Deriving formula for surface area of an ellipsoid 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

 A: 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 $\sinh iz=i\sin z$

