<s>[INST] What is the center of mass of the liquid with a density of $10 -x^{2}$ in a cone? (drawing included) Problem:
A straight circular cone with the height $2$ m and the radius $1$ m has the pointy end downward and is filled with a liquid. In a flat plane parallel with the bottom surface of the cone and at a distance of $x$ m from the pointy end, the density of the liquid is $\rho$ $kg/m^3$ with: 
$\rho = 10- x^2$
Determine the center of mass of the liquid.
Attempted solution:
Not the prettiest picture I have ever drawn:

The basic idea is to get the area of a slice:
$$A(r) = r^2 \pi$$
Congruence gives the radius:
$$\frac{r}{x} = \frac{1}{2} \Rightarrow r = \frac{x}{2}$$
The area becomes:
$$A(x) = \pi \Big(\frac{x}{2}\Big)^2 = \frac{\pi x^2}{4}$$
The volume slice becomes:
$$dV = A(x)dx$$
The dm slice becomes:
$$dm = \rho \pi r^2 dx = \frac{\rho \pi x^2}{4}$$
So left to do is to calculate two integrals from $0$ to $2$ (for $x dm$ and $dm$) and divide the first with the second.
$$\int_0^2 x dm = \frac{\pi}{4} \int_0^2 \rho x^3 dx = \frac{\pi}{4} \int_0^2 (10-x^2) x^3 dx = \frac{\pi}{4} \Big( \frac{10x^4}{4} - \frac{x^6}{6} \Big)  \Big|_0^2 = \frac{22\pi}{3}$$
$$\int_0^2 dm = \frac{\pi}{4} \int_0^2 (10-x^2) dx = \frac{\pi}{4} \Big( 10x- \frac{x^3}{3} \Big)  \Big|_0^2 = \frac{13\pi}{3}$$
Division produces the final answer of:
$$\frac{22}{13}$$
The expected answer is:
$$\frac{55}{38} \approx 1.45$$
length units from pointy end of the cone.
Where did I go wrong and what are some productive ways to finish off this question?
 [/INST] Your work looks good but there is a mistake in the integrand:$$\int_0^2 dm = \frac{\pi}{4} \int_0^2 (10-x^2) \color{red}{x^2}dx = \cdots$$
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