Triple integral volume

  • Thread starter musicmar
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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 lets 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.
 

Answers and Replies

  • #2
1,796
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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:

Code:
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:

[tex]\int_0^1\int_0^{1-x} 1/2(x-y)dzdydz[/tex]
 

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