Triple Integral Volume: Octant x,y,z>=0 Bounded by x+y+1=1 and x+y+2z=1

[musicmar]

Homework Statement

Find the volume of the solid in the octant x,y,z>=0, bounded by x+y+1=1 and x+y+2z=1

The Attempt at a Solution

I've been looking at an example in the textbook that is similar to this problem. First, I found the projection of W onto the xy plane:

x+y-1=(x/2)+(y/2)-(1/2)

(x/2)+(y/2)=(1/2)

This let's you find the bounds of x in terms of y (The bounds of z are just the two functions above).

x=1-y

Here is where I got confused. In the book, with different functions, they used the same process. But the book had 0<=x<=(their function) and 0<=y<=1

and I'm not sure where the lower bound of x or either of the bounds of y came from.

If someone could either clarify where these came from and/or help me continue with my own problem, that would be great.

 

[jackmell]

The plane x+y+1=1 does not go through the first octant. How about drawing them? I mean, I did a quick sketch then just confirmed it in Mathematica:

ContourPlot3D[{x + y + 1 == 1, 2 z == 1 - x - y},
 {x, -5, 5}, {y, -5,  5}, {z, -5, 5},
 PlotRange -> {{-1, 1}, {-1, 1}, {-1, 1}}]

If you wanted just underneath the plane 2z=1-x-y in the first octant, then would it not be:

$$\int_0^1\int_0^{1-x} 1/2(x-y)dzdydz$$