Did I Find the Correct Mass and Center of Mass for the Solid?

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SUMMARY

The discussion centers on calculating the mass and center of mass for a solid bounded by the planes x=0, y=0, z=0, and x+y+z=1, with a density function of δ(x,y,z)=y. The user attempted to find the mass using the triple integral M=∫∫∫δ dV, with limits from 0 to 1 for x, 0 to 1-x for y, and 0 to 1-x-y for z. The user reported a mass of -1/3, which is incorrect, as the integral of a positive density function over a defined region cannot yield a negative result. The correct approach involves verifying the limits of integration and ensuring the density function is applied correctly.

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squeeky
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Homework Statement


Find the mass and center of mass of the solid bounded by the planes x=0, y=0, z=0, x+y+z=1; density[tex]\delta[/tex](x,y,z)=y


Homework Equations


[tex]M=\int\int_D\int\delta dV[/tex]
[tex]M_{yz}\int\int_D\int x \delta dV;M_{xz}\int\int_D\int y \delta dV;M_{xy}\int\int_D\int z \delta dV[/tex]
[tex]C.O.M.=(\bar{x},\bar{y},\bar{z})[/tex]
[tex]\bar{x}=\frac{M_{yz}}{M};\bar{y}\frac{M_{xz}}{M};\bar{z}\frac{M_{xy}}{M}[/tex]


The Attempt at a Solution


I'm not sure if it's right, but I took the limits to be from 0 to 1 for x, 0 to 1-x for y, and 0 to 1-x-y for z. This gave me the equation:
[tex]M=\int^1_0\int^{1-x}_0\int^{1-x-y}_0 \delta dzdydx[/tex]
Solving this, I got a mass of -1/3, Mxy=-17/180, Myz=41/120, Mxz=1/20, and the center of mass at (-41/40, -3/20, 17/60)
Could someone check if I did everything right?
 
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The limits of integration are correct. But I do NOT get "-1/3" as the mass! In fact, it should be obvious that the integral of the function y over a region in the first octant cannot be negative.
 

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