Triple integrals

Why integrate just twice, when you could integrate three times?

Writing it down

If $C = [a, b] \times [c, d] \times [e, f]$ is a cuboid in 3D space, then

∭_{C} g (x, y, z) d V = \int_{e}^{f} \int_{c}^{d} \int_{a}^{b} g (x, y, z) d x d y d z .

Doing it

\begin{aligned} \int_{0}^{3} \int_{0}^{2} \int_{0}^{1} x y^{2} z^{3} d z d y d x = \int_{0}^{3} \int_{0}^{2} x y^{2} \frac{1}{4} z^{4} |_{z = 0}^{1} d y d x \\ = \frac{1}{4} \int_{0}^{3} \int_{0}^{2} x y^{2} d y d x = \frac{1}{4} \int_{0}^{3} \frac{1}{3} x y^{3} |_{y = 0}^{2} d x \\ = \frac{1}{4} \int_{0}^{3} \frac{8}{3} x d x = \frac{1}{4} \times \frac{8}{3} \times \frac{9}{2} = 3. \end{aligned}

Interpreting it

If $E \subset R^{3}$ , then how can we interpret

∭_{E} g (x, y, z) d V ?

If $g (x, y, z) \equiv 1$ , then the integral represents volume of $E$
If $g$ represents density (mass/unit volume) then the integral represents mass.

Example

Suppose that the cuboid $C = [- 3, 3] \times [- 2, 2] \times [0, 1]$ has density

g (x, y, z) = (16 - (x^{2} + y^{2})) (z + 1) .

Use this to determine the mass of $C$

The domain

The domain is a cuboid: $C = [- 3, 3] \times [- 2, 2] \times [0, 1]$ . It looks like so:

A cuboid in space

The function

The function itself may be viewed as a density plot:

A cuboid in space

g (x, y, z) = (16 - (x^{2} + y^{2})) (z + 1)

The mass

The mass may be computed as a triple integral

\begin{aligned} \int_{- 3}^{3} \int_{- 2}^{2} \int_{0}^{1} (16 - (x^{2} + y^{2})) (z + 1) d z d y d x \\ = 4 \int_{0}^{3} \int_{0}^{2} (16 - x^{2} - y^{2}) \frac{3}{2} d y d x \\ = 6 \int_{0}^{3} (32 - 2 x^{2} - \frac{8}{3}) d x \\ = 6 (96 - \frac{2}{3} \times 27 - 8) = 420. \end{aligned}

Tetrahedral domains

A tetrahedron is a solid shape with 4 vertices and 4 triangular faces.

A tetrahedron

The tetrahedron $T$ above has vertices at the origin, $(3, 0, 0)$ , $(0, 2, 0)$ , and $(0, 0, 1)$ .

Plane equation

A tetrahedron with one face labeled

An equation for the plane containing the top is

2 x + 3 y + 6 z = 6 or z = 1 - x / 3 - y / 2.

Tetrahedral integration

If we look at the tetrahedron from above, we see a triangle.

Integrating over a tetrahedron

To integrate with respect to $z$ first, we move from point $(x, y, 0)$ in the triangle to the top face at $(x, y, 1 - x / 3 - y / 2)$ . That yields inner bounds of integration on $z$ of $0 \leq z \leq 1 - x / 3 - y / 2$ .

Triangular integration

Integrating over the remaining triangle

After we've integrated with respect to $z$ , we project down to the $x y$ plane and integrate over the triangle that we see. That yields further bounds of integration

0 \leq y \leq 2 - 2 x / 3 and 0 \leq x \leq 3.

Putting it all together

Finally, to set up an iterated integral of a function $f$ over the tetrahedron $T$ , we should get:

∭_{T} f (x, y, z) d V = \int_{0}^{3} \int_{0}^{2 - 2 x / 3} \int_{0}^{1 - x / 3 - y / 2} f (x, y, z) d z d y d x .

Triple integrals Why integrate just twice, when you could integrate three times?