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I want to calculate: $$\oint_C\frac{1}{x^2 + y^2}dx + 2y\ dy$$ where $C = \{(x,y)\mid x^2 + y^2 = 1\}$. And I want to do it without greens formula.

To calculate the line integral we must then Parameterize the circle $x^2 + y^2 = 1$ into the parameterized path function $f(t) = (\cos(t), \sin(t))$ for $0 \leq t\leq 2 \pi$. So the integral $\oint_C \frac{1}{x^2 + y^2} \, dx + 2y \, dy$ then becomes $\int_0^{2\pi} \frac{1}{\cos(\theta)^2 + \sin(\theta)^2}(-\sin(\theta)) + 2(\sin(\theta)) \cos(\theta)d\theta$ which becomes $\int_0^{2\pi} (-\sin(\theta)) + 2(\sin(\theta)) \cos(\theta)d\theta = \int_0^{2\pi} (\sin(\theta))(2\cos(\theta) -1)\,d\theta =0$?

Can anybody tell me if this is right/close to right? The zero makes me doubt it.

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up vote 8 down vote accepted

Since you're on the curve $x^2+y^2=1$, you can immediately replace $\displaystyle\oint_C \frac{dx}{x^2+y^2}$ with $\displaystyle\oint_C 1\,dx$. As you go around the circle, the sum of all the infinitely many infinitely small increments of $x$ must be $0$ because you return to your starting point. Therefore that integral is $0$. Similarly, $y$ returns to its starting point, so you're integrating $2y\,dy$ from $-1$ to $1$ and from $1$ to $-1$ and adding them together, so there you get $0$ as well.

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Many thanks good sir. – AvatarOfChronos Dec 11 '12 at 13:47

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