Surface area of Elliptic paraboloid

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Homework Help Overview

The discussion revolves around finding the surface area of an elliptic paraboloid defined by its height H and two semiaxes A and B, with specific parametric equations provided for x, y, and z. Participants explore various methods and formulas to approach the problem.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Some participants suggest using specific formulas from external sources, while others propose the use of double integrals to compute the surface area. There is a discussion about the complexity of the resulting integrals and the challenges in solving them.

Discussion Status

The conversation includes a range of ideas, from referencing formulas to discussing the difficulties of evaluating integrals. One participant expresses frustration over the lack of an algebraic formula for the surface area, while others are exploring numerical methods as a potential solution.

Contextual Notes

Participants note that the problem may not have a straightforward algebraic solution and that the evaluation of integrals could lead to elliptic integrals, which are complex to solve. There is also mention of specific limits and conditions under which the surface area can be computed.

tosik
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Homework Statement



Given the elliptic paraboloid of height H and two semiaxes A and B. How to find its surface area?

Homework Equations



x = A * sqrt(u) * cos(v)
y = B * sqrt(u) * sin(v)
z = u

u belongs to [0; H], v belongs to [0; 2*PI)
 
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Formula No. 4 is even better...cant get any easier!
 
I think he means do a double integral chaps.
 
hunt_mat said:
I think he means do a double integral chaps.

Integrating the cross product of the two tangent vectors du*dv IS a double integral. What did you have in mind?
 
I thought about:
<br /> A=\int\int dxdy<br />
But I guess you idea is better.
 
Dear tosik:

You have asked a very hard question. I have immersed myself in this problem for close to a month. Here's my conclusion:

There is no formula for the surface area of an elliptic paraboloid in algebraic form. (At least up to my research.) Using calculus, the resulting integral equation evaluates to elliptic integral which is very very very hard to solve. Mathematica seems to have an indigestion when I run it.

After changing variables and partial integration, the surface area of the elliptic paraboloid is given by the following integral:

(2 a b Sqrt[
2 c^2 b^2 + a^2 (2 c^2 + b^2) +
2 c^2 (-a^2 + b^2) Cos[2 \[Theta]]])/(
3 (a^2 + b^2 + (-a^2 + b^2) Cos[2 \[Theta]])) - (a^4 b^4)/(
3 (a^2 + b^2)^2 c^2 -
3 (a - b)^2 (a + b)^2 c^2 Cos[2 \[Theta]]^2) + (
a^3 b^3 Sqrt[
a^2 b^2 + 2 (a^2 + b^2) c^2 + 2 (-a^2 + b^2) c^2 Cos[2 \[Theta]]])/(
3 (a^2 + b^2)^2 c^2 - 3 (a - b)^2 (a + b)^2 c^2 Cos[2 \[Theta]]^2)

You can paste that in Mathematica and integrate with respect to Theta with limits of zero to 2 Pi.

In the above expression, a and b are the semi axes of the elliptic base of the paraboloid and c is its height.

Now if the the base is circular (a=b) the paraboloid is just a circular paraboloid (paraboloid of revolution) and the parabolic surface is given by:

\[Pi] (-a^4 + a (a^2 + 4 c^2)^(3/2))/(6 c^2)

The above expression is the result of the integral above if a is made equal to b. Notice that in the limit that c --> 0, the paraboloid has no height, the surface area will become the area for the circle. And the limit of the above expression evaluates to just that.

I have started out by saying that there is no algebraic form for the area of an elliptic paraboloid. This doesn't mean that we can't compute for the area of the parabolic surface of an elliptic paraboloid. We can use numerical method to compute for the area to arbitrary accuracy by supplying the values of a, b and c in the above integral.

Respectfully,

Jeruel R. Canales 3/8/2012
 

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