# Center of Mass of an Orange Slice

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?

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.

$$\gamma(\alpha)=\frac{4a}{3\pi}$$ is the centroid of a "flat" semicircle, you cannot use that to "build" the orange slice.
• But doesn’t my method sum the wedged smaller slices contained in the larger slice? Perhaps you mean I cannot take $\gamma(d\alpha)$ to represent a semicircle? Commented Sep 14, 2020 at 9:11
$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ $$\ds{}$$
\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}