Set up triple integral if E is tetrahedron with vertices at

[MarkSA]

Hi,

1) Set up the triple integral in SIX DIFFERENT WAYS $\displaystyle \int\int\int z \ dV$ if E is the tetrahedron with vertices (0,0,0) (1,0,0) (1,2,0) and (0,0,3).

So, i'm having trouble just setting it up one way... I did a slightly similar problem to this before so I think the first step is to find an equation for the plane on the tetrahedron that I can always use for my 'z' bounds. I let one vector be from (1,2,0) to (1,0,0), which is <0,-2,0>, and another be from (1,2,0) to (0,0,3) which is <-1,-2,3>. Cross product of these is: <-6,0,-2>. Equation for the plane is then -6x-2z=0, or z = -3x
So I would begin with:
$\displaystyle \int\int\int^{-3x}_{0} z \ dzdydx$

At this point I am having trouble figuring out what my x and y limits should be though. I thought that if looking in the x-y plane, my y limits would be from 0 to the line made of the (0,0,0) and (1,2,0) points. this is just y=2x. So then I have:
$\displaystyle \int\int^{2x}_{0}\int^{-3x}_{0} z \ dzdydx$
And x limits would be from 0 to the line made from (1,0,0) and (1,2,0) .. this is x = 2
$\displaystyle \int^{2}_{0}\int^{2x}_{0}\int^{-3x}_{0} z \ dzdydx$

This gives me a very wrong answer. Any idea where my problem is?

 

[galactus]

Re: Set up the triple integral

You know what you could do, Mark:

$\displaystyle \frac{1}{3}\int_{0}^{1}\int_{0}^{2x}\int_{0}^{-3x}z \;\ dzdydx$

 

[MarkSA]

Re: Set up the triple integral

Oops, I see I was wrong on the x limits. It should just be 0 to 1. That gives me an answer of 9/4 though, when the correct answer is actually 3/4. Where does the 1/3 that you mutliply the integral by come from?