M408M Learning Module Pages

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Chapter 10: Parametric Equations and Polar Coordinates

Chapter 12: Vectors and the Geometry of Space


Chapter 13: Vector Functions


Chapter 14: Partial Derivatives


Chapter 15: Multiple Integrals


Learning module LM 15.1: Multiple integrals

Learning module LM 15.2: Multiple integrals over rectangles:

Learning module LM 15.3: Double integrals over general regions:

Learning module LM 15.4: Double integrals in polar coordinates:

Learning module LM 15.5a: Multiple integrals in physics:

      Mass, center of mass, and moment of inertia
      Triple integrals in physics

Learning module LM 15.5b: Integrals in probability and statistics:

Learning module LM 15.10: Change of variables:


Triple integrals in physics

Triple integrals in physics

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.

The mass of a little box of volume $dV$ at position $(x,y,z)$ is approximately $\rho(x,y,z) dV$, so the total mass of a solid object $S$ is $$M \ = \ \iiint_S \rho(x,y,z) dV.$$

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:

The location of the center-of-mass is as $(\bar x, \bar y , \bar z)$, where \begin{eqnarray*} \bar x & = & \frac{1}{M} \iiint_S x \,\rho(x,y,z) \, dV, \\ \bar y & = & \frac{1}{M} \iiint_S y \,\rho(x,y,z) \, dV, \\ \bar z & = & \frac{1}{M} \iiint_S z \,\rho(x,y,z) \, dV. \end{eqnarray*}

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.