The **triple integral** (also called the *three-dimensional definite integral* or *volume integral*) ∫∫∫_{D}f(x, y, z) dV is an extension of the “usual” integral to three-dimensions. It is often the preferred choice for solving three-dimensional problems like finding center of mass, moments of inertia, or volume of a solid region.

Watch this introduction to triple integrals, or read on below:

## Formal Definition of a Triple Integral

The definition of a triple integral is very similar to the double integral. A three-variable function *f*(*x*, *y*, *z*) over a three-dimensional domain V can be written as (Bronshtein et al., 2013):

It isn’t as challenging as it looks, as long as you’re comfortable with the rules of integration. That’s because it’s **iterated **integrals: all you have to do is integrate three times, working in layers from the inside out.

## Example

**Example problem**: Find the center of mass for a wooden block with side lengths of 4 m. One corner of the cube is positioned at the origin, with the adjacent corners on the positive x, y and z axes. Assume that the box’s density is proportional to the distance from the x-y plane and that *k* = 1 kg per cubic m per m.

**Solution**

Step 1: Set up the triple integral.

Here, *W* represents the cube, *f*(*x*, *y*, *z*) = *kz* is the density of the cube; *k* is a constant.

Step 2: **Evaluate the integral. **You’ll need to do this three times (once for each integral, with respect to x, y, and then z). **Note**: For a cube-like shape, the integral is independent of the order of integration, as long as f(x,y,z) is a continuous function.

- First, integrate with respect to x (treating y and z like constants):

As the lower limit is “0”, this makes for a fairly easy integration. Usually, you’ll have to evaluate the function at the lower and upper limits of integration, subtracting the two to get your area. However, as the lower bound (x = 0) will always equate to zero, you can basically skip that step and just plug in the value you found for the upper limit. - Evaluate the next layer (with respect to y):

- Evaluate once more and you’re done integrating!

You’re told that the block is measured in meters and that k is 1 kg per cubic m per m. So the mass of the block is 128 kg.

## References

Bronshtein, I. et al., Handbook of Mathematics. 2013. Springer Berlin Heidelberg.

Zill, D. & Wright, W. Calculus: Early Transcendentals. 2009. Jones & Bartlett Learning.

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