# Thread: ∫∫∫ xy dV Triple Integration. Help with bounds!

1. ## ∫∫∫ xy dV Triple Integration. Help with bounds!

∫∫∫ xy dV, where the solid is a tetrahedron with vertices (0,0,0), (1,0,0),(0,2,0), and (0,0,3)

I tried this with 0<x<1, 0<y<-2x+2, and 0<z<3 with no luck =\ Are my bounds wrong in this case?

What would be really helpful is if someone could explain how I am supposed to figure out the bounds. I don't even know how to figure them out which means I can't do any of the problems assigned

2. Originally Posted by crabchef
∫∫∫ xy dV, where the solid is a tetrahedron with vertices (0,0,0), (1,0,0),(0,2,0), and (0,0,3)

I tried this with 0<x<1, 0<y<-2x+2, and 0<z<3 with no luck =\ Are my bounds wrong in this case?

What would be really helpful is if someone could explain how I am supposed to figure out the bounds. I don't even know how to figure them out which means I can't do any of the problems assigned
yes, your limits for z are wrong. see here for the method of finding bounds. post #2 gives a shortcut for special cases (this one). posts #7 and #9 give the longer, more conventional way of tackling it

3. thanks for the link! this is going to take me a while to sort out ^^

ah nevermind, z should equal 3 - 3x - 3y/2

4. Originally Posted by crabchef
∫∫∫ xy dV, where the solid is a tetrahedron with vertices (0,0,0), (1,0,0),(0,2,0), and (0,0,3)

I tried this with 0<x<1, 0<y<-2x+2, and 0<z<3 with no luck =\ Are my bounds wrong in this case?

What would be really helpful is if someone could explain how I am supposed to figure out the bounds. I don't even know how to figure them out which means I can't do any of the problems assigned
The upper surface of the tetrahedron is a plane with points (1,0,0), (0,2,0), and (0,0,3) which has equation $x+ \frac{y}{2}+ \frac{z}{3}= 1$. That can be rewritten $\frac{z}{3}= 1- x- \frac{y}{2}$ and then $z= 3- 3x- \frac{3y}{2}$. That is the upper bound for your tetrahedron.