Stokes Theorem and Circulation
$begingroup$
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
edited Dec 4 '18 at 23:41
Scientifica
6,75141335
asked Dec 4 '18 at 23:23
bandicoot12bandicoot12
10528
$endgroup$
add a comment |
$begingroup$
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
edited Dec 4 '18 at 23:41
Scientifica
6,75141335
asked Dec 4 '18 at 23:23
bandicoot12bandicoot12
10528
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Dec 4 '18 at 23:41
Scientifica
6,75141335
asked Dec 4 '18 at 23:23
bandicoot12bandicoot12
10528
$endgroup$
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
edited Dec 4 '18 at 23:41
Scientifica
6,75141335
asked Dec 4 '18 at 23:23
bandicoot12bandicoot12
10528
edited Dec 4 '18 at 23:41
Scientifica
6,75141335
asked Dec 4 '18 at 23:23
bandicoot12bandicoot12
10528
edited Dec 4 '18 at 23:41
Scientifica
6,75141335
edited Dec 4 '18 at 23:41
Scientifica
6,75141335
edited Dec 4 '18 at 23:41
Scientifica
6,75141335
6,75141335
asked Dec 4 '18 at 23:23
bandicoot12bandicoot12
10528
asked Dec 4 '18 at 23:23
bandicoot12bandicoot12
10528
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Dec 4 '18 at 23:43
whpowell96whpowell96
56615
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Dec 4 '18 at 23:43
whpowell96whpowell96
56615
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 4 '18 at 23:43
whpowell96whpowell96
56615
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 4 '18 at 23:43
whpowell96whpowell96
56615
$endgroup$
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.
answered Dec 4 '18 at 23:43
whpowell96whpowell96
56615
answered Dec 4 '18 at 23:43
whpowell96whpowell96
56615
answered Dec 4 '18 at 23:43
whpowell96whpowell96
56615
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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