Green's Theorem and polar coordinates

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SUMMARY

The discussion focuses on applying Green's Theorem to evaluate the line integral of the vector field defined by the expression \(-y^2 dx + x^2 dy\) over the curve \(C\) parameterized by \(x = \cos t\) and \(y = \sin t\) for \(t\) from \(0\) to \(2\pi\). The user attempts to convert the integral into polar coordinates, resulting in a double integral that evaluates to zero. The conclusion drawn is that the zero result is indeed valid, as it reflects the nature of the vector field and the closed curve used in the evaluation.

PREREQUISITES
  • Understanding of Green's Theorem and its application in vector calculus.
  • Familiarity with polar coordinates and their conversion from Cartesian coordinates.
  • Knowledge of double integrals and their evaluation techniques.
  • Basic proficiency in calculus, particularly in evaluating line integrals.
NEXT STEPS
  • Study the derivation and applications of Green's Theorem in various contexts.
  • Learn the process of converting integrals from Cartesian to polar coordinates.
  • Explore examples of evaluating double integrals over circular regions.
  • Investigate the implications of zero results in line integrals and their geometric interpretations.
USEFUL FOR

Students and educators in mathematics, particularly those studying vector calculus and integral theorems, as well as anyone seeking to deepen their understanding of Green's Theorem and polar coordinate transformations.

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Homework Statement


Using Green's Theorem, (Integral over C) -y^2 dx + x^2 dy=____________
with C: x=cos t y=sin t (t from 0-->2pi)

Homework Equations


(Integral over C) Pdx + Qdy=(Double integral over D) ((partial of Q w.r.t. x)-(partial of P w.r.t. y))dxdy

The Attempt at a Solution



I'm stuck from here. I remember the professor said to use polar coordinates which makes sense to get the limits on D, but how do I convert the integral (-y^2 dx + x^2 dy) to polar?

In my method I go from the original integral over C to (double integral over D) 2x+2y dxdy. I convert this to polar to get limits on D and I get (integral from 0 to 2pi)(integral from 0 to 1) 2rcos(theta)+2rsin(theta)*r*dr*d(theta). Once you calculate all of this you get 0 which I don't believe is correct. Any help?
 
Last edited:
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Why don't you think 0 is correct? It's quite a nice number, one of my favorites, actually.:smile:
 
People sometimes want answers in round numbers. You can't get any rounder than zero.:smile:
 

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