<s>[INST] Area of ​the surface of revolution of the ellipsoid I need to find the surface area of an ellipsoid using the equation of an ellipse. I believe my calculations are correct but the formulas I meet on the Internet are complex and have $\arcsin$ or $\arctan$ in the equation, which makes me believe that there is something wrong in my calculations.
Can anyone help me if there is something wrong? The integration limits are $0$ to $a$ to calculate half of the area but multiplied by $2$ to calculate the whole area.
\begin{align*}
 y(x) &= \sqrt{b^2 a^2-x^2} \\
 y'(x) &= -\frac{x}{\sqrt{b^2a^2-x^2}} \\
 S &= 2\pi \int_a^b f(x)\sqrt{1 + y'^2} dx \\
 &= 4\pi\int_0^a \sqrt{b^2a^2-x^2}
    \sqrt{1 + \left( \frac{-x}{\sqrt{b^2a^2-x^2}} \right)^2} dx \\
 &= 4\pi\int_0^a \sqrt{b^2a^2-x^2}
    \sqrt{1 + \left( \frac{x^2}{b^2a^2-x^2} \right)} dx \\
 &= 4\pi\int_0^a \sqrt{b^2a^2-x^2}
    \sqrt{\frac{b^2a^2}{b^2a^2-x^2}}dx \\
 &= 4\pi\int_0^a \sqrt{b^2a^2} dx \\
 &= 4\pi\sqrt{b^2a^2} \left[ x \right]_0^a \\
 &= 4\pi a\sqrt{b^2a^2}
\end{align*}
 [/INST] The graph $y = \sqrt{a^{2} b^{2} - x^{2}}$ is the upper half of the circle
$$
x^{2} + y^{2} = a^{2} b^{2}
$$
of radius $|ab|$. Revolving gives a portion of a sphere, so it's not entirely a surprise the area element came out to be independent of $x$.
The correct area integral for a spheroid comes from
$$
y = \frac{b}{a}\sqrt{a^{2} - x^{2}},\qquad
y' = -\frac{bx}{a\sqrt{a^{2} - x^{2}}};
$$
this gives
$$
dS = 2\pi y\sqrt{1 + (y')^{2}}\, dx
   = 2\pi b\sqrt{1 - \frac{a^{2} - b^{2}}{a^{4}} x^{2}}.
$$
How you proceed next depends on the sign of $a^{2} - b^{2}$, but in either case you'll need something like a trig or hyperbolic substitution. The result is, as you found on the Internet, a more complicated function of $a$ ad $b$ than your polynomial, though the integral does turn out to be elementary.
By contrast, the area of the ellipsoid
$$
\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}}  = 1,\qquad
0 < a < b < c,
$$
is not an elementary function of $a$, $b$, and $c$.
</s>