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Using calculus, I attempted to find a formula for the surface area of an ellipsoid, which is a solid obtained by rotating the ellipse $\displaystyle\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ around the $x$-axis.

Link: Ellipsoid Surface Area ($C$ represents circumference, $A$ represents area)

Unfortunately, I don't think my formula $A_{ellipsoid} = \pi^2 ab$ is correct because in the case of a sphere with uniform radius $r$, my formula yields $\pi^2 r^2$ even though it is commonly known that a sphere's surface area is given by $4 \pi r^2$.

I would like to know what I did wrong in my derivation, so far I cannot spot any mistakes.

Alexander Bullen
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    You were supposed to find a surface area not a volume. You rotated the function. If you want surface area what were you supposed to rotate? – Ninad Munshi Aug 24 '20 at 00:51
  • @NinadMunshi I have to rotate the ellipse to get a circumference of each individual circle along the ellipsoid shape, so I can multiply that by an infinitesimally small width dx, which yields the infinitesimally small surface area. Click on the link for details. – Alexander Bullen Aug 24 '20 at 01:24
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    First of all, you post a link instead of typing up your work which is expected on this site. Second, despite this inconsiderateness, I still clicked the link and read through all of your work, and pointed out what was wrong, and then you claim I didn't even bother to click your link? That is more than a little rude. You aren't supposed to rotate the ellipse function, that will get you volume. For any function, what do you have to rotate to get surface area? ($2\pi f(x) dx$ is not correct) – Ninad Munshi Aug 24 '20 at 01:31
  • There's no closed form formula for the circumference of an ellipse, just a formula involving elliptic integrals. Maybe it's the same for the surface area of an ellipsoid. – Gerry Myerson Aug 24 '20 at 01:51
  • @NinadMunshi I'm looking for the surface area that surrounds the 3-dimensional shape you can see in the diagram. Hopefully you can understand the procedure I followed, where I add up the areas of each strip along the ellipsoid, which is given by circumference x width (i.e. 2πf(x)dx). – Alexander Bullen Aug 24 '20 at 02:23
  • $2\pi f(x) dx$ is not the correct formula. Don't just repeat that after I just wrote that it was wrong, I might get the impression that you hadn't read my short paragraph. The answer below has the right formula. – Ninad Munshi Aug 24 '20 at 02:26
  • @NinadMunshi Ok I understand now. I didn't know I had to take into account the slope of the surface, my formula assumed the surface of each strip was cylindrical. – Alexander Bullen Aug 24 '20 at 02:41

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Surface area of revolution... uses this formula.

$2\pi\int_{-a}^{a} y\sqrt {1+(\frac {dy}{dx})^2} \ dx$

And you shouldn't need a trig substitution to evaluate it.

Doug M
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