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Calculate $\int_{C} \frac{x}{x^2+y^2} dx + \frac{y}{x^2+y^2} dy~$ where $C$ is straight line segment connecting $(1,1)$ to $(2,2)$

my question is , after calculating the integral using green theorem i got that $\int_{C} \frac{x}{x^2+y^2} dx \frac{y}{x^2+y^2} dy= -\ln(2)$

is it the right answer ? since we are connecting $(1,1)$ to $(2,2) $ AND NOT $(2,2)$ to $(1,1)$

so its question about the sign of the value.

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    $\begingroup$ Green's theorem is for closed curves. Yours isn't. This is a conservative force field, so find the potential function. $\endgroup$ – B. Goddard Jan 9 at 14:36
  • $\begingroup$ I closed it with parametrization $\endgroup$ – Mather Jan 9 at 16:02
  • $\begingroup$ You should add that to your answer so we can see what you did wrong. $\endgroup$ – B. Goddard Jan 9 at 16:28
  • $\begingroup$ Because this is a conservative field, so the integral over a closed curve should be zero. Further, if you use Greens theorem, you get a double integral over a region of area zero, and so, for another reason, you should get zero. $\endgroup$ – B. Goddard Jan 9 at 18:30
  • $\begingroup$ but someone posted that the answer is $\ln(2)$ @B.Goddard $\endgroup$ – Mather Jan 9 at 19:46
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$(1,1),(2,2)$ are joined by the line-segment $C:y=x\in[1,2]$. The integral becomes $$\int_C\frac{xdx+ydy}{x^2+y^2}=\int_C\frac{2xdx}{2x^2}=\int_1^2\frac{dx}x=\ln(2)$$

Alternatively, $$\int_C\frac{xdx+ydy}{x^2+y^2}=\int_C\frac12\cdot\frac{d(x^2+y^2)}{x^2+y^2}=\frac12\int_2^8\frac{dm}m=\frac12\ln(m)\Big|_2^8=\ln(2)$$where $m=x^2+y^2$, that goes from $1^2+1^2\to2^2+2^2$.

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The fundamental theorem of calculus tells you that if ${\bf F} = \nabla f$ in a simply connected region containing the curve, then $$\int_C {\bf F}\cdot d{\bf r}= f(b) - f(a)$$ where the curve $C$ begins at the point $a$ and ends at the point $b$.

Here $${\bf F}(x,y) = \left(\frac{x}{x^2 + y^2}, \frac y{x^2 + y^2} \right). $$ Can you find a function $f(x,y)$ such that $${\bf F}(x,y) = \nabla f(x,y)?$$

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