Could I ask for help to show where I'm going wrong?


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:


(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:


(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:

$a^2\sin\theta\,(a\sin\theta)^2\,\delta\theta\,\delta\phi$ =


Total $2_{nd}$ moment of area = $a^4\int_{\phi=0}^{\pi/2}\,\int_{\theta=0}^{\pi/2}sin^3\theta\,d\theta\,d\phi\,=$




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:


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:


Where have I gone wrong?

  • 1
    $\begingroup$ Why do you think it should be a/2? It would be if it were a full a-by-a-by-a cube but this is clearly less than that. $\endgroup$
    – Deusovi
    Commented Nov 12, 2015 at 11:38
  • $\begingroup$ They're not asking for the second moment about the z-axis, they're asking for the center if gravity, which is an easier caclulation. $\endgroup$ Commented Nov 12, 2015 at 11:54

2 Answers 2


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 $\zeta={a\over2}$.

  • $\begingroup$ +1 for the "abstract nonsense proof", which I much prefer to your calculus proof. $\endgroup$
    – TonyK
    Commented Nov 12, 2015 at 17:03

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 $(\frac{3}{8}a, \frac{3}{8}a, \frac{3}{8}a)$.


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