The center of mass for a wire of constant density making a curve $y=f(x)$ between $x=a$ and $x=b$ is
$$\bar{y} = \frac{\int_a^b dx \: y \sqrt{1+y'^2}}{\int_a^b dx \:\sqrt{1+y'^2}}$$
In the case you describe
$$\bar{y} = \frac{\displaystyle \int_0^2 dx \: \sqrt{x} \sqrt{1+\frac{1}{4 x}}}{\displaystyle \int_0^2 dx \: \sqrt{1+\frac{1}{4 x}}}$$
The top integral is relatively simple and is equal to
$$\int_0^2 dx \: \sqrt{x+\frac{1}{4}} = \frac{2}{3} \left ( \frac{27}{8} - \frac{1}{8} \right ) = \frac{13}{6}$$
The bottom integral may be evaluated by making the substitution $\sec{\theta} = \sqrt{1+\frac{1}{4 x}}$ to get
$$\int_0^2 dx \: \sqrt{1+\frac{1}{4 x}} = \frac{1}{2} \int_{\arccos{\sqrt{8}/3}}^{\pi/2} d\theta \: \csc^3{\theta} = \frac{1}{8} \left(12 \sqrt{2}+\log \left(17+12 \sqrt{2}\right)\right)$$
Therefore your center of mass is
$$\bar{y} = \frac{52/3}{12 \sqrt{2}+\log \left(17+12 \sqrt{2}\right)} \approx 0.846$$
I will assume that you have been supplied with a formula to compute surface area.
We want an expression for the equation of the cone. Really one should use cylindrical coordinates, but let's use the usual Cartesian setup.
Think of the cone as having axix the $z$-axis, and pointing down with the apex at the origin.
We find the radius $r_z$ of the circle of cross-section at height $z$.
By similar triangles we have
$$\frac{r_z}{z}=\frac{a}{h},$$
so $r_x=\frac{a}{h}z$.
Thus the surface of the cone satisfies the equation
$$x^2+y^2=\frac{a^2}{h^2}z^2.$$
Now compute $dS$ as usual, and integrate.
Remarks: $1.$ Actually, this is really a $1$-variable problem, since we are dealing with a surface of revolution. Thus the problem could also be solved using techniques from first-year calculus.
$2.$ In the unlikely case that you do not know a formula for surface area, note that if $(x,y)$ ranges over $R$, and our surface has equation $z=f(x,y)$, then the surface area is given by
$$\iint_{R} \sqrt{f_x^2(x,y)+f_y^2(x,y)+1}\,dx\,dy,$$
where $f_x$ and $f_y$ are the partial derivatives.
Best Answer
I was taking the integral of a rectangle when I should have been taking it about the triangle.
As a result, the inner bounds should be changed and the the correct integral is $\bar x = \frac{\int_{0}^{3}\int_{7x+3}^{36-4x} x(7x+2y+2)\,dydx}{\int_{0}^{3}\int_{7x+3}^{36-4x} (7x+2y+2)\,dydx}$