Center of Mass of an Orange Slice

Question

In a question in the book Advanced Problems in Mathematics by Stephen Siklos, the reader is told to assume that the center of mass of a spherical ‘segment’ (orange slice) of radius $a$ subtending an angle $2\theta$ at the vertical axis passing through the sphere’s center lies at a distance $$\frac{3\pi a\sin\theta}{16\theta}$$ from the axis. I have tried an unusual way to get to this expression, but have ended up at a different expression:

Say the orange slice is such that the axis passing through the center lies along the $X$-axis with the origin at the sphere’s center and the $Y$-axis directed from the axis to orange’s surface. I then consider infinitesimal orange slices of radius $a$ subtending angle $d\alpha$ at the axis, and sum up the product of their masses and the $Y$-coordinate of their centers of mass, as there is no $X$-component of the COM of our original orange slice.

Let $\gamma(\alpha)$ be the function that takes in the angle subtended and returns the distance of COM of any orange slice of radius $a$ from its axis. I arrive at the following: $$\gamma(2\theta) = \frac{\int_{-\theta}^{\theta} \gamma(d\alpha) \cos \alpha \ dm }{\int dm} = \frac{\gamma(d\alpha)}{2\theta} \int_{-\theta}^{\theta} \cos \alpha \ d\alpha$$ I then think of $\gamma(d\alpha)$ as $\lim_{\alpha\to0} \gamma(\alpha)=\frac{4a}{3\pi}$ which is the semicircular disc case, so $$\gamma(2\theta) = \frac{4a\sin\theta}{3\pi \theta} $$ Clearly, something has gone wrong. Can someone guide me on where my error(s) are?

Answer 1

Speaking intuitively, an orange slice has variable thickness, null at the axis and large at the surface. It is a wedge taken from a sphere.

Therefore you can not treat it as a "book", although semi-circular, slightly opened and sum the barycenters of the flat (instead of "wedged") "pages" fanning out.

--- addendum in reply to your comment ---

$\gamma(\alpha)=\frac{4a}{3\pi}$ is the centroid of a "flat" semicircle, you cannot use that to "build" the orange slice.
That of a thin wedged slice will be more external, since there it is relatively more thick.

Answer 2

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\begin{align} &\bbox[5px,#ffd]{\displaystyle% \int_{-\theta}^{\theta}\int_{0}^{\pi}\int_{0}^{a} r\sin\pars{\xi}\cos\pars{\phi}\, r^{2}\sin\pars{\xi}\,\dd r\,\dd\xi\,\dd\phi \over \displaystyle \int_{-\theta}^{\theta}\int_{0}^{\pi}\int_{0}^{a} r^{2}\sin\pars{\xi}\,\dd r\,\dd\xi\,\dd\phi} \\[5mm] = &\ {\pars{a^{4}/4}\pars{\pi/2}\bracks{2\sin\pars{\theta}} \over \pars{a^{3}/3}\pars{2}\pars{2\theta}} = \bbx{{3\pi \over 16}\,{\sin\pars{\theta} \over \theta}\, a} \\ & \end{align}