What is the value of the line integral?

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SUMMARY

The discussion centers on evaluating the line integral using Green's theorem for the integral \(\oint_{\gamma} \frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy\). The integral is computed along two curves: the parabola defined by \(x = -y^2 + 4\) and a square with vertices at (−1, 0), (1, 0), (0, 1), and (0, −1). The conclusion is that the integral evaluates to zero, as it represents a line integral of a conservative vector field, confirmed by the equality of the partial derivatives.

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  • Understanding of Green's theorem
  • Knowledge of line integrals in vector calculus
  • Familiarity with conservative vector fields
  • Ability to compute partial derivatives
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  • Study the applications of Green's theorem in various contexts
  • Learn about conservative vector fields and their properties
  • Explore examples of line integrals in different coordinate systems
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Students of calculus, particularly those studying vector calculus, as well as educators and anyone looking to deepen their understanding of line integrals and Green's theorem.

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Homework Statement


Use Green’s theorem to find the integral
\oint_{\gamma} \frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy
along two different curves γ: first where γ is the simple closed curve which goes along x = −y2 + 4 and x = 2, and second where γ is the square with vertices (−1, 0), (1, 0), (0, 1), (0, −1).


Homework Equations





The Attempt at a Solution


I'm bit confused b/c
d(\frac{-y}{x^2+y^2})/dy = \frac{y^2-x^2}{(x^2+y^2)^2}
d(\frac{x}{x^2+y^2})/dx = \frac{y^2-x^2}{(x^2+y^2)^2}

Then by Green's theorem one gets

\int_{A}\int (d(\frac{x}{x^2+y^2})/dx-d(\frac{-y}{x^2+y^2})/dy) dx dy = \int_{A}\int 0 dx dy = 0

What am I missing?
 
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You are not missing anything. The answer IS 0.

Here you have a line integral of a conservative field (you can tell it's conservative from the equality of the partial derivatives).
A line integral of a conservative field will always be zero for any path that begins and ends at the same point, i.e any closed curve.
 

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