Centre of mass of a cone with variable density

dfsgasfa

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I want to find the centre of mass of a cone with variable density
the equation of the cone is:
0<z<1
0<phi<pi/4

I found the mass of the cone = 1/3
Myz=1/20 pi
Mxz=0
and Mxy=0
is that correct?
 
I want to find the centre of mass of a cone with variable density
the equation of the cone is:
0<z<1
0<phi<pi/4

I found the mass of the cone = 1/3
Myz=1/20 pi
Mxz=0
and Mxy=0
is that correct?
How did you arrive at your answers? What are the definitions of the various variables (x, y, z, and \(\displaystyle \phi\))? Where do you account for the density, \(\displaystyle \rho\), and what was the function for its variability?

Please be complete. Thank you! ;)
 
How did you arrive at your answers? What are the definitions of the various variables (x, y, z, and \(\displaystyle \phi\))? Where do you account for the density, \(\displaystyle \rho\), and what was the function for its variability?

Please be complete. Thank you! ;)

The density of the cone is density=x
As phi=pi/4 radius of the cone at any point is equal to the height of the cone, so z=r
To find the mass of the cone I integrated xdV with 0<theta<pi/2, 0<r<1, r<z<1 and then multiplied the result by 4 to obtain the mass of the complete cone

To obtain the value for Myz I integrated x^2dV from 0<theta<2pi, 0<r<1, r<z<1
To obtain the value for Mxz I integrated xydV from 0<theta<2pi, 0<r<1, r<z<1
To obtain the value for Mxy I integrated xzdV from 0<theta<2pi, 0<r<1, r<z<1
 
I want to find the centre of mass of a cone with variable density
the equation of the cone is:
0<z<1
0<phi<pi/4
A cone is three dimensional so is described by three coordinates. I assume the third is \(\displaystyle 0\le \theta\le 2\pi\). But what is the density function?

I found the mass of the cone = 1/3
Myz=1/20 pi
Mxz=0
and Mxy=0
is that correct?
It is impossible to say without knowing the density function.
 
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