Centre of mass of a hollow cone with open base Consider a hollow cone ( whose base is open) of Mass $M$ , radius $R$ slant height $l$ and height $h$ . I need to find the centre of mass of this cone in terms of $h$ only.
My work :

I thought that the cone is made up of a large (infinite) number of rings, each of unequal radii.
My approach  :
By symmetry, the centre of mass of cone must lie along the axis joining the centre of cone ( means the centre of the base circle) and it's vertex. Let this axis be called as $x-$axis.
Consider a ring at a height $x$
from the centre of cone ( I assumed the centre to be the origin) . Now, by similarity of triangles, the radius of this ring in terms of it's distance $x$ from centre is as ( see figure also):
$$\rm Radius = R - \frac {R×x}{h}$$
Now, the mass of cone is $M$ .
Hence, it's superficial mass density is :  $\frac {M} { π × R × l}$ , because it's mass is uniformly distributed among it's curved surface area.
Now, of a ring at distance $x$ from the centre, radius is already obtained above. Now, flatten the ring so as to form a rectangle of breadth $dx$ and length as same as circumference of the ring calculated as below :
$$2×\pi × \left [ R - \frac {R×x}{h} \right ]$$
Thus, surface area of ring as a rectangle is :
Area $=$ length $×$ breadth
Area $= 2×π× \left [ R - \frac{R×x}{h} \right ] × dx$
Multiplying above area by superficial mass density (which is calculated above) , we finally get the mass of one ring (denoted by $dM$) .
Now, the centre of mass of each ring is at it's centre (i.e. at a distance $x$ from centre of cone).
Now, the main problem arises from here.
Case 1 :

If I calculate the centre of mass as :
$$\frac { \int _ 0^h  xdM}{\int_ 0^h dM}$$

That is :
$$\frac { \int _ 0^h  xdM}{\int_ 0^h dM}= 
\frac { \int _ 0^h  x \frac {2 \pi R}{h} 
\  (h-x) \frac {M}{\pi R l} dx}{\int_ 0^h 
\frac {2 \pi R}{h} 
\  (h-x) \frac {M}{\pi R l} dx } $$
$$=\frac {h}{3} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$
I get, centre of mass $= h / 3$ ; which is of course correct as I checked from internet that in actual, it's centre of mass is at a height $h/3$ from the centre of cone.
Case 2 :

If I put integral of $dM$ in denominator of case $(1)$ directly as $M$

Then I get $h^2/3l$ , that is :
$$\frac { \int _ 0^h  xdM}{M} = \frac { \int _ 0^h  x \frac {2 \pi R}{h} 
\  (h-x) \frac {M}{\pi R l} dx}{M}$$
$$= \frac {h^2}{3l} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$
Clearly, case $(1)$ yields correct answer for centre of mass of hollow cone while case $(2)$ don't.
The problem arises in case $(2)$ because I put the denominator ( integral of $dM$ ) as equal to $M$. Can anyone tell me that why I can't put so and why putting so results in incorrect result.
 A: Consider the cone to be composed of circles with uniform linear density i.e. circles of increasing radius stacked one on top of the other. Each circle has a mass proportional to its radius, so clearly the circles at the tip of the cone contribute the least mass.
Because the cone is symmetrical about an axis, we know the center of mass must lie on this axis. Thus, we only care about the distribution of mass along this axis, and we only have to worry about how much mass is in each circle, and then weight each point along the axis by the corresponding mass in each circle.
To be more mathematical, your cone has surface area
$$
A =\int2\pi r\sqrt{{\rm d}h^2 + {\rm d} r^2} = \int_0^H {\rm d}h \,2\pi R\left(\frac{h}{H}\right)\sqrt{1 + \left(\frac{R}{H}\right)^2}\equiv\int{\rm d}h \,\ell(h) = \pi R\sqrt{R^2 + H^2}
$$
where $\ell(h)$ is the circumference of a given cross-section of the circle weighted by the appropriate measure. The surface density is
$$
\sigma = \frac{M}{A}\,.
$$
and the corresponding linear mass density
$$
m(h) \equiv \sigma\ell(h)
$$
Thus, the center of mass is located at
$$
H_{cm} = \frac{\int{\rm d}h\, m(h) h}{\int{\rm d}h\, m(h)} = \frac{2}{3} H\,.
$$
Note, in my coordinates, $h = H$ corresponds to the base of the cone, so the center of mass is closer to the base as one would expect.
