# Question about orientation of a line integral of 2 points of a circle

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

• 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)
– Alan
Sep 14 at 22:53
• @Alan I supposed "from A to B" (thus counterclockwise) is the orientation Sep 14 at 22:55
• $B = (-1, \frac{\sqrt{3}}{2})$ is not on the circle. Did you mean $B = (-1, \sqrt{3})$? Sep 15 at 0:47
• @MathLover yes! Sorry for the typo, gonna check it now. Sep 15 at 0:55

$$\displaystyle 2\int_0^\frac{2\pi}{3}(\cos^2{t}-\sin^2{t}) ~ dt = -\frac{\sqrt{3}}{2} ~$$ 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.