Volume of a tetrahedron of a function

[Liquidxlax]

Homework Statement

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)

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

 

[Char. Limit]

So you're dealing with the volume integral:

[tex]\int \int \int z^2 dz dy dx[/tex]

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:

[tex]\int \int \int_0^{1-x-y} z^2 dz dy dx[/tex]

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

 

[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

 

[Char. Limit]

No worries. Have a great day!