Double integral in polar coordinate

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SUMMARY

The discussion focuses on converting a double integral over an elliptical region defined by \(\frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1\) into polar coordinates. The correct substitutions are \(x = a r \cos \theta\) and \(y = b r \sin \theta\), leading to the integral transformation \(\int_0^{2\pi} \int_0^{r_1(\theta)} (r^2) r \, dr \, d\theta\), where \(r_1(\theta) = \frac{ab}{\sqrt{(b \cos \theta)^2 + (a \sin \theta)^2}}\). The discussion emphasizes the geometric interpretation of these substitutions and the necessity of understanding the mapping from Cartesian to polar coordinates for ellipses.

PREREQUISITES
  • Understanding of double integrals and their applications
  • Familiarity with polar coordinates and transformations
  • Knowledge of ellipses and their equations
  • Basic calculus concepts, including Riemann sums
NEXT STEPS
  • Study the derivation of polar coordinates for non-circular shapes
  • Learn about the Jacobian determinant in coordinate transformations
  • Explore advanced techniques in double integrals, particularly in elliptical regions
  • Investigate applications of double integrals in physics and engineering contexts
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Students and educators in calculus, particularly those focusing on multivariable calculus, as well as mathematicians interested in coordinate transformations and integral calculus applications.

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Homework Statement


With a > 0, b > 0, and D the area defined by

[tex]D: \frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1[/tex]

Change the integral expression below:

[tex]\iint\limits_D (x^2+y^2) dx\,dy[/tex]

by using x = a r cos θ, y = b r sin θ. After that evaluate the integral.

The Attempt at a Solution



First of all, I can already visualize that the area is an ellipse, and because of that if we want to express the outside boundary in polar coordinates, it is

[tex]r_1(\theta) = \frac{ab}{\sqrt{(b \, cos \theta)^2+(a \, \sin \theta)^2}}[/tex]

Therefore, since x^2 + y^2 = r^2, the integral can be expressed in terms of θ and r as:

[tex]\int_0^{2\pi} \int_0^{r_1(\theta)} \! (r^2)r \, dr\,d\theta[/tex]

I've also understood how this construct comes from the Riemann sums.

However, I don't understand the substitution hinted from the problem. For every random point (r, θ), clearly x = r cos θ and not x = a r cos θ! Can someone give a geometric interpretation/explanation, and from that show how one changes the integral expression?

Thanks a lot.
 
Physics news on Phys.org
x = r cos θ and y = r sin θ maps out a circle, not an ellipse.

The equation of an ellipse is:

x = a cos θ
y = b sin θ

Substitute these into your integral.
 

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