Mass and Moment via Integration

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Mass and Moments

Based (mostly) on “Center of Mass in Two Dimensions” in section 4.6 of the textbook.

Key Ideas

The formula for calculating mass from density:

Moment is mass weighted by distance from an axis or balance point; the formula for moment captures that:

Center of mass, in 2 dimensions, is the point ( M_y/m, M_x/m ).

All of the above scale to 3 dimensions.

Examples

All of these examples are good places to use Mathematica to keep track of the values you need and to do your calculations. You can download a notebook with the class solutions to the examples from Canvas.

2 Dimensions

Imagine a thin plate lying between the x axis and the curve y = 1 - x². If the density of the plate is given by ρ(x,y) = x²y, what is the total mass of the plate? What are its moments, and where is the center of mass?

One of the implicit but important parts of solving this was figuring out where the endpoints of the plate on the x axis are. Do this by solving y = 1 - x² = 0 to find the points where the upper side of the plate intersects the bottom (namely x = -1 and x = 1). See the Mathematica notebook for the rest of the solution.

3 Dimensions

A pyramid is balanced on its tip at the origin of a coordinate system. The pyramid has height 1, and its sides lie in the planes z = x, z = -x, z = y, and z = -y. The density of the pyramid is ρ(x,y,z) = e^z - xy². Which side of the pyramid should you stick a shim under to keep it from falling?

This problem immediately runs into the question of what the bounds of the integrals for mass, moment, etc. should be. A good way to figure out bounds for any multidimensional integral is to start by identifying a dimension in which the bounds are simple constants. Make that the outermost integral. In this problem the only such dimension is z. Then work in through the integrals, identifying dimensions where the bounds can be calculated from the variables defined by outer integrals. In this problem, both y and x depend on z, namely both range from -z to +z. This thinking leads to integrals that look like…

Once again, see the Mathematica notebook for calculations of the relevant integrals.

At the end of those calculations, moment about the xz plane is 0, meaning that the pyramid is perfectly balanced in the y dimension. But moment about the yz plane is negative, meaning that the pyramid is unbalanced towards its negative-x side. So putting a shim under that side would be a good idea.

Next

We don’t necessarily have to integrate a multivariable function over a region in its domain. We could also integrate along a curve through the domain. This is called a “line integral,” and makes a nice transition from multivariable functions to the final module of this course, vector calculus.

To learn something about line integrals, please read “Scalar Line Integrals” in section 5.2 of the textbook.

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