an iterated integral with bounds that match the solid.

What a triple integral means

There are three pieces to read:

• $f(x,y,z)$ is the quantity being added up.
• $E$ is the solid region where you add it.
• $dV$ means a tiny piece of volume.

So $\iiint_E f(x,y,z)\,dV$ means "add the values of $f$ over all the tiny volume pieces in $E$."

That interpretation changes with the integrand:

• If $f(x,y,z)=1$, the result is volume.
• If $f(x,y,z)=\rho(x,y,z)$ is density, the result is mass.
• If $f$ is temperature, charge density, or another distributed quantity, the result is the total amount of that quantity over the solid.

The result is not automatically a volume. That depends on what the integrand represents.

How to write a triple integral as an iterated integral

Most course problems are evaluated one variable at a time. Under the usual conditions used in calculus, you rewrite the triple integral as an iterated integral such as

$$\iiint_E f(x,y,z)\,dV = \int_a^b \int_{g_1(x)}^{g_2(x)} \int_{h_1(x,y)}^{h_2(x,y)} f(x,y,z)\,dz\,dy\,dx.$$

The exact bounds depend on the region. The main idea is simple: the inner limits describe the innermost slice, the next limits describe the stack of those slices, and the outer limits describe the full sweep through the solid.

You can often change the order of integration, but the bounds must change with it. That is where many setup errors happen.

Which coordinate system makes the setup easier

Cartesian coordinates

Use Cartesian coordinates when the solid is naturally described by planes or rectangular bounds, such as boxes and simple regions cut by graphs like $z = x+y$.

Then

$$dV = dx\,dy\,dz$$

up to the order you choose.

Cylindrical coordinates

Use cylindrical coordinates for regions with circular symmetry around an axis, such as cylinders or cones. With

$$x=r\cos\theta, \quad y=r\sin\theta, \quad z=z,$$

the volume element becomes

$$dV = r\,dr\,d\theta\,dz.$$

The extra factor $r$ is not optional. It comes from the coordinate change.

Spherical coordinates

Use spherical coordinates when spheres or spherical symmetry make the region easier to describe. One common convention is

$$x=\rho\sin\phi\cos\theta, \quad y=\rho\sin\phi\sin\theta, \quad z=\rho\cos\phi,$$

with volume element

$$dV = \rho^2 \sin\phi \, d\rho\,d\phi\,d\theta.$$

Conventions for angles can vary by course, so it is worth checking which one your class uses.

Worked example: mass on the unit cube

Find the mass of the unit cube

$$E = \{(x,y,z) : 0 \le x \le 1,\ 0 \le y \le 1,\ 0 \le z \le 1\}$$

if the density is

$$\rho(x,y,z)=1+x+y+z.$$

Because the region is a box, Cartesian coordinates are the natural choice:

$$M=\iiint_E (1+x+y+z)\,dV = \int_0^1 \int_0^1 \int_0^1 (1+x+y+z)\,dz\,dy\,dx.$$

Integrate with respect to $z$ first:

$$\int_0^1 (1+x+y+z)\,dz = \left[(1+x+y)z + \frac{z^2}{2}\right]_0^1 = 1+x+y+\frac{1}{2}.$$

So the integral becomes

$$\int_0^1 \int_0^1 \left(\frac{3}{2}+x+y\right)\,dy\,dx.$$

Now integrate with respect to $y$:

$$\int_0^1 \left(\frac{3}{2}+x+y\right)\,dy = \left[\left(\frac{3}{2}+x\right)y + \frac{y^2}{2}\right]_0^1 = 2+x.$$

Then integrate with respect to $x$:

$$\int_0^1 (2+x)\,dx = \left[2x+\frac{x^2}{2}\right]_0^1 = 2+\frac{1}{2} = \frac{5}{2}.$$

So the mass is

$$M=\frac{5}{2}.$$

This example shows the difference between volume and mass clearly. If the density had been $1$ everywhere, the same region would have volume $1$. Because the density is larger than $1$ on much of the cube, the mass comes out larger.

Common triple integral mistakes

1. Using bounds that do not actually describe the solid.
2. Forgetting that the order of integration controls which limits can depend on which variables.
3. Treating the outer variables as active during the inner step instead of holding them constant.
4. Dropping the Jacobian factor in cylindrical or spherical coordinates.
5. Calling the answer "volume" when the integrand is not $1$.

When triple integrals are used

Triple integrals appear when a quantity is distributed through a volume rather than along a line or across a surface.

• In geometry, they give volume.
• In physics and engineering, they give mass when density varies through a solid.
• In electromagnetism and fluid models, they add charge, energy, or other quantities over a 3D region.
• In probability, they can integrate a density over a 3D domain when three continuous variables are involved.

The setup depends on the region. The interpretation depends on the integrand.

Quick check before you integrate

Before doing the algebra, ask:

1. What does the integrand represent here: volume density, mass density, or something else?
2. What solid am I integrating over, exactly?
3. Would a different coordinate system make the bounds simpler?

Those three checks usually catch more mistakes than any symbolic trick.

Try a similar problem

Try your own version with $\rho(x,y,z)=2+x$ on the same unit cube and compute the mass. Then explore a cylinder and decide whether cylindrical coordinates make the bounds simpler before you start integrating.