Stokes Theorem and Circulation












1












$begingroup$


Find the circulation of $F = frac{-y}{x^2+y^2}i + frac{x}{x^2+y^2}j$ along the unit circle.



I decided to express $F$ in cylindrical coordinates:
$$F = frac{1}{r}{hat{theta}}.$$
However, I found $nabla times F = 0$
So, I assume that would imply,
$$oint Fcdot dr = iint_S (nabla times F)cdot n,dA = iint_Svec{0} cdot n,dA = 0.$$
Later, I attempted to solve this by parametizing F and the path along the circle and found a non-zero answer.



Why does the double integral yield a different answer?






multivariable-calculus vector-fields curl







share|cite|improve this question











$endgroup$












  • $begingroup$
    You got the right answer 0. Now check your work in the second method and if you don't find the mistake, post it in your question so that others verify it.
    $endgroup$
    – Scientifica
    Dec 4 '18 at 23:40
















1












$begingroup$


Find the circulation of $F = frac{-y}{x^2+y^2}i + frac{x}{x^2+y^2}j$ along the unit circle.



I decided to express $F$ in cylindrical coordinates:
$$F = frac{1}{r}{hat{theta}}.$$
However, I found $nabla times F = 0$
So, I assume that would imply,
$$oint Fcdot dr = iint_S (nabla times F)cdot n,dA = iint_Svec{0} cdot n,dA = 0.$$
Later, I attempted to solve this by parametizing F and the path along the circle and found a non-zero answer.



Why does the double integral yield a different answer?






multivariable-calculus vector-fields curl







share|cite|improve this question











$endgroup$












  • $begingroup$
    You got the right answer 0. Now check your work in the second method and if you don't find the mistake, post it in your question so that others verify it.
    $endgroup$
    – Scientifica
    Dec 4 '18 at 23:40














1












1








1





$begingroup$


Find the circulation of $F = frac{-y}{x^2+y^2}i + frac{x}{x^2+y^2}j$ along the unit circle.



I decided to express $F$ in cylindrical coordinates:
$$F = frac{1}{r}{hat{theta}}.$$
However, I found $nabla times F = 0$
So, I assume that would imply,
$$oint Fcdot dr = iint_S (nabla times F)cdot n,dA = iint_Svec{0} cdot n,dA = 0.$$
Later, I attempted to solve this by parametizing F and the path along the circle and found a non-zero answer.



Why does the double integral yield a different answer?






multivariable-calculus vector-fields curl







share|cite|improve this question











$endgroup$




Find the circulation of $F = frac{-y}{x^2+y^2}i + frac{x}{x^2+y^2}j$ along the unit circle.



I decided to express $F$ in cylindrical coordinates:
$$F = frac{1}{r}{hat{theta}}.$$
However, I found $nabla times F = 0$
So, I assume that would imply,
$$oint Fcdot dr = iint_S (nabla times F)cdot n,dA = iint_Svec{0} cdot n,dA = 0.$$
Later, I attempted to solve this by parametizing F and the path along the circle and found a non-zero answer.



Why does the double integral yield a different answer?






multivariable-calculus vector-fields curl




multivariable-calculus vector-fields curl






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 23:41









Scientifica

6,75141335




6,75141335










asked Dec 4 '18 at 23:23









bandicoot12bandicoot12

10528




10528












  • $begingroup$
    You got the right answer 0. Now check your work in the second method and if you don't find the mistake, post it in your question so that others verify it.
    $endgroup$
    – Scientifica
    Dec 4 '18 at 23:40


















  • $begingroup$
    You got the right answer 0. Now check your work in the second method and if you don't find the mistake, post it in your question so that others verify it.
    $endgroup$
    – Scientifica
    Dec 4 '18 at 23:40
















$begingroup$
You got the right answer 0. Now check your work in the second method and if you don't find the mistake, post it in your question so that others verify it.
$endgroup$
– Scientifica
Dec 4 '18 at 23:40




$begingroup$
You got the right answer 0. Now check your work in the second method and if you don't find the mistake, post it in your question so that others verify it.
$endgroup$
– Scientifica
Dec 4 '18 at 23:40










1 Answer
1






active

oldest

votes


















2












$begingroup$

This contradiction is pretty subtle and is usually used as a learning exercise for exactly this reason. You are correct that $nablatimes F=0$, but we have two contradictory statements:
$$oint Fcdot dr = 2pi, text{ and} \ iint_{R}nablatimes FdA=0.$$



Green's Theorem says that these two integrals should be equal, but a careful look at the assumptions for Green's Theorem will show that $F$ must be defined over the entire interior of the curve, even if the curl is identically $0$ everywhere.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ah, I see. Could you give me a physical idea of what's happening at (0,0)? Like, why does the fact that $F$ is undefined at the origin make the double integral equal 0?
    $endgroup$
    – bandicoot12
    Dec 4 '18 at 23:46










  • $begingroup$
    @bandicoot12 the fact that $F$ is undefined at the origin is not what's making the double integral zero, it's why green's theorem wasn't contradicted in this example
    $endgroup$
    – qbert
    Dec 5 '18 at 0:28








  • 2




    $begingroup$
    @bandicoot12 the fact that $F$ is undefined at the origin is what makes the two integrals not equal, not what makes the double integral $0$. The double integral is $0$ because the curl is $0$ and the line integral is $2pi$ since we can just compute it. For nice vector fields, Green's Theorem says they are equal, but the singularity at $0$ makes this vector field not nice so Green's Theorem does not apply here
    $endgroup$
    – whpowell96
    Dec 5 '18 at 0:40











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

This contradiction is pretty subtle and is usually used as a learning exercise for exactly this reason. You are correct that $nablatimes F=0$, but we have two contradictory statements:
$$oint Fcdot dr = 2pi, text{ and} \ iint_{R}nablatimes FdA=0.$$



Green's Theorem says that these two integrals should be equal, but a careful look at the assumptions for Green's Theorem will show that $F$ must be defined over the entire interior of the curve, even if the curl is identically $0$ everywhere.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ah, I see. Could you give me a physical idea of what's happening at (0,0)? Like, why does the fact that $F$ is undefined at the origin make the double integral equal 0?
    $endgroup$
    – bandicoot12
    Dec 4 '18 at 23:46










  • $begingroup$
    @bandicoot12 the fact that $F$ is undefined at the origin is not what's making the double integral zero, it's why green's theorem wasn't contradicted in this example
    $endgroup$
    – qbert
    Dec 5 '18 at 0:28








  • 2




    $begingroup$
    @bandicoot12 the fact that $F$ is undefined at the origin is what makes the two integrals not equal, not what makes the double integral $0$. The double integral is $0$ because the curl is $0$ and the line integral is $2pi$ since we can just compute it. For nice vector fields, Green's Theorem says they are equal, but the singularity at $0$ makes this vector field not nice so Green's Theorem does not apply here
    $endgroup$
    – whpowell96
    Dec 5 '18 at 0:40
















2












$begingroup$

This contradiction is pretty subtle and is usually used as a learning exercise for exactly this reason. You are correct that $nablatimes F=0$, but we have two contradictory statements:
$$oint Fcdot dr = 2pi, text{ and} \ iint_{R}nablatimes FdA=0.$$



Green's Theorem says that these two integrals should be equal, but a careful look at the assumptions for Green's Theorem will show that $F$ must be defined over the entire interior of the curve, even if the curl is identically $0$ everywhere.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ah, I see. Could you give me a physical idea of what's happening at (0,0)? Like, why does the fact that $F$ is undefined at the origin make the double integral equal 0?
    $endgroup$
    – bandicoot12
    Dec 4 '18 at 23:46










  • $begingroup$
    @bandicoot12 the fact that $F$ is undefined at the origin is not what's making the double integral zero, it's why green's theorem wasn't contradicted in this example
    $endgroup$
    – qbert
    Dec 5 '18 at 0:28








  • 2




    $begingroup$
    @bandicoot12 the fact that $F$ is undefined at the origin is what makes the two integrals not equal, not what makes the double integral $0$. The double integral is $0$ because the curl is $0$ and the line integral is $2pi$ since we can just compute it. For nice vector fields, Green's Theorem says they are equal, but the singularity at $0$ makes this vector field not nice so Green's Theorem does not apply here
    $endgroup$
    – whpowell96
    Dec 5 '18 at 0:40














2












2








2





$begingroup$

This contradiction is pretty subtle and is usually used as a learning exercise for exactly this reason. You are correct that $nablatimes F=0$, but we have two contradictory statements:
$$oint Fcdot dr = 2pi, text{ and} \ iint_{R}nablatimes FdA=0.$$



Green's Theorem says that these two integrals should be equal, but a careful look at the assumptions for Green's Theorem will show that $F$ must be defined over the entire interior of the curve, even if the curl is identically $0$ everywhere.






share|cite|improve this answer









$endgroup$



This contradiction is pretty subtle and is usually used as a learning exercise for exactly this reason. You are correct that $nablatimes F=0$, but we have two contradictory statements:
$$oint Fcdot dr = 2pi, text{ and} \ iint_{R}nablatimes FdA=0.$$



Green's Theorem says that these two integrals should be equal, but a careful look at the assumptions for Green's Theorem will show that $F$ must be defined over the entire interior of the curve, even if the curl is identically $0$ everywhere.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 4 '18 at 23:43









whpowell96whpowell96

56615




56615












  • $begingroup$
    Ah, I see. Could you give me a physical idea of what's happening at (0,0)? Like, why does the fact that $F$ is undefined at the origin make the double integral equal 0?
    $endgroup$
    – bandicoot12
    Dec 4 '18 at 23:46










  • $begingroup$
    @bandicoot12 the fact that $F$ is undefined at the origin is not what's making the double integral zero, it's why green's theorem wasn't contradicted in this example
    $endgroup$
    – qbert
    Dec 5 '18 at 0:28








  • 2




    $begingroup$
    @bandicoot12 the fact that $F$ is undefined at the origin is what makes the two integrals not equal, not what makes the double integral $0$. The double integral is $0$ because the curl is $0$ and the line integral is $2pi$ since we can just compute it. For nice vector fields, Green's Theorem says they are equal, but the singularity at $0$ makes this vector field not nice so Green's Theorem does not apply here
    $endgroup$
    – whpowell96
    Dec 5 '18 at 0:40


















  • $begingroup$
    Ah, I see. Could you give me a physical idea of what's happening at (0,0)? Like, why does the fact that $F$ is undefined at the origin make the double integral equal 0?
    $endgroup$
    – bandicoot12
    Dec 4 '18 at 23:46










  • $begingroup$
    @bandicoot12 the fact that $F$ is undefined at the origin is not what's making the double integral zero, it's why green's theorem wasn't contradicted in this example
    $endgroup$
    – qbert
    Dec 5 '18 at 0:28








  • 2




    $begingroup$
    @bandicoot12 the fact that $F$ is undefined at the origin is what makes the two integrals not equal, not what makes the double integral $0$. The double integral is $0$ because the curl is $0$ and the line integral is $2pi$ since we can just compute it. For nice vector fields, Green's Theorem says they are equal, but the singularity at $0$ makes this vector field not nice so Green's Theorem does not apply here
    $endgroup$
    – whpowell96
    Dec 5 '18 at 0:40
















$begingroup$
Ah, I see. Could you give me a physical idea of what's happening at (0,0)? Like, why does the fact that $F$ is undefined at the origin make the double integral equal 0?
$endgroup$
– bandicoot12
Dec 4 '18 at 23:46




$begingroup$
Ah, I see. Could you give me a physical idea of what's happening at (0,0)? Like, why does the fact that $F$ is undefined at the origin make the double integral equal 0?
$endgroup$
– bandicoot12
Dec 4 '18 at 23:46












$begingroup$
@bandicoot12 the fact that $F$ is undefined at the origin is not what's making the double integral zero, it's why green's theorem wasn't contradicted in this example
$endgroup$
– qbert
Dec 5 '18 at 0:28






$begingroup$
@bandicoot12 the fact that $F$ is undefined at the origin is not what's making the double integral zero, it's why green's theorem wasn't contradicted in this example
$endgroup$
– qbert
Dec 5 '18 at 0:28






2




2




$begingroup$
@bandicoot12 the fact that $F$ is undefined at the origin is what makes the two integrals not equal, not what makes the double integral $0$. The double integral is $0$ because the curl is $0$ and the line integral is $2pi$ since we can just compute it. For nice vector fields, Green's Theorem says they are equal, but the singularity at $0$ makes this vector field not nice so Green's Theorem does not apply here
$endgroup$
– whpowell96
Dec 5 '18 at 0:40




$begingroup$
@bandicoot12 the fact that $F$ is undefined at the origin is what makes the two integrals not equal, not what makes the double integral $0$. The double integral is $0$ because the curl is $0$ and the line integral is $2pi$ since we can just compute it. For nice vector fields, Green's Theorem says they are equal, but the singularity at $0$ makes this vector field not nice so Green's Theorem does not apply here
$endgroup$
– whpowell96
Dec 5 '18 at 0:40


















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