Center of mass of a hemisphere with uniform surface density

[JJfortherear]

Homework Statement

A hemisphere of radius R, covering +/- R in the x and y directions and 0 to R in the Z direction (only the top half of a sphere centered at the origin)
Find the center of mass.

Homework Equations

Cm=(1/m)$$\int(z dm)$$

The Attempt at a Solution

dm is sigma dA, and dA is x dz, where x = $$\sqrt{}R^2-z^2$$. The limits are 0 to R, and the answer should be 3/8 R, but when I evaluate the integral I get nothing like that.

 

[vik10622]

since you are trying to find for a hemisphere not for a hemispherical shell, your equation for 'dm' is wrong.
You should consider not the surface density, but instead the volume density ρ since it is a solid object
$\rho = \frac{M}{\frac{2}{3} \pi R^3}$

You then have

$dm = \rho A dz$ where A is the area of the circular disc centered on the z-axis with radius $r = \sqrt{R^2 - z^2}$.

The final integral then becomes

$CM (z-comp) = \frac{1}{M} \int_0^R \ dz\ \pi (R^2 - z^2) \rho\ z$

which on solving gives the result $$\frac{3 R}{8}$$