Calculus: Integral along a curve.

  • Thread starter Herculi
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  • #1
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Homework Statement:
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Relevant Equations:
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Let $F = (P(x,y),Q(x,y))$ a field of vector class 1 in the ring $A={(x,y): 4<x²+y²<9}$ and $x,y$ reals.

I am having trouble to understand why this alternative is wrong:

If $ \partial P /\partial y = \partial Q /\partial x$ for every x,y inside A, so $\int_{C} Pdx + Qdy = 0$ for every circumference $\epsilon $ A.

I mean, the condition implies that $Curl F = 0$, and we have that the field of vector is C1, so we don't need to worry with anomalies or problems that could appear as $(...)/0$. In fact, $\int_{C} Pdx + Qdy = \int_{S} \nabla \times F \space ds = \int_{S} 0 \space ds = 0$

Where is my error? If i am wrong, could you give me an example of a vector field that does not satisfies the implication? Where is the error in my demonstration?
 

Answers and Replies

  • #2
ergospherical
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The annulus ##A## is not simply connected, therefore ##\mathbf{F}(x,y) = (P(x,y), Q(x,y))^T## is not a conservative vector field even if ##\dfrac{\partial P}{\partial y} = \dfrac{\partial Q}{\partial x}##. For example, consider the line integral of a vector field ##\mathbf{F}(x,y) = \dfrac{1}{x^2+y^2} (-y,x)^T## around a closed curve ##C: \mathbf{r}(t) = a(\cos{t}, \sin{t})^T## for ##t \in [0, 2\pi]## and with ##a \in (2,3)##.
 

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