Consider a body of revolution of constant density that in cylindrical coordinates can be written as $$V=[(r,\theta,z):0\leq\theta\leq2\pi,0\leq z \leq a, 0\leq r \leq h(z)]$$
Given that $$\bar{z}=\frac{\int_{0}^{a}z(h(z))^2 dz}{\int_{0}^{a}(h(z))^2dz}$$
Use the result above to find the centre of mass of the top which lies above the upper half of the cone $x^2+y^2=3z^2$ and below the sphere $x^2+y^2+z^2=2z$
I do not know how to proceed without understanding what is $h(z)$, can anyone explain it to me ? Thanks in advance.
$h(z)$ refers to the radius of the cross-section of the body at a given height $z$. If $h(z)$ was constant, then the body would be a cylinder. For a cone, the radius of the cross-section has a linear relation with the height. For your example of $x^2 + y^2 = 3z^2$, this can be written in polar as $$r^2 = 3z^2 \rightarrow r = z\sqrt{3}$$ (assuming the top half implied a positive $z$ so the sign of the square root is positive). The sphere is a bit more tricky. Given your equation of $x^2 + y^2 + z^2 = 2z$, this can be written as \begin{align*} x^2 + y^2 + z^2 &= 2z \\ x^2 + y^2 + (z^2 - 2z + 1) &= 1 \\ x^2 + y^2 + (z-1)^2 &= 1. \end{align*} Written in polar, this is equivalent to $r^2 = 1 - (z-1)^2$, or $r = \sqrt{1-(z-1)^2}$. Combining these should give you insight into how you can solve your integral.
PS: Realized I didn't mention why $h(z)$ must be the radius. Looking at the definition of the $V$ in Cylindrical Coordinates displays the constraint $0 \leq r \leq h(z)$. This means that at the $z$, polar coordinates can only go from $0$ to $h(z)$, so because there are not any constraints on the angle $\theta$, the cross section is a disk (not a circle) of radius $h(z)$. Hope this helps.