Calculating using Direct and Green's Theorem

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SUMMARY

This discussion focuses on calculating a line integral using both direct evaluation and Green's Theorem for the function defined by the integral \(\int (x-y) dx + (x+y) dy\) over the curve \(C: x^2+y^2=4\). The direct evaluation yields a result of \(8\pi\), while the application of Green's Theorem initially led to confusion regarding the limits of integration, ultimately revealing an error in the limits used, which was corrected upon further review. The correct evaluation using Green's Theorem should align with the direct calculation, confirming the result of \(8\pi\).

PREREQUISITES
  • Understanding of line integrals and their applications
  • Familiarity with Green's Theorem and its conditions
  • Knowledge of polar coordinates and transformations
  • Basic calculus skills, particularly in evaluating double integrals
NEXT STEPS
  • Review the derivation of Green's Theorem and its applications in vector calculus
  • Practice converting Cartesian coordinates to polar coordinates in integrals
  • Explore common pitfalls in evaluating line integrals and double integrals
  • Study examples of vector fields and their circulation using Green's Theorem
USEFUL FOR

Students and educators in calculus, particularly those studying vector calculus, as well as anyone seeking to deepen their understanding of line integrals and Green's Theorem applications.

bugatti79
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Homework Statement


Calculate the folowing directly and with greens theorem

Homework Equations



[itex]\int (x-y) dx + (x+y) dy[/itex]

C= x^2+y^2=4

The Attempt at a Solution



Directly

[itex]x= r cos \theta, y=r sin \theta, r^2=4, dx = -r sin \theta d \theta, dy= r cos \theta d \theta[/itex]

Substituting I get

[itex]\displaystyle \int_0^{2 \pi} (-r^2 sin \theta cos \theta +r^2 sin^2 \theta) d \theta+(r^2 cos^2 \theta +r^2 sin \theta cos \theta) d \theta[/itex]

[itex]=4 \int_0^{2 \pi} d \theta= 8 \pi[/itex]

Greens theorem

[itex]\displaystyle \int \int_R (G_x -F_y)dA= \int_0^{2 \pi}\int_0^2 2 r dr d \theta = 2 \pi[/itex]...? I can't spot the error!
 
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bugatti79 said:
[itex]\displaystyle \int \int_R (G_x -F_y)dA= \int_0^{2 \pi}\int_0^2 2 r dr d \theta = 2 \pi[/itex]...? I can't spot the error!

I can't spot the error either because you didn't show your [incorrect] work to get ##2\pi##.
 
LCKurtz said:
I can't spot the error either because you didn't show your [incorrect] work to get ##2\pi##.

I have spotted it this morning. Just used wrong limits in calculation although shown correctly above. Late night concentration I guess.

Thanks LCKurtz
 

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