Vtyjkyuk

I am finding the center of mass of sphere $x^2+y^2+z^2=2z$ when $delta(x,y,z)=sqrtx^2+y^2+z^2$. I did the mass as follows (I hope it is right):

$$M=int_0^2pidthetaint_0^pi/2int_0^2cosphirho^3sinphi,drho ,dphi=8pi/5.$$

Now I did the $x_0$ of the center:

$$x_0=frac1M int_0^2pi cos(theta) , dtheta int_0^pi/2 int_0^2cosphi rho^4sin^2 phi , drho , dphi=0.$$

Here, I suspect the rest for $y_0$ and $z_0$ is obvious. Indeed, I think that I would get $y_0=0, z_0=1$ after evaluating the corresponding integrals because the center of the sphere is $(0,0,1)$. What do you think?

I think that it is more natural to assume that the sphere is $x^2+y^2+z^2=1$ and that $delta(x,y,z)=sqrtx^2+y^2+(z+1)^2$. ThenbeginalignM&=int_0^piint_0^2piint_0^1r^2sin(varphi)sqrtr^2+2rcos(varphi)+1,mathrm dr,mathrm dtheta,mathrm dvarphi\&=2piint_0^piint_0^1r^2sin(varphi)sqrtr^2+2rcos(varphi)+1,mathrm dr,mathrm dvarphiendalignand, yes, it is equal to $frac8pi5$.

By symmetry, the center of mass is a point of the type $(0,0,z_0)$. In factbeginalignz_0&=frac1Mint_0^piint_0^2piint_0^1r^3sin(varphi)cos(varphi)sqrtr^2+2rcos(varphi)+1,mathrm dr,mathrm dtheta,mathrm dvarphi\&=frac2piMint_0^piint_0^1r^3sin(varphi)cos(varphi)sqrtr^2+2rcos(varphi)+1,mathrm dr,mathrm dvarphi\&=frac17.endalign

Of course, this means that the center of mass of the original version of the problem is $left(0,0,1+frac17right)=left(0,0,frac87right)$.