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Surface area of Elliptic paraboloid

by tosik
Tags: elliptic, paraboloid, surface
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tosik
#1
Feb17-09, 06:02 AM
P: 5
1. The problem statement, all variables and given/known data

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

2. Relevant 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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Dick
#2
Feb17-09, 07:33 AM
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Try using formula 2 in http://mathworld.wolfram.com/SurfaceArea.html
LuisVela
#3
Jul19-10, 10:53 PM
P: 33
Formula No. 4 is even better......cant get any easier!

hunt_mat
#4
Jul21-10, 03:54 AM
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Surface area of Elliptic paraboloid

I think he means do a double integral chaps.
Dick
#5
Jul21-10, 07:49 AM
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Quote Quote by hunt_mat View Post
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?
hunt_mat
#6
Jul21-10, 07:56 AM
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I thought about:
[tex]
A=\int\int dxdy
[/tex]
But I guess you idea is better.
Jeruel
#7
Mar7-12, 10:40 PM
P: 1
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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