I'm asked to evaluate this:

What is the surface area of the surface defined by $\frac{x^2}{3} + \frac{y^2}{3} + \frac{z^2}{4} = 1$?

I first parameterized it with spherical coordinates and then I took the cross product and then the magnitude of the cross product of $r(\phi) \times r(\theta)$. \begin{equation} \left. \begin{aligned} x &= \sqrt{3} \cos\theta \sin\phi, \\ y &= \sqrt{3} \sin\theta \sin\phi, \\ z &= 2\cos\phi; \end{aligned}\right\}\qquad 0 < \theta < 2\pi,\quad 0 < \phi < \pi. \end{equation} I took the partials with respect to $\phi$ and $\theta$ and arrived at \begin{align*} r(\theta) &= (-\sqrt{3} \sin\theta \sin\phi, \sqrt{3} \cos\theta \sin\phi, 0), \\ r(\phi) &= (\sqrt{3} \cos\theta \cos\phi, \sqrt{3} \sin\theta \cos\phi, -2 \sin\phi). \end{align*}

Once I found the partials, I took the cross product between them and then took its magnitude to get $$ \int \sin\phi \sqrt{12\sin^2\phi + 9\cos^2\phi}\, d\phi. $$

Now, I've tried plenty of trig things but I just can't solve this integral. Can someone help me?

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    $\begingroup$ Since the arclength of a noncircular ellipse has no simple formula it's likely also not simple to get the value of the surface area of your nonspherical ellipsoid. The integral seems like it might be a transform of an elliptical integral... $\endgroup$ – coffeemath May 17 '15 at 5:54
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    $\begingroup$ The soultion is non elementary. You have been asked to compute the surface of an ellipsoid, and that is given in terms of the first and second incomplete elliptic integrals. $\endgroup$ – Rogelio Molina May 17 '15 at 5:55
  • $\begingroup$ Have a look at en.wikipedia.org/wiki/Ellipsoid for the general case. Interesting are he approximate formulas for the general case and the exact formulas for the oblate and prolate cases. $\endgroup$ – Claude Leibovici May 17 '15 at 6:32
  • $\begingroup$ You changed the equation (which is no more readable) but you don't tell how you arrived to this integral which by the way results in a complex result. Please, fix the equation and explain. $\endgroup$ – Claude Leibovici May 17 '15 at 8:17
  • $\begingroup$ Hi, I edited it. Hopefully it is more clear. $\endgroup$ – Alex May 17 '15 at 8:38

Note that two of the axes of the ellipsoid are the same, and the third is longer. Thus it is actually a prolate spheroid. The formula for the surface area of the prolate spheroid $\frac{x^2 + y^2}{a^2} + \frac{z^2}{c^2}=1$ is: $$ S=2\pi a^2 + 2\pi\frac{ac}{e}\sin^{-1}e $$ where $e$ is the ellipticity $\sqrt{1-\frac{a^2}{c^2}}$.

This formula is derived by considering the spheroid as a surface of revolution about the $z$-axis and doing the usual integration to find its area:

$$S = 2\pi \int_{-c}^{c}{r(z) \sqrt{1+(r'(z))^2}\,dz}$$

where $r(z) = a\sqrt{1-\frac{z^2}{c^2}}$ is the radius of the circular cross section at height $z$.


  • $\begingroup$ Hi! Thank you for your comment. Could you set up the integral for deriving the formula? I'm not quite understanding what to do. Do you disregard the "z" value? $\endgroup$ – Alex May 17 '15 at 7:58
  • $\begingroup$ @Alex The link gives the details, although you can also do this using the integral you originally set up, per user86418's comments. $\endgroup$ – augurar May 17 '15 at 19:49

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