Triple Integrals - Finding Mass

1. Nov 15, 2007

Hells_Kitchen

The problem is the following:
I need to find the mass, moments along the axis and the center of mass of the tetrahydron (centroid) with vertecies (-1,0,0) (1,0,0) (0,1,0) and
(0,-1,0) and (0,0,2) basically it has a square base with an area of 4 and height 2 units.

You are also given the density function of the solid:

P(x,y,z) = absX + absY +absZ where (X,Y,Z) are variables

Using triple integrals its really easy to find the volume or you can just find it at sight area of base times height however when integrating the density function over that volume I suspect that i have to divide the integral in four different parts since x and y can be either negative or positive while z is always positive so its absolute value will be z at all times.
I was wondering if this approach is right and if it is would the limits of integration for all four different integrals be different and how would i go by finding them?

Plus once the integral of the mass is found i suspect that the moments along each axis Mxy Myz and Mxz will be easier to find by just plugging in each corresponding varialbe in the integrand however i am not quite sure if i would plug in each corresponding variable in all four integrals or just in some of them....?????

I hope someone can help!

2. Nov 15, 2007

HallsofIvy

Staff Emeritus
I would be inclined to do it this way: imagine the solid divided into thin slabs on top of one another- each a square. The line from (1, 0, 0) to (0, 0, 2) is given by z= -2x-2 or x= 1- z/2 so the vertex of the "slab" at height z is at (1-z/2, 0, z). It should be clear that the other four vertices are at (-1+z/2, 0, z), (0, 1- z/2, z), and (0, -1+z/2, z). You can integrate a function over the entire pyramid by integrating the function over that "slab" and then integrating with respect to z.

By the way, it seems obvious from symmetry that the center of mass is on the z-axis.

3. Nov 15, 2007

Hells_Kitchen

So are you saying that in order to find the mass you would take the triple integral of the positive values of the density function

P(x,y,z) = x + y + z
and having limits of integration the following:
x from -1 +z/2 to 1- z/2
y from -1 +z/2 to 1- z/2
z from 0 to 2
Of course you would have to integrate first either with respect to x or y but LAST with respect to z.
If this would be the case should also be the case the if you take the triple integral of just the volume dV with the above limits of integration you should get 8 units^3. Since the volume of the tetrahydron is just 8 at sight. But if i do the math out for it comes out to be 14/3 cubic units

and you are right its pretty obvious that the centroid will have coordinates (0,0,z(bar))

Last edited: Nov 15, 2007
4. Nov 16, 2007

Hells_Kitchen

Anyone can shoot some help to this one?