What is the solution to this volume integral problem?

[johnwalton84]

I'm have trouble trying to evaluate the volume integral (shown in question.gif).

I've attempted integrating it a few different ways, either achieveing an answer of 3 or 5.75, and I'm not sure where I'm going wrong. (Some of what I've done is in attempted_solution.gif)

Any comments gratefully received...

 

[arildno]

Your upper limits are not meaningful!
Let's look at this closely:
a) 0<=y<=3
These bounds should be obvious.
b) 0<=x and 0<=z
Okay?
c) Now, to the last limit indicated by the plane x+z=1
Since, by b) neither x or z can be negative, we can choose the following limits:
0<=x<=1
0<=z<=1-x

Okay with this?

 

[dextercioby]

Your upper limits are not meaningful!
Let's look at this closely:
a) 0<=y<=3
These bounds should be obvious.
b) 0<=x and 0<=z
Okay?
c) Now, to the last limit indicated by the plane x+z=1
Since, by b) neither x or z can be negative, we can choose the following limits:
0<=x<=1
0<=z<=1-x

Okay with this?

Since we're speaking about the triorthogonal tetrahedron determined by the intersection of the plane $3x+y+3z=3$ with the coordinate axes,maybe a graphical representation might help.
Chosing the integration limits correctly (the way Arildno showed you),u find after easy calculations that the integral should yield 1.

 

[johnwalton84]

Yes, I've got it now. Thanks.

:rofl: