Multivariable Calc Problem (Surface Area/Integral)

  • Thread starter Thread starter nietzsche
  • Start date Start date
  • Tags Tags
    Multivariable
Click For Summary
SUMMARY

The discussion focuses on finding the surface area of the ellipsoid defined by the equation (x/a)² + (y/a)² + (z/b)² = 1. The surface area can be calculated using a parametrization G(u,v) and the double integral of the norm of the cross product of ∂uG and ∂vG. Participants suggest using polar coordinates and transforming the problem into a sphere, while emphasizing the importance of understanding elliptic integrals. The incomplete elliptic integral of the first kind is highlighted as a common solution approach, alongside series solutions derived from the integral of sin²n(θ).

PREREQUISITES
  • Understanding of parametrization in multivariable calculus
  • Familiarity with elliptic integrals and their properties
  • Knowledge of double integrals and cross products
  • Experience with polar coordinates and transformations
NEXT STEPS
  • Study the properties of elliptic integrals, particularly the incomplete elliptic integral of the first kind
  • Learn how to derive series solutions for integrals involving sin²n(θ)
  • Explore numerical methods for calculating elliptic integrals, including the arithmetic-geometric mean technique
  • Investigate applications of elliptic integrals in engineering, particularly in solid mechanics
USEFUL FOR

Students and professionals in mathematics, engineering, and physics who are working on problems involving surface area calculations, elliptic integrals, and multivariable calculus techniques.

nietzsche
Messages
185
Reaction score
0

Homework Statement



Find the surface area of the ellipsoid (x/a)^2 + (y/a)^2 + (z/b)^2 = 1.

Homework Equations



If G(u,v) is a map from R2 to R3 that parametrizes the surface, then the area of the surface is equal to the double integral over the domain of G of the norm of the cross product of ∂uG and ∂vG.

The Attempt at a Solution



Well, I can see that we can parametrize the surface with a G whose domain is the circle x^2 + y^2 = a^2. I've tried using polar coordinates, I've tried first transforming it into a sphere and then into polar coordinates, but I just can't seem to get an integral that I can work with.

If someone could perhaps give me a push in the right direction (maybe a parametrization to try out), I would really appreciate it.
 
Physics news on Phys.org
You need to understand elliptic integrals. Wolfram site and wikipedia entry on the elliptic have sufficient information for a solution and you can get some general elliptic manipulation strategies + history from http://everything2.com/title/elliptic+integral+standard+forms".

To derive the actual series solution of the elliptic you will have to derive the series solution to the integral of sin^(2n) theta between 0 to pi/2, transform the elliptic integral to a series representation in terms of sin^(2n) theta, then use the previously derived expression, finally sum. This might take some time the first time. However, most people just use the incomplete elliptic integral of the first kind solution directly. That is also very reasonable.

Edit: If I remember correctly, Einstein played with this problem (with his first wife), while in school. I don't think he ended up making any particularly noteworthy contributions. Most of the current formulation is still from Legendre, Weistrauss, Gauss, Jacobi. Everyone still relies on either the series form or the numerical arithmetic geometric mean technique (its a cool algorithm for calculating values to arbitrary precision) for quoting solutions to the elliptic. The series form is used extensively in the derivation of analytical solutions in engineering (I have seen a lot appear in solids).
 
Last edited by a moderator:
Thanks, I'll have a look!
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
11
Views
3K
  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
3
Views
2K