Help setting up triple integral

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SUMMARY

The discussion focuses on calculating the triple integral Mxz=\int\int\int y dV for the solid W bounded by the coordinate planes and the plane defined by the equation 4x + 2y + 3z = 16. The user initially set up the integral incorrectly, particularly in defining the limits for x, y, and z. Correct boundaries are established as 0 ≤ z ≤ (16 - 4x - 2y)/3, 0 ≤ y ≤ 8, and 0 ≤ x ≤ (16 - 2y)/4, ensuring accurate integration over the defined solid.

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Homework Statement


W is the solid bounded by the three coordinate planes and the surface 4x+2y+3z=16, Calculate Mxz=[tex]\int[/tex][tex]\int[/tex][tex]\int[/tex] y dV


Homework Equations





The Attempt at a Solution


the surface 4x +2y +3z=16 is a plane that crosses boundries at (4,0,0), (0,8,0) and (0,0,16/3)

This is how I set up my integral but I am getting an incorrect answer...
[tex]\int[/tex][tex]\int[/tex][tex]\int[/tex] y dzdxdy
0[tex]\leq[/tex]z[tex]\leq[/tex]16-4x-2y
0[tex]\leq[/tex]x[tex]\leq[/tex]1/2(y-8)
0[tex]\leq[/tex]y[tex]\leq[/tex]8

can anyone tell what I'm doing wrong
 
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your boundary.
4x+2y+3z=16
To get the y-range, set x=0, z=0, so y is between 0 and 8
for x-range, set z=0. then x is between 0 and (16-2y)/4
z goes between 0 and (16-4x-2y)/3
the crossing points are irrelavant.
try and see what u get.
 

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