Centre of gravity of eighth of a sphere Could I ask for help to show where I'm going wrong?
Question
Find the position of the center of gravity of that part of a thin spherical shell x^2 + y^2 + z^2 = a^2 which exists in the first octant.
My Answer
Working in spherical coordinates:
The $2_{nd}$ moment of area of a small area element $dA$ about the z-axis is:
$dA=(a\sin\theta)^2\,\delta{A}$
(This is because in spherical coordinates $(r,\theta,\phi)$ the perpendicular distance of an area element of the shell from the z-axis is $a\sin\theta$ and the product of area and this distance squared is the 2nd moment of area)
Now, in this problem, in spherical coordinates the elemental area is given by:
$\delta{A}=a^2\,\sin\theta\,\delta\theta\,\delta\phi$
(this comes from the general result that in spherical coordinates $(r,\theta,\phi)$ the elemental area is given by $\delta{A}=r^2\,\sin\theta\,\delta\theta\,\delta\phi\,$ but in this problem r = a is constant)
So we have that the elemental $2_{nd}$ moment of area is given by:
$a^2\sin\theta\,(a\sin\theta)^2\,\delta\theta\,\delta\phi$ = 
$a^4sin^{3}\theta\,\delta\theta\,\delta\phi\,\therefore$
Total $2_{nd}$ moment of area = $a^4\int_{\phi=0}^{\pi/2}\,\int_{\theta=0}^{\pi/2}sin^3\theta\,d\theta\,d\phi\,=$
$a^4\int_{\theta=0}^{\pi/2}\frac{2}{3}\,d\phi\,=$
$a^4\left[\frac{2}{3}\phi\right]_0^{\pi/2}\,=$
$a^4(\frac{2}{3}\frac{\pi}{2})\,=\,\frac{a^4\pi}{3}$
Now, we can equate this with the radius of gyration (k) of the eighth-of-a-sphere and its total area (A) as follows:
$\frac{a^4\pi}{3}=Ak^2=\frac{4\pi a^2}{8}k^2$
(Dividing by eight as we only have one eighth of a sphere)
And so, this leads to:
$k=\frac{\sqrt{2}a}{\sqrt{3}}$
And as the radius of gyration k is the distance at which a point mass of the same mass as the eighth-of-a-sphere will produce an equivalent moment as the whole eighth-of-a-sphere we have that the x-compnent of the centre of mass of the whole object is $\frac{\sqrt{2}a}{\sqrt{3}}$ and so by symmetry the answer to the whole problem is $(\frac{\sqrt{2}a}{\sqrt{3}},\frac{\sqrt{2}a}{\sqrt{3}},\frac{\sqrt{2}a}{\sqrt{3}})$
However, I believe the answer should be:
$(a/2,a/2,a/2)$
Where have I gone wrong?
 A: I'm using spherical coordinates with $\theta=0$ at the equator. A latitude zone of your spherical triangle  $S$ of width $d\theta$ and at the geographical latitude $\theta$ is a quarter of an "infinitesimal lampshade". It's $z$-level is $z=a\>\sin\theta$, and its  area is given by
$$dA={\pi\over 2}\>a\cos\theta\cdot a\>d\theta\ .$$
Weighing in this infinitesimal piece of surface with its vertical height $z$ results in
$$dM_z=z\cdot dA={\pi\over2}a^3\>\sin\theta\>\cos\theta\>d\theta\ .$$
It follows that the $z$-coordinate of the centroid (the "average $z$-value" of the area elements) computes to
$$\zeta={M_z(S)\over A(S)}={1\over{\pi\over2} a^2}\>{\pi\over2} a^3\int_0^{\pi/2}\sin\theta\cos\theta\>d\theta={a\over2}\ .$$
Due to symmetry the coordinates of the centroid then are
$$(\xi,\eta,\zeta)=\left({a\over2},{a\over2},{a\over2}\right)\ .$$
An abstract nonsense proof of this could go as follows: It suffices to compute $\zeta$. The $\zeta$ of your $S$ is the same as the $\zeta$ of the full upper hemisphere. It is well known that the spherical area between two horizontal planes $z={\rm const.}$ is equal to the corresponding area of a circumscribed cylinder. But for the cylinder it is obvious that $\zeta={a\over2}$.
A: edit: Never mind this answer. I mistakenly though you were talking about a solid eight of a sphere.
Why use spherical coordinates at all? 
Let $A$ be the one-eight sphere located in the positive octant. The total volume of $A$ is just $\frac{a^3\pi}{6}$. So to calculate the $z$ component of the center of mass, and by symmetry the others, we need to calculate $$\frac{6}{a^3\pi}\int_AzdV$$
But for a fixed $z$ the horizontal slice through A at height $z$ is just a quarter circle of radius $\sqrt{a^2-z^2}$ which has a very convenient area of simply $\frac{\pi}{4}(a^2-z^2)$. So we can compute as follows
$$\begin{align}
\frac{6}{a^3\pi}\int_AzdV &= \frac{6}{a^3\pi} \int_0^a z\cdot \frac{\pi}{4}(a^2-z^2)dz\\
&=\frac{3}{2a^3}\int_0^a a^2z-z^3dz \\
&=\frac{3}{2a^3}\left[\frac{a^2z^2}{2}-\frac{z^4}{4}\right]_0^a \\
&=\frac{3}{2a^3}(\frac{a^4}{2}-\frac{a^4}{4}) \\
&=\frac{3}{2a^3}\cdot\frac{a^4}{4} = \frac{3}{8}a
\end{align}$$
So the center of mass is at $(\frac{3}{8}a, \frac{3}{8}a, \frac{3}{8}a)$.
