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Find $\int_C{F \cdot \hat n ds}$ where $F= (2xy,-y^2)$ and $\hat n$ is the unit outward normal to the curve C in the xy-plane and C is the ellipse $\frac{x^2}{a^2}+ \frac{y^2}{b^2}=1$ traversed in the anticlockwise direction.

Its the $\hat n$ that is stuffing me up. I have successfully parameterized it as:
$c(t) = (a\cos(t),b\sin(t))$ and i differentiated it and I know from there I normally have to take the magnitude of it as this is a path integral not a line integral.

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up vote 3 down vote accepted

If I understood right, you have a $z=f(x,y)$ in the problem such that $z=0$ is the following flat parametrized curve: $$C: r(t)=(a\cos(t),b\sin(t),0), 0\le t\le2\pi$$ For example $z$ could be $$z+1=\frac{x^2}{a^2}+\frac{y^2}{b^2}$$ By the way, according to the assumptions we'd like to have the following integral. Of course, you can evaluate it by using Stokes' Theorem as well.

$$\oint_C \textbf{F}\big(a\cos(t),b\sin(t),0 \big)\cdot (-a\sin(t),b\cos(t),0)~dt\\\\ =\int_0^{2\pi}(2ab\cos(t)\sin(t),-b^2\sin^2(t))\cdot (-a\sin(t),b\cos(t),0)~dt=...$$

Please check if these points are what you have been looking for or not. I think the rest is easy.

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Nice work, Babak, as usual! ;-) – amWhy Jun 15 '13 at 16:38
Thanks Babak, could you explain though what has happened to the $\hat n$? It gives the right answer – Jesse Ross Jun 16 '13 at 1:00
@JesseRoss: I used Stockes's theorem. Instead of evaluating the integral you noted. – Babak S. Jun 16 '13 at 4:28

I just use (dy/ds, -dx/ds) as the unit outward normal vector. And I hope it helps you.

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So what would u do for this one and i dont fully understand this line of reasoning is that just a formula? – Jesse Ross Jun 15 '13 at 12:44
I think it is one way to switch between two kinds of line integrals. – eccstartup Jun 15 '13 at 13:30

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