On the last slide, we learned how to work with 2-dimensional laminas. Of course, the world isn't 2 dimensional! In the 3-dimensional world, density is mass per unit volume. Liquid water has a density of 1 gram per cubic centimeter, while osmium, the densest metal, has a density of 22.6 g/cc. Air at sea level has a density of $1.2 \times 10^{-3}$ g/cc, or roughly 1/800 that of water, while air on top of Mt. Everest is only a third as dense as air at sea level. To get a mass per unit area, you have to integrate the mass per unit volume over the third variable. For instance, the atmospheric pressure (force per unit area) that you feel at ground level is the weight of all the air above you.

This is usually computed in rectangular coordinates, with $dV = dx\, dy\, dz$, but sometimes it's easier to use cylindrical coordinates, with $dV = r \,dr \,d\theta\, dz$, or spherical coordinates. Center of mass works almost the same as in 2 dimensions:

You have to be careful with moment of inertia, since that depends on which axis you are rotating around. For rotations around the $z$ axis, the moment of inertia is $$I_3 = \iiint_S (x^2+y^2) \rho(x,y,z) dV,$$ since the distance from $(x,y,z)$ to the $z$ axis is $r=\sqrt{x^2+y^2}$. However, if you are rotating around the $x$ axis, then the moment of inertia is $I_1=\iiint_S (y^2+z^2) \rho(x,y,z) dV$, and if you are rotating around the $y$ axis, then the moment of inertia is $I_2=\iiint_S (x^2+z^2) \rho(x,y,z) dV.$

Rotating around diagonal axes is even more complicated. To deal with such cases, physicists define a moment of inertia tensor. This is a $3 \times 3$ matrix whose $ij$ entry is $\iiint_S x_i x_j \rho(x,y,z) dV$, where $x_1$ means $x$, $x_2$ means $y$, and $x_3$ means $z$. The moment of inertia around any axis can be computed from this matrix.