<s>[INST] Finding the $z$ coordinate of center of mass for $C:= \{ (x,y,z) \in \mathbb{R^3}: \sqrt{x^2+y^2}\leq z \leq 1$ The Problem

Let $C$ be the cone
$$C:= \{ (x,y,z) \in \mathbb{R^3}: \sqrt{x^2+y^2}\leq z \leq 1$$
Assume that $C$ has a constant mass density and find the z coordinate of the center of mass.

The work I have done:
\begin{align*}
M_{xy}&=\iint_{R}\int_{\sqrt{x^2+y^2}}^{1} z \,\delta \,dx \,dy \,dx\\
&=\iint_{R} \delta \,\frac{z^2}{2} \Bigg|_{\sqrt{x^2+y^2}}^{1} \,dy \,dx\\
&=\iint_{R} \delta \,\Big(\frac{1}{2}-\frac{\sqrt{x^2+y^2}}{2} \Big) \,dy \,dx
\end{align*}
Switching to polar coordinates now and pulling out the constants:
\begin{align*}
&=\frac{\delta}{2}\int_{0}^{2\pi}\int_{0}^{1} \big(r-r^3 \big) \,dr \,d\theta\\
&=\frac{\delta}{2} (\frac{\pi}{2})=\frac{\pi}{4}\,\delta\,.\\
\\
M&=\iint_{R}\int_{\sqrt{x^2+y^2}}^{1} \,\delta \,dx \,dy \,dx\\
&=\iint_{R} \delta (1-\sqrt{x^2+y^2}) \,dy \,dx
\end{align*}
Switching again to polar:
\begin{align*}
&=\delta \int_{0}^{2\pi} \int_{0}^{1} (1-r) \,r \,dr \,d\theta\\
&=\delta \int_{0}^{2\pi} \frac{1}{6} d\theta = \frac{\pi}{3}\,\delta
\end{align*}
Therefore:
$$\bar{z}=\frac{M_{xy}}{M} = \frac{\pi\delta}{4}\cdot\frac{3}{\pi\delta}=\frac{3}{4}$$
Is my answer correct?
Edit: I also know that I could have switched to cylindrical from the beginning. If I did, would the integral I need to evaluate be $$M=\int_{0}^{2\pi}\int_{0}^{1}\int_{r}^{1} \,r \,dz \,dr\,d\theta$$
and
$$M_{xy}=\int_{0}^{2\pi}\int_{0}^{1}\int_{r}^{1} \,rz \,dz \,dr\,d\theta\,?$$
 [/INST] This is a cone of base radius $1$, and height $1$.  The center of mass formula is
$ \overline{z} = \displaystyle \dfrac{ \iiint z dV } {\iiint dV} $
The numerator is
$ \displaystyle \iiint z dV = \int_{z=0}^1 (2 \pi z^3)dz = \dfrac{\pi}{2} $
The denominator is
$ \displaystyle \iiint z dV = \int_{z=0}^1 (2 \pi z^2)dz = \dfrac{2\pi}{3} $
Hence,
$ \overline{z} = \dfrac{ \dfrac{1}{2} }{ \dfrac{ 2}{3}} = \dfrac{3}{4} $
</s>