# Triple integral volume

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.

ContourPlot3D[{x + y + 1 == 1, 2 z == 1 - x - y},
PlotRange -> {{-1, 1}, {-1, 1}, {-1, 1}}]
$$\int_0^1\int_0^{1-x} 1/2(x-y)dzdydz$$