# This strange triple integral

## Homework Statement

$${\int}{\int}{\int}ydV$$ over the region E, where E is bounded by x=0, y=0, z=0, and 2x+2y+z=4

n/a

## The Attempt at a Solution

Assuming that x and y must both be positive, which the boundary conditions seem to require, then the most either one can be is 2, so I have the bounds for x and y being from 0 to 2. Then for z, I guessed that 4-2x-2y is above the xy-plane(webMathematica's Three-Dimensional Visualizer doesn't make the picture any clearer), so the integration looks like so

$${\int_0^2}{\int_0^2}{\int_0^{4-2x-2y}}ydzdxdy$$, which becomes
$${\int_0^2}{\int_0^2}4y-2xy-2y^{2}dxdy$$ and then
$${\int_0^2}4y-4y^{2}dy$$ and finally integrating to
$$(2y^{2}-\frac{4}{3}y^{3})_{0}^{2}=-\frac{8}{3}$$, which puzzles me to no end. If I integrate above the xy-plane, I get a negative answer, and if I integrate below the xy-plane, I get a positive answer. Can somebody illuminate me on this nonsense?(I struggled to identify properly the bounds as may be clear to you)

tiny-tim
Homework Helper
hi planck42! Assuming that x and y must both be positive, which the boundary conditions seem to require, then the most either one can be is 2, so I have the bounds for x and y being from 0 to 2.

nooo you can't integrate both x and y from 0 to 2 …

that would include x = y = 2, and then your z integral would be from 0 to -4, giving you a negative contribution! hi planck42! nooo you can't integrate both x and y from 0 to 2 …

that would include x = y = 2, and then your z integral would be from 0 to -4, giving you a negative contribution! That makes sense; but now I feel even more lost.

tiny-tim
Homework Helper

for each fixed value of z, what does y go between? …

then for each fixed values of z and y, what does x go between? for each fixed value of z, what does y go between? …

then for each fixed values of z and y, what does x go between? I don't find it possible to answer both questions as there is only one boundary function.

Let's just suppose that x is going from 0 to 2(it should be equivalent to taking z from 0 to 4 if I do this right). Then for a fixed value of x, y should be $$\frac{4-z-2x}{2}$$. For a fixed x and y, z should be 4-2x-2y. The problem I have with this is that if I find z for fixed x, I get a z that depends on y, and if I find y for fixed x, then that y is still dependent on z, so there's no acceptable way to arrange the integrals. There must be something that I am missing.

tiny-tim
Homework Helper
hi planck42! just pick 'em off one at a time …

x from 0 to 2;

for fixed x, y from 0 to 2 - x;

for fixed x and y, z from … ? 0 to 4-2x-2y. But how did you get y being from 0 to 2-x for fixed x only?

tiny-tim
if you haven't fixed z, then it's for any value of z (and that includes z = 0, so the maximum for y is 2 - x) 