# Homework Help: Equation of the plane, keep getting z=0 but can't be

1. Dec 3, 2008

### briteliner

1. The problem statement, all variables and given/known data
ok so the problem is to evaluate (triple integral)x^2 dV of the solid tetrahedron with vertices (0,0,0) (1,0,0) (0,1,0) and (1,0,1)

2. Relevant equations

3. The attempt at a solution
so (triple integral)x^2 dzdydx
i made the x-y plane be the base with 0<x<1 and 0<y<1-x
for the z limits, i know they should be 0 to the plane but i am having trouble getting the equation of the plane. i need to use the points (1,0,0) (0,1,0) and (1,0,1) right?
but when i do the cross products of the two vectors <-1,1,0> and <0,0,-1> i get that the k component is 0. which makes no sense looking at the picture because z is not always zero. what am i doing wrong?

2. Dec 3, 2008

### Dick

You aren't drawing the right picture. The plane defined by those three points is perpendicular to the x-y plane. I think you want to use (0,0,0), (0,1,0) and (1,0,1) to define the plane that lies above the region in the x-y plane. Try the picture again.

3. Dec 3, 2008

### briteliner

now i'm getting a plane equation of z=x which makes no sense when i integrate

4. Dec 3, 2008

### Dick

That's correct. But how can that make no sense when you integrate? It's a perfectly reasonable limiting plane. One variable in your integral is a dummy, one is the actual limit.

5. Dec 3, 2008

### briteliner

got it, thanks!