# Find the center mass of right circular conic shell of base radius $a$ and height $h$ and constant density

We were assigned the question for homework: Find the center of mass of a right circular conic shell, radius $a$, height $h$ and constant density $\rho$. This is a multi-variable calculus class and we just learned about surface integrals.

To find the CM you compute: $$x=\frac{M_{x=0}}{m}= \frac{\int\int\int_R x \rho dV}{\int\int\int_R \rho dV}$$

And similarly for $y$ and $z$. Now I know you can assume $x=0$ and $y=0$ by symmetry. Therefore the CM of the cone will lie somewhere on the $z$- axis. My major issue here is that I don't understand how to compute the SA of the cone. If I can get this I understand how to do the rest of the question.

If anyone could help me I would appreciate it!

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.

• Ah, yes ok. I was forgetting about the relation including the radius and height. Thanks so much for your help! – user68203 Apr 6 '13 at 18:33
• You are welcome. The substitution $x=r\cos\theta$, $y=r\sin\theta$ may be useful once you have set up the integral. – André Nicolas Apr 6 '13 at 19:03