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Calculate $$\int_C \frac{x^2-y^2}{x^2+y^2}ds$$ where C is the circle $x^2 + y^2 = 4$ from $A = ({2,0})$ to $B = (-1, \sqrt{3})$

I calculated $\,\,\,\,r = (2\cos{t},2\sin{t})\,\,\,\,$ and $\,\,\,\,||r'(t)|| = 2\,\,\,\,$ so we have: $$2\int_0^\frac{2\pi}{3}(\cos^2{t}-\sin^2{t})dt = -\frac{\sqrt{3}}{2}$$

Since the author gave the points and explicitly said "from A to B" I suppose I got right, however the result is negative, that seems wrong

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    $\begingroup$ If it doesn't specify direction, one would default to the counterclockwise direction from A to B (Since otherwise there are two equally valid arcs between the two points on the circle) $\endgroup$
    – Alan
    Sep 14 at 22:53
  • $\begingroup$ @Alan I supposed "from A to B" (thus counterclockwise) is the orientation $\endgroup$ Sep 14 at 22:55
  • $\begingroup$ $B = (-1, \frac{\sqrt{3}}{2})$ is not on the circle. Did you mean $B = (-1, \sqrt{3})$? $\endgroup$
    – Math Lover
    Sep 15 at 0:47
  • $\begingroup$ @MathLover yes! Sorry for the typo, gonna check it now. $\endgroup$ Sep 15 at 0:55
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$ \displaystyle 2\int_0^\frac{2\pi}{3}(\cos^2{t}-\sin^2{t}) ~ dt = -\frac{\sqrt{3}}{2} ~ $ and that is the correct answer.

$\cos^2t - \sin^2t = \cos2t, 0 \leq t \leq \frac{2\pi}{3}$ and if we rewrite $u = 2t$,

we have $~ \cos u, 0 \leq u \leq \frac{4 \pi}{3}$. We know $\cos u$ is positive in the first quadrant and negative in the second and given $|\cos u| = |\cos(\pi-u)|$, the integral in the first and second quadrant will cancel out. So the integral only depends on third quadrant $\pi \leq u \leq \frac{4\pi}{3}$ and $\cos$ function is negative in the third quadrant. So clearly, the answer should be negative.

Finally, note that in line integral of a scalar function, orientation of the curve is not relevant. You have to integrate over the curve, just like you integrate to find area of a region. The only difference being that in this case, you have an integrand that may be negative over the arc.

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    $\begingroup$ Thanks for confirming my suspicions, I was thinking there was a mistake during substitution or parametrization. Seeing it was just the "nature" of this integral makes it easier to understand. $\endgroup$ Sep 15 at 2:14

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