Question regarding Greens theorem

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SUMMARY

The discussion centers on the application of Green's Theorem to the vector field defined by f = x/sqrt(x^2+y^2) dx + y/sqrt(x^2+y^2) dy around the unit circle. It is established that the integral cannot be evaluated using Green's Theorem because the components P = x/sqrt(x^2+y^2) and Q = y/sqrt(x^2+y^2) lack continuous first-order partial derivatives in the vicinity of the curve due to the singularity at the origin (0,0). This violates the necessary condition that P and Q must have continuous partials on an open region containing the curve and its interior.

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Kuma
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Homework Statement



I have some questions similar to this one. I have to just provide reasoning as to why this can or cannot be evaluated using greens theorem.

given f = x/sqrt(x^2+y^2) dx + y/sqrt(x^2+y^2) dy, and the curve c is the unit circle around the origin. Why can/cannot the integral of f around c be evaluated using greens theorem?

Homework Equations


The Attempt at a Solution



So I said that because taking P as the first term and Q as the second term of f, they both don't have continuous first order partials on c (which is a condition for greens theorem) because there are restrictions on what x and y can be (ie x^2 + y^2 > 0). I'm not sure if that is right though if someone can clarify.
 
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Kuma said:

Homework Statement



I have some questions similar to this one. I have to just provide reasoning as to why this can or cannot be evaluated using greens theorem.

given f = x/sqrt(x^2+y^2) dx + y/sqrt(x^2+y^2) dy, and the curve c is the unit circle around the origin. Why can/cannot the integral of f around c be evaluated using greens theorem?


Homework Equations




The Attempt at a Solution



So I said that because taking P as the first term and Q as the second term of f, they both don't have continuous first order partials on c (which is a condition for greens theorem) because there are restrictions on what x and y can be (ie x^2 + y^2 > 0). I'm not sure if that is right though if someone can clarify.

That's the right idea. P and Q must have continuous partials on an open region containing C and its interior, not just on C. And the problem with the partials is at (0,0).
 

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