# Calculating the center of mass of a body in $\mathbb{R^3}$

I want to calculate the center of mass of the body defined by $$\begin{cases} x^2+4y^2+9z^2\leq 1 \\x^2+4y^2+9z^2\leq 6z \end{cases}$$ where the density of mass is proportional to the distance to the plane $$xy$$. First of all I will have to calculate the mass, meaning I have to solve $$M=\iiint dm=\iiint_V \lambda z dV$$ My problem is that I don't know what to do when the body is given in that way, I've done similar problems where it was clear I had to do a change of coordinates (spherical, cylindrical...) however in this case I don't know how to set the integral.

• How did you come to this conclusion? – John Keeper Dec 15 '18 at 23:01
• The ellipsoids are obtained from the unit sphere by scaling along the coordinate axes. – amd Dec 16 '18 at 0:08
• I wrote the constants wrongly. The right transformation simplifies the first inequality to $r^2\leq1.$ It is $x=r\cos t\cos \phi, 2y=r\sin t \cos \phi, 3z=r\sin \phi.$ – user376343 Dec 16 '18 at 0:20
• math.stackexchange.com/questions/3043727/… – Hans Lundmark Dec 17 '18 at 10:05