Find the centre of gravity of an elliptic arc.

Question

Find, without using Pappus' theorem, the centre of gravity of an arc of an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ which situated in the first quadrant.

Answer 1

For $a>b$,

\begin{align*} (x,y) &= (a\sin t,b\cos t) \\[5pt] e &= \sqrt{1-\frac{b^2}{a^2}} \\[5pt] ds &= \sqrt{a^2\cos^2 t+b^2\sin^2 t} \, dt \\[5pt] &= a\sqrt{1-e^2\sin^2 t} \, dt \\[5pt] \int_{0}^{\frac{\pi}{2}} ds &= aE(e) \\[5pt] \int_{0}^{\frac{\pi}{2}} x \, ds &= \int_{0}^{\frac{\pi}{2}} a^2\sin t \sqrt{1-e^2\sin^2 t} \, dt \\[5pt] &= \frac{a^2}{2}+\frac{ab^2}{2\sqrt{a^2-b^2}} \sinh^{-1} \frac{\sqrt{a^2-b^2}}{b} \\[5pt] \int_{0}^{\frac{\pi}{2}} y \, ds &= \int_{0}^{\frac{\pi}{2}} ab\cos t \sqrt{1-e^2\sin^2 t} \, dt \\[5pt] &= \frac{b^2}{2}+\frac{a^2b}{2\sqrt{a^2-b^2}} \sin^{-1} \frac{\sqrt{a^2-b^2}}{a} \\[5pt] \begin{pmatrix} \bar x \\ \bar y \end{pmatrix} &= \frac{1}{2E(e)} \begin{pmatrix} a+\dfrac{b^2}{\sqrt{a^2-b^2}} \sinh^{-1} \dfrac{\sqrt{a^2-b^2}}{b} \\[3pt] \dfrac{b^2}{a}+\dfrac{ab}{\sqrt{a^2-b^2}} \sin^{-1} \dfrac{\sqrt{a^2-b^2}}{a} \end{pmatrix} \\ \end{align*}

Compare the results with surface areas of spheroids here, an implication of Pappus' centroid theorem.