Calculus: Integral along a curve.

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SUMMARY

The discussion centers on the evaluation of line integrals for vector fields in the context of a non-simply connected region, specifically the annulus defined by \( A = \{(x,y): 4 < x^2 + y^2 < 9\} \). The user incorrectly assumes that if \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \), then the line integral \( \int_{C} Pdx + Qdy \) equals zero for any closed curve \( C \) within \( A \). The error lies in the fact that the vector field \( F(x,y) = (P(x,y), Q(x,y)) \) is not conservative due to the annulus not being simply connected, which invalidates the application of Green's Theorem in this case. An example provided illustrates a vector field that satisfies the condition but is not conservative.

PREREQUISITES
  • Understanding of vector calculus, specifically line integrals and curl.
  • Familiarity with Green's Theorem and its conditions.
  • Knowledge of conservative vector fields and simply connected regions.
  • Basic proficiency in multivariable calculus, including partial derivatives.
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  • Study Green's Theorem and its application to simply connected regions.
  • Learn about conservative vector fields and the implications of non-simply connected domains.
  • Explore examples of vector fields that are not conservative despite satisfying the curl condition.
  • Investigate the properties of line integrals in different topological spaces.
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Students and professionals in mathematics, particularly those focusing on vector calculus, as well as educators seeking to clarify concepts related to line integrals and vector fields in non-simply connected regions.

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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?
 
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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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