# Thread: mass and center of a tetrahedron

1. ## mass and center of a tetrahedron

I am to fond the mass of the tetrahedron bound by the xy, yz, and xz planes and x + 2y + 3z = 1 THe density is 1 gm/cm^3 and x,y,z are in cm Then I am to find the center of this.

The bounds of course are my problem!

I think that they all three would be bound from 0 to 1 but I am not sure if it is that easy. Or would x be bound from 0 to 1 and then just solve the equation x + 2y + 3z = 1 for both y and z and use 0 as the lower bounds for all????

2. One way to write it is $x=1-2y-3z$.

Then $0 \leq x \leq 1-2y-3z$

The base is $0\leq y \leq \frac{1}{2},\ 0\leq z \leq \frac{1}{3}(1-2y)$

To find the center of mass:
$mx_T=\iiint_K x\varrho(x,y,z)dxdydz$
$my_T=\iiint_K y\varrho(x,y,z)dxdydz$
$mz_T=\iiint_K z\varrho(x,y,z)dxdydz$

where $m$ is the mass $=\iiint_K \varrho(x,y,z)dxdydz$ and $\varrho(x,y,z)$ is the density.

3. If you want the base in the xy-plane: $0\leq z\leq \frac{1}{3}(1-x-2y),\ 0\leq y\leq \frac{1}{2}(1-x),\ 0\leq x\leq 1$

4. There is, by the way, a simple formula that says that the volume of the tetrahedron with vertices at (0, 0, 0), (a, 0, 0), (0, b, 0), and (0, 0, c) is abc/6.

The plane x+ 2y+ 3z= 1 cuts the axes at (1, 0, 0), (0, 1/2, 0), and (0, 0, 1/3) so the volume of this tetrahedron is (1)(1/2)(1/3)/6= 1/36.

Since the density is just 1, the mass is the volume.

To get the x coordinate of the centroid, integrate x over that tetrahedron and divide by the volume.