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I am asked to find the line integral of $\vec{F}(x,y) = \frac{1}{x^2+y^2} \left< y, -x \right>$ on the path given by the circle $x^2+y^2=a^2$ anti-clockwise.

My parametrization looks like this:

$$ x = a \cos(t)\\ y = a \sin(t) $$

so that $\vec{r}(t) = \left< a \cos(t), a \sin(t) \right>$.

Then, $$\vec{r}'(t) = \left< -a \sin(t) , a \cos(t) \right>$$

for $0 \leq t \leq 2\pi$.

Continuing,

$$ \vec{F}(\vec{r}) = \left< \frac{\sin(t)}{a}, \frac{\cos(t)}{a} \right>\\ \vec{F}(\vec{r}) \cdot \vec{r}' (t) = 2cos^2(t) - 1 $$

So

$$\int_{0}^{2\pi} 2cos^2(t) - 1 = 0$$

Which does not match the textbook's answer:

The textbook's answer is $-2\pi$.

Am I doing something wrong?

Thank you.

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1 Answer 1

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Your $F$ is wrong. There should be a minus sign on the second component. And you get the wanted result.

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  • $\begingroup$ Ah I see, thank you! Best regards Rugh. $\endgroup$
    – bru1987
    Commented Aug 28, 2016 at 12:56

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