# Volume of a tetrahedron of a function

1. Sep 14, 2011

### Liquidxlax

1. The problem statement, all variables and given/known data
Calculate the volume integral of the function T = z^2 over the tetrahedron with corners at
(0,0,0), (1,0,0) , (0,1,0), and (0,0,1)

3. The attempt at a solution

z to x (1,0,-1)
z to y (0,1,-1)

Then i crossed them to get (1,1,1)

Found the plane n dot (x-1, y , z) = x+y+z-1=0

Normally i'd have no problem from here, but i'm not sure how i'm supposed to incorporate the T=z^2

2. Sep 14, 2011

### Char. Limit

So you're dealing with the volume integral:

$$\int \int \int z^2 dz dy dx$$

First thing is to find the bounds. You know that this is the volume contained in the first octant and bound by x+y+z-1=0. This can be rearranged to z=1-x-y, and the lower bound of z is zero (because it's in the first octant). So we can get our first bounds:

$$\int \int \int_0^{1-x-y} z^2 dz dy dx$$

From there can you see the other two sets of bounds?

3. Sep 14, 2011

### Liquidxlax

okay, so when z is 0 y= 1-x and x goes from 0 to 1. If that is right i did this initially and i need to have more confidence in my answers...

thank you for the clarification

4. Sep 14, 2011

### Char. Limit

No worries. Have a great day!