Center of Mass, cones, polar coordinates, etc

Hey guys, its been over a year since I've taken Calc 3 and forget some things.

I'm trying to find the Center of Mass of an arbitrary half cone, and I figure I'm supposed to be doing triple integration with polar coordinates.
I just can't remember how I should be setting up my integrals.
Assuming the maximum radius is a, and the height is h, should it go something like this?
V=int(0,pi)int(0,a)int(r,a)r^3(dh)(dr)(dtheta)
and the CoM (along the y axis) is something along the lines of (int(0,pi)int(0,a)int(r,a)rsin(theta)r^3(dh)(dr)(dtheta))/V
Then I do similarly for the x and z axes?

You seem to be using cylindrical coordinates.

V = int(0,2pi)int(0,a)int(0,h[1-(r/a)])r(dz)(dr)(dtheta)
 = 2pi h int(0,a){(1-(r/a))r}
 = 2pi h (a^2)/6
 = pi (a^2) h/3 as usual

Your integral gives (a^5)/20, which doesn't even have units of volume...

You're assuming constant density?  The x and y coordinates of the CoM should be trivially 0 by symmetry.

CoM_y = int(0,2pi)int(0,a)int(0,h[1-(r/a)])[rsin(theta)]r(dz)(dr)(dtheta)
= 0 due to the theta integral

Find the volume of the cone and halve it. Cone volume = (1/3) * pi^2 * h

half-cone, so x would =0,but y would not.  But that just means I change the 2pi into pi, I'm assuming...  Anyway, thanks.  I think I've got it now.