Triple Integral for Volume in Rectangular Coordinates

[hackmonkey7]

Homework Statement

The region R in 3D is cut from the first octant (x,y,z >= 0) by the plane X+Z = 1, Y+2Z = 2.
Set up the volume in all 6 ways in rectangular coordinates.
Then evalute the volume in two of these ways.
Make sure to specify limits of integration in every case.

Homework Equations

I figure Y+2Z = 2 is the upper bound for Z and Z=0 is the lower bound.

Here's a rough sketch I drew of the base on X-Y plane, and the 3-D of what I imagined it to look like:
[PLAIN]http://img576.imageshack.us/img576/2599/calc3.png [/PLAIN]

The Attempt at a Solution

So I put the triple integration in:
$$V(R) = \int_0^1{\mathrm \int_(2X)^(2-2X){\mathrm \int_0^(1-(1/2)Y){\mathrm }{\mathrm d}Z}{\mathrm d}Y}{\mathrm d}X$$

When I work this out, I keep getting a negative value!
I know the answer is suppose to be 2/3 because a rectangular pyramid's volume = 1/3 * height * base area, which the base area is 1x2 times 1 height * 1/3 = 2/3.
But I can't seem to manipulate it anyway to get it. Where am I going wrong? Am I setting the wrong plane as the Z bound?
Thanks for any help.

 

[LCKurtz]

Homework Statement

The region R in 3D is cut from the first octant (x,y,z >= 0) by the plane X+Z = 1, Y+2Z = 2.
Set up the volume in all 6 ways in rectangular coordinates.
Then evalute the volume in two of these ways.
Make sure to specify limits of integration in every case.

Homework Equations

I figure Y+2Z = 2 is the upper bound for Z and Z=0 is the lower bound.

No, that isn't correct. Both of your slanted planes are part of the "roof". You have to break it up into two different integrals if you integrate in the z direction first taking separate xy regions under each section of the corresponding z plane.