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Find $$\int_{C} \frac{x dy - y dx}{x^{2}+y^{2}}$$ along the oriented broken line $C$ with vertices $(2,-2)$, $(4,4)$, $(-5,5)$ oriented counterclockwise.

I noted that $C$ is a closed curve which passes through the origin, so Green's theorem cannot be applied here. Also, the vector field is not conservative, so the integral is nonzero.

jimjim
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nchar
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    Line-segments are pretty easy to parametrize, and the integrand is rational, so there's no reason you can't just calculate this integral by brute force. – David H Dec 31 '14 at 10:49
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    @DavidH: $2^{32}\equiv ? \ mod \ 3$, No reason for not to calculate by brute force but we don't calculate Why ? – mesel Dec 31 '14 at 12:29

3 Answers3

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Hint: If you take $C$ as a circle centered at origin with radius $r$, the result of integral does not change. (Why ?)

mesel
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    If $C_1$ and $C_2$ are two simple, closed curves that enclose origin, without intersection or touching, then we can apply Green's theorem to the region between the two curves? – Nighty Dec 31 '14 at 10:00
  • @LeeKM: Yes, exactly. Then that means that you can take any closed curve including the singulur point. – mesel Dec 31 '14 at 10:01
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    Then it allows me to evaluate the line integral over $x^2 + y^2 = r^2$, in this case the problematic denominator will become $r^2$ and I can calculate the line integral explicitly! – Nighty Dec 31 '14 at 10:13
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Set $F(x,y)=\left(\frac{x}{x^2+y^2},\frac{-y}{x^2+y^2}\right)$, therefore $$...=\int_C F\cdot d\ell\underset{green}{=}\iint_{\text{int}(C)}\text{rot}(F)dxdy$$

which it work because $F$ is $\mathcal C^1(\text{int}(C))$

idm
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Let $C_1$ be the segment from $(2,-2)$ to $(4,4)$. Let $C_2$ be the segment from $(4,4)$ to $(-5,5)$. Let $C_3$ be the segment from $(-5,5)$ to $(2,-2)$.

Over the simply connected region $\{(x,y) \in \mathbb{R}^2 \ | \ x > 0\}$, the gradient of $f_1(x,y) = \arctan \dfrac{y}{x}$ is the vector field $\left(\dfrac{-y}{x^2+y^2},\dfrac{x}{x^2+y^2}\right)$, which is continuous over this region.

Since the curve $C_1$ lies entirely in this region, we have $\displaystyle\int_{C_1}\dfrac{x\,dy-y\,dx}{x^2+y^2} = f_1(4,4)-f_1(2,-2)$.

Over the simply connected region $\{(x,y) \in \mathbb{R}^2 \ | \ y > 0\}$, the gradient of $f_2(x,y) = \text{arccot} \dfrac{x}{y}$ is the vector field $\left(\dfrac{-y}{x^2+y^2},\dfrac{x}{x^2+y^2}\right)$, which is continuous over this region.

Since the curve $C_2$ lies entirely in this region, we have $\displaystyle\int_{C_2}\dfrac{x\,dy-y\,dx}{x^2+y^2} = f_2(-5,5)-f_2(4,4)$.

The curve $C_3$ is a straight line segment, which is parameterized by $x = t$, $y = -t$, for $t \in [-5,2]$. This is easy enough to evaluate (you should get that the integrand is $0$ on $C_3$).

Finally, add the three pieces together to get the answer.

JimmyK4542
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