My professor gave us this question on a calculus II quiz. One of my calculus III pals suggested I use surface integrals, but that tool is not available to us (I don't know how to use it yet, nor do my classmates). We only know how to use the surface area formula for a curve rotated about a line or axis, which is

S= $\int_a^b 2π*f(x)*\sqrt{1+f'(x)^2} dx$

Surface area of the ellipsoid $\frac{x^2}{16}+\frac{y^2}{8}+z^2=1$

We have tried to find an equation for the curve of the ellipse and rotate it about the x-axis, removing the variable $z^2$ and solving $\frac{x^2}{16}+\frac{y^2}{8}=1$ for y, which yielded $y=\sqrt{8-\frac{x^2}{2}}$ We then solved to find where the curve crossed the x-axis, at -4 and 4. Then, we took the derivative of that and plugged all components into the formula above. Somehow, we are not getting a correct answer. If anyone else knows of a different approach we could try, please let us know!

Additionally, if you can modify the equation so that there is a coefficient of $z^2$ that makes the problem solvable, I may be able to try it from there.

Thank you!

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    $\begingroup$ The first problem is that your ellipsoid has three different length axes, so you can't form it by rotating a curve about one of the axes (unless you change scale to make two of the axes the same length, but you'd have to examine what that did to the surface area). $\endgroup$ – Chappers Jun 22 '15 at 15:44
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    $\begingroup$ If you look at the formula for the surface area of a general ellipsoid this seems to be beyond calculus II. As @Chappers noted, it would be easier if it were an ellipsoid of revolution, but it isn't. $\endgroup$ – Rory Daulton Jun 22 '15 at 15:52
  • $\begingroup$ @RoryDaulton Wow, that's even worse than I thought it might be. Although not surprising that elliptic integrals appear! $\endgroup$ – Chappers Jun 22 '15 at 15:54
  • $\begingroup$ Right, I emailed my professor to ask whether it was a curve $\frac{x^2}{16}+\frac{y^2}{8}=1$ rotated about some line (as that was the original equation we were given), but he sent me back a "correction" which was just to add the $z^2$ which served to only further complicate the problem. $\endgroup$ – Carly Sawatzki Jun 22 '15 at 15:56
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    $\begingroup$ Either $\frac{1}{8}$ or $\frac{1}{16}$ as coefficient for $z^2$ will turn the problem into a surface of revolution. $\endgroup$ – vadim123 Jun 22 '15 at 16:28

let $x=4\sin{u}\cos{v},y=2\sqrt{2}\sin{u}\sin{v},z=\cos{u}$. then $$E=x''_{u}+y''_{u}+z''_{u}=16\cos^2{u}\cos^2{v}+8\cos^2{u}\sin^2{v}+\sin^2{u}$$ $$F=x''_{v}+y''_{v}+z''_{v}=16\sin^2{u}\sin^2{v}+8\sin^2{u}\cos^2{v}$$ $$G=x'_{u}x'_{v}+y'_{u}y'_{v}+z'_{u}z'_{v}=-16\sin{u}\cos{u}\sin{v}\cos{v}+8\sin{u}\cos{u}\sin{v}\cos{v}$$ so \begin{align*}EG-F^2&=(128\cos^2{u}+16\sin^2{u}\sin^2{v}+8\sin^2{u}\cos^2{v})\sin^2{u}\\ &=8\sin^2{u}+120\sin^2{u}\cos^2{u}+8\sin^4{u}\cdot \sin^2{v} \end{align*} so Surface area is $$I=\int_{0}^{\pi}\int_{0}^{2\pi}\sqrt{8\sin^2{u}+120\sin^2{u}\cos^2{u}+8\sin^4{u}\cdot \sin^2{v}}dvdu$$ then use Elliptic integral

  • $\begingroup$ This seems to be a good answer, but it is not: the OP clearly asks for an answer not involving surface integrals (that he has not yet studied). Why don't people read the questions attentively? Why do they post answers only do get votes and reputation? $\endgroup$ – Alex M. Jun 22 '15 at 17:03
  • $\begingroup$ @AlexM.: The answers are using surface integrals because there doesn't seem to be another way. I don't see a way that it can be changed to a surface of revolution. The answers may be useful to someone other than the OP, or possibly to the OP when they do learn surface integrals. $\endgroup$ – robjohn Jun 22 '15 at 17:12
  • $\begingroup$ @math110: Numerically, our answers seem to agree. $\endgroup$ – robjohn Jun 22 '15 at 17:20
  • $\begingroup$ @math110: is one of the $F$s supposed to be a $G$? $\endgroup$ – robjohn Jun 22 '15 at 20:01
  • $\begingroup$ @robjohn,Oh,Yes, $\endgroup$ – math110 Jun 23 '15 at 1:27

The normal to the surface is $$ \left(\frac x8,\frac y4,2z\right) $$ The ratio of the normal component to the $z$-component is $$ \frac{\sqrt{\frac{x^2}{64}+\frac{y^2}{16}+4z^2}}{2z} $$ Thus, using $z^2=1-\frac{x^2}{16}-\frac{y^2}8$, the surface area is $$ \begin{align} &2\iint_{\frac{x^2}{16}+\frac{y^2}{8}\lt1}\frac{\sqrt{4-\frac{15x^2}{64}-\frac{7y^2}{16}}}{2\sqrt{1-\frac{x^2}{16}-\frac{y^2}8}}\,\mathrm{d}y\,\mathrm{d}x\\ &=2\int_0^4\int_0^{\sqrt{8-\frac{x^2}2}}\sqrt{\frac{256-15x^2-28y^2}{16-x^2-2y^2}}\,\mathrm{d}y\,\mathrm{d}x\\ &=\sqrt2\int_0^4\int_0^{\sqrt{16-x^2}}\sqrt{\frac{256-15x^2-14y^2}{16-x^2-y^2}}\,\mathrm{d}y\,\mathrm{d}x \end{align} $$ In the last step, we substituted $y\mapsto y/\sqrt2$.

Numerically, this is $83.974845470544$, but I don't have a closed form for the integral.

Unless I am missing some key simplification, this seems a bit advanced for Calc II.

  • $\begingroup$ To the downvoter: unless you say otherwise, I must assume that the reason is that this answer uses surface integrals. As I've mentioned in a comment to another answer, there just does not seem to be any other way to handle the question. So we have a choice of simply saying, "sorry, it can't be done that way," or saying, "here is a way it can be done. When you get to the appropriate section, come back and look again." I chose the one that offers a bit more help. $\endgroup$ – robjohn Jun 22 '15 at 19:40

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