1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
3
$\begingroup$

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 $z$.

Source:

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.