Center of mass of semi-sphere I haven't calculated center of mass before and I'd like to know how I can do it in practise. 
I want to find the center of mass of a semi-sphere. Could you explain me, step by step, what I have to do?
Many thanks
 A: We calculate the centre of mass of a half-ball of radius $1$.  Without loss of generality we may assume that the ball is made of material with density $1$.  
Imagine that the ball is sitting on a table, flat side down.  By symmetry the centre of mass is on the vertical line through the centre of the ball. The only question is: How far up? 
We will calculate the moment of the ball about the plane of the table, and divide by the mass of the half-ball.  By  a standard formula, the mass of the half-ball is $\dfrac{2\pi}{3}$.
Imagine now that the half-ball is an industrial ham. Imagine a very thin slice of that ham, sliced parallel to the table, but left in place.  Let the slice be taken from height $z$ to height $z+dz$, where $dz$ is extremely small.  The slice is almost a cylinder of very small height $dz$. 
We first calculate the radius $r=r(z)$ of the slice. By the Pythagorean Theorem, we have $r^2+z^2=1$, so $r=\sqrt{1-z^2}$.
Thus the area of the slice is $\pi r^2=\pi(1-z^2)$.  The thickness is $dz$, so the volume, and therefore the mass, of the slice is approximately $\pi (1-z^2)\,dz$.
The slice is at perpendicular distance $z$ from the table. So the moment of the slice about the plane of the table is approximately $\pi (1-z^2)(z)\,dz$.
"Add up" (integrate) from $z=0$ and $z=1$. The full moment of the ball is
$$\int_0^1 \pi (1-z^2)(z)\,dz.$$
Calculate. We get $\dfrac{\pi}{4}$.
Finally, divide by the mass $\dfrac{2\pi}{3}$. We get $\dfrac{3}{8}$.
For a ball of radius $R$, just multiply by $R$. The centre of mass is $\dfrac{3 R}{8}$ above the centre of the half-ball.   
A: Let $f(x) = \sqrt{r^2-x^2}$ and model the hemisphere as $H=\{(x,y) | x \in [0,r], \, |y| \le f(x) \}$.
Then compute $\overline{x} = \frac{\int_0^r 2 \pi xf(x)^2 dx}{\int_0^r 2 \pi f(x)^2 dx}$.
These integrals are straightforward to evaluate:
$\int_0^r 2 \pi xf(x)^2 dx = 2 \pi \int_0^r x ( r^2-x^2) dx = 
2 \pi \frac{r^4}{4}$.
$\int_0^r 2 \pi f(x)^2 dx = 2 \pi \int_0^r ( r^2-x^2) dx = 
2 \pi \frac{2 r^3}{3}$.
$\overline{x} = \frac{3}{8} r$.
A: 1- Assume the flat surface is located on a horizontal plane with the bowl (semi-sphere) below)
2- X = left/right
3- Z = frt/back
4- Y = vertical offset
Both X & Z coordinates will be on the axis of rotation, i.e. at 0 offset. The center of mass will be .1875 * D from the flat surface of the semi-sphere.
RJ
A: $a > 0$: radius.
\begin{align}
\vec{r}_{\rm cm}
&\equiv
{1 \over V}\int_{V}\vec{r}\,{\rm d}V
=
{1 \over \left(4\pi a^{3}/3\right)/2}\int_{V}{1 \over 2}\nabla r^{2}\,{\rm d}V
=
{3 \over 4\pi a^{3}}\int_{S}r^{2}\,\hat{r}\,{\rm d}S
\\[3mm]&=
{3 \over 4\pi a^{3}}\left[%
\int_{\Huge\frown}a^{2}\,{z \over a}\,\hat{z}\,{\rm d}S\
+\
\int_{\Huge\_}\left(x^{2} + y^{2}\right)\left(-\hat{z}\right)\,{\rm d}x\,{\rm d}y
\right]
\\[3mm]&=
{3 \over 4\pi a^{3}}\,\hat{z}\left\lbrack%
a^{2}\int_{0}^{2\pi}
\int_{0}^{\pi/2}\cos\left(\theta\right)\
a^{2}\sin\left(\theta\right)\,{\rm d}\theta\,{\rm d}\phi\
-\
\int_{0}^{a}\rho^{3}\,{\rm d}\rho\int_{0}^{2\pi}{\rm d}\phi
\right]
\end{align}
$$
\begin{array}{|c|}\hline\\
\color{#ff0000}{\large\quad%
\vec{r}_{\rm cm} = {3 \over 8}\,a\;\hat{z}
\quad}
\\ \\ \hline
\end{array}
$$
