I need to find the centre of gravity of a half ball?

Question

Previously I had successfully calculated the C.O.G. of a Tetrahedron but after several attempts, I am not able to do the half ball one.

the half ball $H = \{(x, y, z) \in \Bbb R^3| z\ge 0, x^2 + y^2 + z^2 \le 1\} $

I know I have to use triple integral into this, but I think I am not able to initiate it the right way.

Answer 1

First, the volume of the half ball $B$ with radius $R$ is $\frac23\pi R^3$. In spherical coordinates $(R,\theta,\varphi)$, the coordinates of the centre of gravity $G$ are given by the integrals \begin{align}\newcommand\d{\mathop{}\!\mathrm{d}} x_G&=\frac3{2\pi R^3}\iiint_B x\d x\d y\d z=\frac3{2\pi R^3}\iiint_{\substack{0\le r \le R\\[0.35ex]0\le\theta\le 2\pi\\[0.2ex]0\le\varphi\le\tfrac\pi2}}(r\sin\varphi\cos\theta)\,r^2\sin\varphi\d r\d\theta\d\varphi \\ y_G&=\frac3{2\pi R^3}\iiint_B y\d x\d y\d z=\frac3{2\pi R^3}\iiint_{\substack{0\le r \le R\\[0.35ex]0\le\theta\le 2\pi\\[0.2ex]0\le\varphi\le\tfrac\pi2}}(r\sin\varphi\sin\theta)\,r^2\sin\varphi\d r\d\theta\d\varphi\\ z_G&=\frac3{2\pi R^3}\iiint_B z\d x\d y\d z=\frac3{2\pi R^3}\iiint_{\substack{0\le r \le R\\[0.35ex]0\le\theta\le 2\pi\\[0.2ex]0\le\varphi\le\tfrac\pi2}}(r\cos\varphi)\,r^2\sin\varphi\d r\d\theta\d\varphi \end{align}

Answer 2

By symmetry I can tell that it is on the line $x=y=0$

Now to get $z$ we need to find few things:

the volume of the shape is: $\frac23\pi r^3=\frac23\pi$

Now the integral: we are searching for an integral of $dz$ so we need a function that "have" the change of $z$ in it, let's take the function of the radius of a circle parallel to $z=0$. By Pythagorean theorem $z^2+r^2=1\implies r(z)=\sqrt{1-z^2}$ now the area of the circle is $\pi r^2=\pi (1-z^2)$, we need to add the thickness of the circle, so we get $\pi (1-z^2)dz$

Now there is one last thing before the integration, we need to multiply the circle by its "hight", by how far it is from the circle at $z=0$, because all the circles are parallel this "hight" is just $z$ so we get $\int_0^1\pi(1-z^2)z\, dz=\frac14\pi$

Now the last part, the average of a shape in $\Bbb R^3$ is the integral/the volume so we get $\frac{\frac{\pi}4}{\frac{2\pi}3}=\frac38$ and because I know that $x=y=0$ I have the point $(0,0,\frac38)$