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In the following example calculate the line integral once directly from the definition of the line integral and once by using Green's Theorem.

Consider the region S bounded between the square with corners at the points (4,4),(-4,4),(-4,-4) and (4,-4) (oriented counterclockwise), and the circle of radius 1 centered at (-1,0) (oriented clockwise) and $F(x,y)=\bigg(\frac{-y}{(x+1)^2+y^2}, \frac{x+1}{(x+1)^2+y^2}\bigg)$ and calculate $\int_{ds}F dx$

(Hint for calculating the line integral: Use the definition $tan^{-1}$a +$tan^{-1}a^{-1}$=$\frac{pi}{2}$.

So when I first use the Green's theorem method, I found out $\frac{dF_2}{dx_1}$=$\frac{dF_1}{dx_2}$, so what should I do? the integral is just =0?

And then if I do the direct line integral ,how can I link to the hint given about $tan^{-1}$a +$tan^{-1}a^{-1}$=$\frac{pi}{2}$???

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You cannot use Green's Theorem because there is a singularity at the point $(-1,0)$ in the vector field. This problem is probably specifically designed to illustrate that sometimes Green's Theorem gives different answers from line integrals when the hypotheses are not met.

The line integral for the circle should not be hard - parameterize the curve, plug and chug, you will probably get $2\pi$ as your answer if I'm seeing what's going on here correctly.

The square one will be a bit tedious. I'll help you with one side (I really don't know why they assign these questions, they're points and time-consuming, but here goes...)

The right edge of the square is $$(4,t),\hspace{3mm}-4\le t\le t$$ The derivative is $(0,1)$, and the field evaluates to $$F(4,t)=\left(\frac{-t}{25+t^2},\frac{5}{25+t^2}\right)$$ We get for the integral $$\int_{-4}^4{\frac{5}{25+t^2}}=\arctan\left({t\over 5}\right)=\arctan(4/5)-\arctan(-4/5)=2\arctan(4/5)$$ I suppose when you add the four sides up, you'll get two pairs like $\arctan(4/5)$ and $\arctan(5/4)$ which, using the hint, will eventually give you the same $2\pi$.

Note: The field is still path-independent, even though Green's Theorem does not apply!

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  • $\begingroup$ Thanks so much!! I did something similar to you for the direct line integral. So there is no way to use the Green's theorem? Can we break up the region into parts and apply it? I remember my prof said something about breaking parts during lecture... $\endgroup$ – user108297 Mar 27 '14 at 18:07
  • $\begingroup$ so the answer is 0 right? add the four sides up and then minus the circle part? $\endgroup$ – user108297 Mar 27 '14 at 18:08
  • $\begingroup$ You can only use Green's Theorem over curves which do not contain that point of singularity. As to the answer, I do not think it would be 0, I'm pretty sure it would be a positive number since everything is oriented positively (CCW on the outside, CW on the inside). $\endgroup$ – user138335 Mar 27 '14 at 18:19
  • $\begingroup$ ok forget about Green's Theorem then, The line integral for the circle is 2pi and the square is 2pi too, should we add them up or subtract to get the answer? $\endgroup$ – user108297 Mar 27 '14 at 18:28
  • $\begingroup$ It depends how you oriented them when computing the line integrals. If they really are the same, you will add them if you have followed the instructions (CW for circle, CCW for square) or otherwise you will need to throw negative signs in if you have switched orientations when doing the integrals. $\endgroup$ – user138335 Mar 27 '14 at 18:39

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