Elliptic Line Integral: Solving for Circulation Around an Ellipse

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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...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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