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

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

$$\iiint_C g(x,y,z)\, dV = \int_e^f \int_c^d \int_a^b g(x,y,z) \, dx \, dy \, dz.$$If $E\subset \mathbb R^3$, then how can we interpret

$$\iiint_E g(x,y,z)\, dV?$$- If $g(x,y,z)\equiv1$, then the integral represents volume of $E$
- If $g$ represents density (mass/unit volume) then the integral represents mass.

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

The function itself may be viewed as a density plot:

$$g(x,y,z) = (16-(x^2+y^2))(z+1)$$The mass may be computed as a triple integral

$$\begin{align} & \int_{-3}^3 \int_{-2}^2 \int_0^1 (16-(x^2+y^2))(z+1) \, dz \, dy \, dx \\ &= 4 \int_0^3 \int_0^2 (16-x^2-y^2) \, \frac{3}{2} \, dy \, dx \\ &= 6 \int_0^3 \left(32 - 2x^2-\frac{8}{3}\right) \, dx \\ &= 6 \left(96 - \frac{2}{3}\times27 - 8\right) = 420. \end{align}$$A tetrahedron is a solid shape with 4 vertices and 4 triangular faces.

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

An equation for the plane containing the top is

$$2 x + 3 y + 6 z = 6 \text{ or } z=1-x/3-y/2.$$If we look at the tetrahedron from above, we see a triangle.

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 $0\leq z \leq 1-x/3-y/2$.

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

$$0\leq y \leq 2-2x/3 \text{ and } 0 \leq x \leq 3.$$Finally, to set up an iterated integral of a function $f$ over the tetrahedron $T$, we should get:

$$\iiint_T f(x,y,z) \, dV = \int_0^3 \int_0^{2-2x/3} \int_0^{1-x/3-y/2} f(x,y,z)\,dz\,dy\,dx.$$