Elliptic Line Integral: Solving for Circulation Around an Ellipse

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Homework Help Overview

The discussion revolves around evaluating a line integral around an ellipse, specifically the integral \oint_C xdy - ydx, where C is defined by the ellipse's parameters. The problem involves concepts from vector calculus and the application of Green's theorem.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss parameterizing the ellipse and applying Green's theorem to evaluate the integral. There are questions about the discrepancy between the calculated result and the answer provided in a textbook, with some participants exploring the implications of the area of the ellipse.

Discussion Status

The discussion is ongoing, with participants expressing confusion about the results and seeking clarification on the relationship between the line integral and the area of the ellipse. Some guidance has been offered regarding the interpretation of the integral in relation to the area.

Contextual Notes

There is mention of differing results from various textbooks, raising questions about potential errors or misunderstandings in the interpretation of the integral's value. The context includes the specific parameters of the ellipse and the mathematical properties being discussed.

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Homework Statement


Let C be the ellipse with center (0,0), major axis of length 2a, and minor axis of length 2b. Evaluate \oint_C xdy - ydx.

Homework Equations


I solved this two ways. First I parameterized x and y as x=a \cos \theta and similarly for y. I also applied Green's theorem, which yielded \oint_C xdy - ydx = 2 \int \int_D dA where D is the area enclosed by C (ie an ellipse.) In both cases I got the answer 2\pi a b.

The Attempt at a Solution


My only question is, the book I am using says the answer is \frac{\pi a b}{2}. This is an ETS book and they don't usually have typos, especially when it's the answer key to a previously administered exam. What am I missing?
 
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The answer is pi*a*b/2. If a=b=r then it's a circle and the area is pi*r^2. So the contour is half that.
 
I apologize for being so dense, but I'm still confused. A couple different books I have print the result

\frac{1}{2}\oint_C -ydx + xdy = \iint_{R} dA = A

If the area of the ellipse is A=\pi a b then I would think that the value of the line integral is 2A.
 
Sorry, yes, I think you are right. Don't know what I was thinking...
 

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