We use integrals to model mass.
1 Mass
We learned some time ago that if the density of an object is uniform,
\[ \text {mass} = \text {density}\times \text {volume}. \]
When the
density of an object is not uniform, we define a density function \(\rho (x,y,z)\) and mass \(m\), to
write:
\[ \d m = \rho (x,y,z) \d V \]
Summing these together with an integral, we find the mass is equal to
\[ \iiint _R \rho (x,y,z) \d V. \]
Find the mass of the solid defined by the region
\[ R = \{(x,y,z):0\le x\le 3, 0\le y\le 6-2x, 0\le z<2-y/3-2x/3\} \]
with density function
\(\rho (x,y,z)=3x\).
Write with me
\begin{align*} M &= \iiint _R \rho (x,y,z)\d V \\ &= \int _0^3\int _0^{\answer [given]{6-2x}}\int _0^{\answer [given]{2-y/3-2x/3}} \big (\answer [given]{3x}\big )\d z\d y\d x\\ &= \int _0^3\int _0^{\answer [given]{6-2x}} (\answer [given]{6x-xy-2x^2})\d y\d x\\ &= \int _0^3\eval {\answer [given]{6xy-xy^2/2-2x^2y}}_0^{\answer [given]{6-2x}}\d x\\ &= \int _0^3 (\answer [given]{18x-12x^2+2x^3}) \d x\\ &= \answer [given]{27/2} \end{align*}
2 Moments and center of mass
A moment is a scalar quantity describing how mass is distributed in relation to a
point, line, or plane.
Let a region
\(R\subset \R ^3\) define a solid with density function
\(\rho (x,y,z)\).
-
The moment about the \((x,y)\)-plane is given by
\[ M_{xy} = \iiint _R z \rho (x,y,z) \d V \]
-
The moment about the \((x,z)\)-plane is given by
\[ M_{xz} = \iiint _R y \rho (x,y,z) \d V \]
-
The moment about the \((y,z)\)-plane is given by
\[ M_{yz} = \iiint _R x \rho (x,y,z) \d V \]
The moments are directly related to the center of mass of an object.
Let a region
\(R\subset \R ^3\) define a solid with density function
\(\rho (x,y,z)\). The
center of mass is defined to
be the point
\[ (\bar {x},\bar {y},\bar {z}) = \left (\frac {M_{yz}}{M},\frac {M_{xz}}{M},\frac {M_{xy}}{M}\right ). \]
Consider the solid defined by the region
\[ R= \{ (x,y,z) : -1\le x \le 1, -\sqrt {1-x^2}<y<0, 0\le z\le -y\} \]
with density function
\[ \rho (x,y,z) = 10 + x^2 + 5y -5z. \]
Find the center of mass of this solid.
At this point we need to
compute
four triple integrals. Each computation will require a number of careful
steps. Get out several sheets of paper and take a deep breath. First we’ll compute the
mass. Write with me:
\begin{align*} M &= \iiint _R \big (10+x^2+5y-5z\big )\d V \\ &= \int _{-1}^1\int _{-\sqrt {1-x^2}}^0\int _0^{-y} \big (10+x^2+5y-5z\big )\d z\d y \d x\\ &= \answer [given]{\frac {34}{5}-\frac {15\pi }{16}} \end{align*}
Now we’ll compute \(M_{yz}\). Write with me:
\begin{align*} M_{yz} &= \iiint _R \answer [given]{x(10+x^2+5y-5z)}\d V \\&=\answer [given]{0} \end{align*}
Now we’ll compute \(M_{xz}\). Write with me:
\begin{align*} M_{xz} &= \iiint _R \answer [given]{y(10+x^2+5y-5z)}\d V\\ &= \answer [given]{2-\frac {61\pi }{48}} \end{align*}
Now we’ll compute \(M_{xy}\). Write with me:
\begin{align*} M_{xy} &= \iiint _R \answer [given]{z(10+x^2+5y-5z)}\d V \\ &= \answer [given]{\frac {61\pi }{96}-\frac {10}{9}} \end{align*}
The center of mass is
\[ \begin{bmatrix} \bar {x}\\ \bar {y}\\ \bar {z} \end{bmatrix} = \begin{bmatrix} \answer [given]{0}\\ \answer [given]{\left (2-\frac {61\pi }{48}\right )/\left (\frac {34}{5}-\frac {15\pi }{16}\right )}\\ \answer [given]{\left (\frac {61\pi }{96}-\frac {10}{9}\right )/\left (\frac {34}{5}-\frac {15\pi }{16}\right )} \end{bmatrix} \]
As stated before, there are many uses for triple integration beyond finding
volume. When \(h(x,y,z)\) describes a rate of change function over some space region
\(R\), then
\[ \iiint _R h(x,y,z)\d V \]
gives the total change over \(R\). Our example of this was computing
mass via a density function. Here a density function is simply a “rate of
mass change per volume” function. Thus, integrating density gives total
mass.
While knowing how to integrate is important, it is arguably much more important to
know how to set up integrals. It takes skill to create a formula that describes a
desired quantity; modern technology is very useful in evaluating these formulas
quickly and accurately.
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