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Consider $\textbf{F}(x,y)=\frac{-y}{x^2+y^2}\textbf{i}+\frac{x}{x^2+y^2}\textbf{j}$.

Let $C_1$ be the upper half of the unit circle oriented counterclockwise, and let $C_2$ be the lower half of the unit circle oriented clockwise. Compute $\int_{C_1}\textbf{F} \dot \ d\textbf{r}$ and $\int_{C_2}\textbf{F} \dot \ d\textbf{r}$.

$C_1:x^2+y^2=1, x=\cos(\theta), y=\sin(\theta), 0 \leq \theta \leq \pi$

$dx = -\sin(\theta), dy = \cos(\theta)$

$\int_{C_1}\textbf{F} \dot \ d\textbf{r} = \int_0^{\pi} \langle \frac{-\sin(\theta)}{\sin^2(\theta)+\cos^2(\theta)}, \frac{\cos(\theta)}{\sin^2(\theta)+\cos^2(\theta)} \rangle \langle -\sin(\theta), \cos(\theta) \rangle \ d\theta$

$=\int_0^{\pi} \sin^2(\theta)+\cos^2(\theta) \ d\theta$

$= \left. \theta \right|_{0}^{\pi}$

$= \pi.$

$C_2: x=\cos(\theta), y=-\sin(\theta), 0 \leq \theta \leq \pi$

$dx=-\sin(\theta), dy=-\cos(\theta)$

$\int_{C_2}\textbf{F} \dot \ d\textbf{r} = \int_0^{\pi} \langle \sin(\theta), \cos(\theta) \rangle \langle -\sin(\theta), -\cos(\theta) \rangle \ d\theta$

$=\int_0^{\pi} -\sin^2(\theta)-\cos^2(\theta) \ d\theta$

$=\left. -\theta \right|_0^{\pi}$

$=-\pi$

Is $\int_C\textbf{F} \dot \ d\textbf{r}$ always independent of path? Why or why not?

No, $\int_C\textbf{F} \dot \ d\textbf{r}$ is not always independent of path. Because the vector field is undefined at (0,0), it is not conservative. Therefore, it depends on the path taken.

Could someone please tell me if I computed the line integrals correctly? Also, I'm not sure about my last answer about path independence. Is this correct?

Thank you.

user7000
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  • The cosines in the denominators are squared, it seems you left that off. –  Jul 18 '14 at 22:50
  • But it looks correct to me. –  Jul 18 '14 at 22:55
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    Just because the vector field is undefined at (0,0) doesn't necessarily mean it can't be conservative in some regions that don't include that point (in fact, this field is conservative in simply connected regions that don't include (0,0)). But the fact that you have two paths going from (1,0) to (-1,0) over which the field integrates to different values is precisely the definition of it being path dependent. – Ned Jul 18 '14 at 23:08
  • @Ned If the question doesn't explicitly mention the region, shouldn't he just take it to be all of $\Bbb R^2$ (or at the very least, in the unit disc centered at the origin)? That would be my interpretation of the question. –  Jul 18 '14 at 23:15
  • Can't be all of RxR since the vector field isn't defined at (0,0). But yes, it probably should be interpreted as "conservative in RxR - {(0,0,)}", the refutation of which is the calculation of the two line integrals. – Ned Jul 18 '14 at 23:19
  • OK, not sure what you mean by region being continuous, but the question was whether F is "always path independent", and the answer to that is contained in the calculation of the line integrals. In this case, the region is not simply connected so you "can't tell" by general principles, but the two given line integrals do tell you the answer, which is no. – Ned Jul 18 '14 at 23:36
  • Thanks for catching the typo and thanks for discussing my question! ps-I'm a girl – user7000 Jul 18 '14 at 23:43
  • The necessary and sufficient condition for path independence is that the function $\mathbf F$ be the gradient of some scalar field (that is $\mathbf F = \nabla \phi$ for some scalar field $\phi$). However it can be difficult to find that scalar field in practice so there are several tests that have been devised which are sufficient. One of them is to just check a few different paths and if any of them are not the same, then it's not path independent. This method has the disadvantage of not being able to verify a field IS path independent -- only that it's not (if you chose wisely). –  Jul 19 '14 at 00:13
  • Another test is IF your vector field is defined over an open, simply connected region and is continuously differentiable (technically Frechet differentiability is sufficient, but that's not important), then you can verify that $\nabla \times \mathbf F = 0$ (this is what Mary Star does below). This does not necessarily hold if the vector field $\mathbf F$ is NOT defined over a simply connected region (for instance $\Bbb R^2 - {0}$ is not simply connected). That's why Mary is telling you it is path independent even though your result says it isn't -- she's forgotten this condition. –  Jul 19 '14 at 00:18
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    I guess Mary deleted her answer, but anyway, your professor is surely wanting you to say that because the integrals over the different 2 paths have different results, $\mathbf F$ must not be a conservative vector field (i.e. it's not path independent). –  Jul 19 '14 at 00:37

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