# Finding the center mass when density is not constant or uniform throughout

I am struggling on this problem: Suppose that an object in three space has six sides and vertices at $(0, 0, 0), (4, 0, 0),(0, 6, 0), (4, 6, 0), (0, 0, 5), (4, 0, 8), (0, 6, 12),$ and $(4, 6, 15)$ where the units are given in cm. The density of this object is not uniform and is given by $ρ(x, y, z) = 0.2e^{−0.2x+0.5y+0.1z}$ grams per square m. Find the center of mass of this object.

So I would totally understand how to complete this problem is the density of the object was uniform. I would simply develop the total mass as:

$\int_0^{12} \int_0^6 \int_0^4 ρ(x, y, z) = 0.2e^{−0.2x+0.5y+0.1z}\,dx\,dy\,dz$.

And then the center of mass would simply be $\bar{x}$ and $\bar{y}$. But with non-uniform density, how would I work this problem out?

• You're formula is incomplete. The formula you have to use is $$\vec R =\frac{1}{M}\int_V \vec r\rho(\vec r)\,\mathrm{d}V$$ where $\vec R$ are the coordinates of the center of mass and $M=\int_V \rho(\vec r)\,\mathrm{d}V$ is the mass of the object. – dinosaur Nov 5 '14 at 7:31
• The limits of your integration are not consistent with the vertices that you listed. – copper.hat Nov 5 '14 at 7:31
• What are the bounds for the total mass then? – user3472296 Nov 5 '14 at 13:19
• See my answer below for appropriate limits. – copper.hat Nov 5 '14 at 15:41

The total mass is $M=\int \int \int \rho(x,y,z) dx dy dz$.
The centre of mass is computed as $\begin{bmatrix} \bar{x} \\ \bar{y} \\ \bar{z}\end{bmatrix} = {1 \over M} \int \int \int \begin{bmatrix} x \\ y \\ z \end{bmatrix} \rho(x,y,z) dx dy dz$.
Note that if we let $f(x,y) = 5+{3 \over 4} x + {7 \over 6} y$, then the body can be described by $\{(x,y,z) | x \in [0,4], y \in [0,6], z \in [0,f(x,y)] \}$, and so the limits of integration become $\int_{x=0}^4 \int_{y=0}^6 \int_0^{f(x,y)}$.
Hence $M= \int_{x=0}^4 \int_{y=0}^6 \int_0^{f(x,y)} \rho(x,y,z) dx dy dz$.
You would need to check this, but I get $(\bar{x}, \bar{y} , \bar{z}) \approx (1.88, 4.63, 7.06)$.