Never mind. I figured out the answer. I used the same technique as above but I changed my y boundary to 1-x < y < 0 however when I look back I still don't understand how that is. Can anyone help me understand this part.
I have been working on this problem for a while and I have tried about 5 or six different ways and can't get the answer.
If you could help that would be awesome.
Find triple integral ZZZT x dV, where T is the tetrahedron bounded by the planes x=1, y=1, z=1 and x+y+z=2. I can't seem to get the boundaries right.
For my latest attempt I tried:
2-y-x < z < 1
1-x < y < 1
0 < x < 1
I was trying to find the missing piece (like the corner of the square) then multiply the answer by three to get the part of the tetrahedron within the boundary but I got 1/4 and the answer is 1/8.