Help setting up triple integral

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The discussion focuses on setting up a triple integral to calculate Mxz for a solid defined by the plane 4x + 2y + 3z = 16 and the coordinate planes. The user initially sets up the integral but receives an incorrect answer, prompting a request for help. Key points include the identification of the boundaries for x, y, and z based on the intersections with the coordinate planes. The correct ranges for integration are clarified, emphasizing that y ranges from 0 to 8, x from 0 to (16 - 2y)/4, and z from 0 to (16 - 4x - 2y)/3. The discussion concludes with a suggestion to re-evaluate the integral setup based on these boundaries.
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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=\int\int\int 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...
\int\int\int y dzdxdy
0\leqz\leq16-4x-2y
0\leqx\leq1/2(y-8)
0\leqy\leq8

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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