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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)