One finds the centre of mass for these bodies just as one
does for any body, by integration.
The $x_i$-coordinate of the centre of mass of a region $A$ in $\mathbb{R}^n$
is
$$\frac{\int_A x_i\ dx_1\cdots dx_n}{\int_A\ dx_1\cdots dx_n}.$$
Here the region might as well be that between the planes $x_n=a$ and $x_n=b$
in the sphere of radius $r$ centred at the origin. Using the symmetry
of $A$ this ratio of integrals equals
$$\frac{\gamma_{n-1}\int_a^b x_n(r^2-x_n^2)^{(n-1)/2} dx_n}
{\gamma_{n-1}\int_a^b (r^2-x_n^2)^{(n-1)/2} dx_n}$$
(in the $x_n$ direction) where $\gamma_{n-1}$ is the volume of the $(n-1)$-dimensional
unit ball (and obligingly cancels). These integrals can be attacked by
trig substitutions.
Edited
It's now clear that my original interpretation of your question
was wrong. However it's still not clear whether your centre of mass
is for a solid orthant or its curved surface. In any case if your orthant
is defined by the conditions $x_1,\ldots,x_k\ge0$ then its centre
of mass has the form $(a,\ldots,a,0\ldots,0)$ where $a$ depends on $r$
and $n$ but not on $k$. One sees this from the symmetry of the problem.
Thus the problem reduces to the hemispheric case. In the solid case the
answer is
$$a=\frac{\int_0^r x(r^2-x^2)^{(n-1)/2} dx}
{\int_0^r (r^2-x^2)^{(n-1)/2} dx}$$
while in the "shell" case it is
$$a=\frac{\int_0^r x(r^2-x^2)^{(n-3)/2} dx}
{\int_0^r (r^2-x^2)^{(n-3)/2} dx}$$
(if I've done my sums right).
