Deriving formula for surface area of an ellipsoid

Question

I am doing some research on ellipsoids. I am not sure where the formula for the surface area of a prolate ellipsoid comes from. Can anyone please help me with how to derive the formula. I have the formula below

Answer 1

The method is very standard and appears in most calculus texts.

Let $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ be the ellipse such that $a>b$.

• For prolate spheroid, it rotates about the $x$-axis.

\begin{align*} y &= \frac{b}{a} \sqrt{a^2-x^2} \\ \frac{dy}{dx} &= -\frac{bx}{a\sqrt{a^2-x^2}} \\ ds &= \sqrt{1+\left( \frac{dy}{dx} \right)^2} \, dx \\ &= \sqrt{1+\frac{b^2x^2}{a^2(a^2-x^2)}} \, dx \\ &= a\frac{\sqrt{1-\left( 1-\frac{b^2}{a^2} \right) \frac{x^2}{a^2}}} {\sqrt{a^2-x^2}} \, dx \\ S &= \int_{-a}^{a} 2\pi y \, ds \\ &= 4b\pi \int_{0}^{a} \sqrt{1-\left( 1-\frac{b^2}{a^2} \right)\frac{x^2}{a^2}} \, dx \\ &= 4b\pi \left[ \frac{x}{2} \sqrt{1-\left( 1-\frac{b^2}{a^2} \right) \frac{x^2}{a^2}}+ \frac{a^2}{2\sqrt{a^2-b^2}} \sin^{-1} \frac{x\sqrt{a^2-b^2}}{a^2} \right]_{0}^{a} \\ &= 2\pi b \left( b+\frac{a^2}{\sqrt{a^2-b^2}} \sin^{-1} \frac{\sqrt{a^2-b^2}}{a} \right) \\ \end{align*}

• For oblate spheroid, it rotates about the $y$-axis.

\begin{align*} x &= \frac{a}{b} \sqrt{b^2-y^2} \\ \frac{dx}{dy} &= -\frac{ay}{b\sqrt{b^2-y^2}} \\ ds &= \sqrt{1+\left( \frac{dx}{dy} \right)^2} \, dy \\ &= \sqrt{1+\frac{a^2y^2}{b^2(b^2-y^2)}} \, dy \\ &= b\frac{\sqrt{1+\left( \frac{a^2}{b^2}-1 \right) \frac{y^2}{b^2}}} {\sqrt{b^2-y^2}} \, dy \\ S &= \int_{-b}^{b} 2\pi x \, ds \\ &= 4a\pi \int_{0}^{b} \sqrt{1+\left( \frac{a^2}{b^2}-1 \right)\frac{y^2}{b^2}} \, dy \\ &= 4a\pi \left[ \frac{y}{2} \sqrt{1+\left( \frac{a^2}{b^2}-1 \right) \frac{y^2}{b^2}}+ \frac{b^2}{2\sqrt{a^2-b^2}} \sinh^{-1} \frac{y\sqrt{a^2-b^2}}{b^2} \right]_{0}^{b} \\ &= 2\pi a \left( a+\frac{b^2}{\sqrt{a^2-b^2}} \sinh^{-1} \frac{\sqrt{a^2-b^2}}{b} \right) \\ \end{align*}

The two cases are interchangeable by flipping the roles of $a$ and $b$ together with $\sinh iz=i\sin z$