What Went Wrong with Finding the Area of an Ellipse?

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SUMMARY

The area of an ellipse defined by the equation x²/a² + y²/b² = 1 is conclusively shown to be πab. The error in the initial attempt arose from incorrect integration limits and transformations. The correct integration involves using the Jacobian determinant J(u,v) = ab and integrating over the area A, leading to the correct area calculation of πab. This highlights the importance of proper setup in double integrals when working with transformations.

PREREQUISITES
  • Understanding of double integrals in calculus
  • Familiarity with Jacobian transformations
  • Knowledge of polar coordinates
  • Basic properties of ellipses and their equations
NEXT STEPS
  • Study the application of Jacobian determinants in coordinate transformations
  • Learn about double integrals in polar coordinates
  • Explore the derivation of area formulas for conic sections
  • Practice solving integrals involving transformations and area calculations
USEFUL FOR

Students studying calculus, particularly those focusing on multivariable calculus, as well as educators teaching geometry and integration techniques.

Knissp
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Homework Statement


Show that the area of [tex]x^2/a^2+y^2/b^2=1[/tex] is [tex]\pi ab[/tex]

Homework Equations


Given transformations:
[tex]x=au[/tex]
[tex]y=bv[/tex]

The Attempt at a Solution



[tex]J(u,v) = a*b[/tex]

[tex]\int\int ((au)/a)^2+((bv)/b)^2 J(u,v) dudv[/tex]

[tex]\int\int u^2+v^2 J(u,v) dudv[/tex]

[tex]\int_0^{2\pi}\int_0^1 r^2 J(u,v) r drd\theta[/tex]

[tex]\int_0^{2\pi}\int_0^1 a b r^3 drd\theta[/tex]

[tex]\int_0^{2\pi} 1/4 a b d\theta[/tex]

=[tex]\frac{\pi a b}{2}[/tex]

But that's obviously wrong. Where did I mess up?
 
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Right from the start lies your mistake!
You are to integrate:
[tex]\int_{A}dA=\int_{A}dxdy=\int_{A}abdudv=\int_{0}^{2\pi}\int_{0}^{1}abrdrd\theta=\pi{ab}[/tex]
 
:eek:

I can't believe I even had to ask this! Thanks so much!
 

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