Integral of function over ellipse

Click For Summary
SUMMARY

The discussion focuses on calculating the integral of the function \(\iint_S \sqrt{1-\left(\frac{x}{a}\right)^2 -\left(\frac{y}{b}\right)^2} dS\) over the surface of an ellipse defined by \(\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2 = 1\). Nick suggests using Green's Theorem and a change of coordinates involving the Jacobian determinant to simplify the problem. The transformation leads to the integral being expressed as \(a\cdot b\cdot\iint_C\sqrt{1-m^2-n^2}\mathrm{d}S'\), where \(C\) represents the unit circle in the new coordinates.

PREREQUISITES
  • Understanding of double integrals and surface integrals
  • Familiarity with Green's Theorem
  • Knowledge of coordinate transformations and Jacobian determinants
  • Basic concepts of elliptical geometry
NEXT STEPS
  • Study Green's Theorem applications in multivariable calculus
  • Learn about Jacobian determinants in coordinate transformations
  • Explore the properties of elliptical integrals
  • Investigate the use of polar coordinates in double integrals
USEFUL FOR

Mathematicians, physics students, and anyone involved in advanced calculus or integral geometry who seeks to understand the integration of functions over elliptical domains.

nickthequick
Messages
39
Reaction score
0
Hi,

I'm trying to find
\iint_S \sqrt{1-\left(\frac{x}{a}\right)^2 -\left(\frac{y}{b}\right)^2} dS

where S is the surface of an ellipse with boundary given by \left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2 = 1.

Any suggestions are appreciated!

Thanks,

Nick
 
Physics news on Phys.org
Do you mean the interior of an ellipse?

Anyway, the first thing I though of is Green's Theorem for some reason. Probably since then we can make the substitution ##\left(\dfrac xa\right)^2+\left(\dfrac yb\right)^2=1##.

The second thing I thought of was a change of coordinates and a multiplication by the Jacobian determinant, then we have it reduced to

$$a\cdot b\cdot\iint_C\sqrt{1-m^2-n^2}\mathrm{d}S'$$

where C is the unit circle wrt m and n and S' should have a fairly obvious definition.
 
Last edited:
Try the substitution$$
\frac x a = r\cos\theta,\,\frac y b = r\sin\theta$$
 
Got it!

Thanks
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K