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This strange triple integral

  • Thread starter planck42
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  • #1
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Homework Statement


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


Homework Equations


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

[tex]{\int_0^2}{\int_0^2}{\int_0^{4-2x-2y}}ydzdxdy[/tex], which becomes
[tex]{\int_0^2}{\int_0^2}4y-2xy-2y^{2}dxdy[/tex] and then
[tex]{\int_0^2}4y-4y^{2}dy[/tex] and finally integrating to
[tex](2y^{2}-\frac{4}{3}y^{3})_{0}^{2}=-\frac{8}{3}[/tex], 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)
 

Answers and Replies

  • #2
tiny-tim
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hi planck42! :smile:
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 :redface:

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! :rolleyes:
 
  • #3
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hi planck42! :smile:


nooo :redface:

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! :rolleyes:
That makes sense; but now I feel even more lost.
 
  • #4
tiny-tim
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ok, start with integrating z from 0 to 4 …

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? :wink:
 
  • #5
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ok, start with integrating z from 0 to 4 …

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? :wink:
I don't find it possible to answer both questions as there is only one boundary function.
 
  • #6
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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 [tex]\frac{4-z-2x}{2}[/tex]. 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.
 
  • #7
tiny-tim
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hi planck42! :wink:

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 … ? :smile:
 
  • #8
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0 to 4-2x-2y. But how did you get y being from 0 to 2-x for fixed x only?
 
  • #9
tiny-tim
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oh now i see what's been worrying you!

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) :wink:
 

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