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The elliptic paraboloid has height h, and two semiaxes a, b. How to find its surface area? Does it possible to use a direct formula without integrals?

Ivan Bunin
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2 Answers2

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The elliptic paraboloid is represented parametrically as follows:

$$x=a \sqrt{u} \cos{v}$$ $$y=b \sqrt{u} \sin{v}$$ $$z=u$$

The surface area of this object is given by

$$\int_0^h du \: \int_0^{2 \pi} dv \: \sqrt{E \,G - F^2}$$

where the 1st fundamental form is given by

$$E=1+\frac{a^2 \cos^2{v} + b^2 \sin^2{v}}{4 u}$$ $$F=\frac{1}{4} (b^2-a^2) \sin{2 v}$$ $$G = (a^2 \sin^2{v}+b^2 \cos^2{v}) u$$

The integral simplifies to

$$\int_0^h du \: \left [\sqrt{b^2 \left(a^2+4 u\right)} E\left(\frac{4 \left(b^2-a^2\right) u}{b^2 \left(a^2+4 u\right)}\right)+\sqrt{a^2 \left(b^2+4 u\right)} E\left(\frac{4 (a-b) (a+b) u}{a^2 \left(b^2+4 u\right)}\right)\right ]$$

where $E$ is the elliptic integral defined as

$$E(m) = \int_0^{\pi/2} dx \sqrt{1-m \sin^2{x}}$$

This is about the best you'll do as far as I can see.

Ron Gordon
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  • The best but still very complex :) – Ivan Bunin Mar 24 '13 at 15:51
  • @IvanBunin: surface area is a very complex topic in general. I didn't do SA calculations for surfaces other than figures of revolution (and simple objects like cubes and prisms) until I took differential geometry. – Ron Gordon Mar 24 '13 at 15:54
  • @IvanBunin: you should at this point accept one of the solutions if you found it useful. I know you think mine is complex, but I challenge you to find something simpler. (Arc length and surface area of the ellipse are the reason the elliptic integrals have their names.) – Ron Gordon Apr 02 '13 at 13:19
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    When one takes the parameter $m$ as argument, you usually don't square it before multiplying by $\sin^2$; similarly, one uses the square if one is taking the modulus $k$ as the argument for the complete elliptic integral. See this for instance. – J. M. ain't a mathematician Apr 03 '13 at 06:20
  • @J.M.: Thank you for usefull answer and reference! – Ivan Bunin Apr 03 '13 at 09:45
  • @J.M.: thanks for pointing that out. I do recall Mathematica using a different definition than other standard references. – Ron Gordon Apr 03 '13 at 09:48
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    At the very least, @Ron, Mathematica and A&S both use the parameter convention (which also happens to be my preference), while Maple and the DLMF use the modulus convention. So yes, care ought to be taken... :) – J. M. ain't a mathematician Apr 03 '13 at 09:51
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The function in question is $z=\frac{x^2}{a^2}+\frac{y^2}{b^2}$.

It is more convenient to work in polar coordinates, so that $x=a r \cos \theta,y=b r\sin \theta,z=r^2$, and the parameters vary like so: $0\leq \theta \leq 2\pi,0 \leq r \leq \sqrt{h}$.

You can find the surfrace area according the the formula $A=\int_0^{2 \pi} d \theta \int_0^\sqrt{h} dr \| {\bf x}_r \times {\bf x}_\theta \|$, where ${\bf x}=(x,y,z)$. I couldn't evaluate this integral analytically.

user1337
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