Applications of Green's theorem
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Let $fin C^1$. Prove that for every $x_*$ : $(nabla times f)(x_*)=lim_{epsilon rightarrow 0} frac{1}{pi epsilon ^2} int_{partial B_epsilon (x_*)}f(x)dx$
I know that
$$
lim_{epsilon rightarrow 0} frac{1}{pi epsilon ^2} int_{partial B_epsilon (x_*)}f(x)dx =lim_{epsilon rightarrow 0} frac{1}{pi epsilon ^2} int_{B_epsilon (x_*)}nabla times f(x)dx
$$
But how can I continue from here? As I obviously can't calculate explicitly the right side of the equation...
calculus integration multivariable-calculus surface-integrals
edited Nov 17 at 17:11
rafa11111
893317
asked Nov 17 at 16:56
ChikChak
769418
add a comment |
up vote
1
down vote
favorite
Let $fin C^1$. Prove that for every $x_*$ : $(nabla times f)(x_*)=lim_{epsilon rightarrow 0} frac{1}{pi epsilon ^2} int_{partial B_epsilon (x_*)}f(x)dx$
I know that
$$
lim_{epsilon rightarrow 0} frac{1}{pi epsilon ^2} int_{partial B_epsilon (x_*)}f(x)dx =lim_{epsilon rightarrow 0} frac{1}{pi epsilon ^2} int_{B_epsilon (x_*)}nabla times f(x)dx
$$
But how can I continue from here? As I obviously can't calculate explicitly the right side of the equation...
calculus integration multivariable-calculus surface-integrals
edited Nov 17 at 17:11
rafa11111
893317
asked Nov 17 at 16:56
ChikChak
769418
Shouldn't it be something like $int_{B_epsilon (x_*)} nabla times f cdot mathbf{n} dS$ in the integral on RHS?
– rafa11111
Nov 17 at 17:15
@rafa11111 Why? And also if so, how to continue?
– ChikChak
Nov 17 at 17:17
I suppose you used Kelvin-Stokes theorem to get the second expression. However, it states that the integral of $f$ along a closed path $C$ is equal to the integral of $nabla times f$ in the surface enclosed by $C$. Therefore, the integral on RHS must be evaluated in a surface. I think that you can get the original equation from the second one by using the fact that the area of $B_epsilon(x_*)$ is proportional to $epsilon^2$, which is also dividing the RHS. It's just a guess.
– rafa11111
Nov 17 at 17:23
1
For the 3D case I have recently posted this answer to the question "How do I use Stokes' theorem to find curl?.
– Daniele Tampieri
Nov 17 at 17:27
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $fin C^1$. Prove that for every $x_*$ : $(nabla times f)(x_*)=lim_{epsilon rightarrow 0} frac{1}{pi epsilon ^2} int_{partial B_epsilon (x_*)}f(x)dx$
I know that
$$
lim_{epsilon rightarrow 0} frac{1}{pi epsilon ^2} int_{partial B_epsilon (x_*)}f(x)dx =lim_{epsilon rightarrow 0} frac{1}{pi epsilon ^2} int_{B_epsilon (x_*)}nabla times f(x)dx
$$
But how can I continue from here? As I obviously can't calculate explicitly the right side of the equation...
calculus integration multivariable-calculus surface-integrals
edited Nov 17 at 17:11
rafa11111
893317
asked Nov 17 at 16:56
ChikChak
769418
Let $fin C^1$. Prove that for every $x_*$ : $(nabla times f)(x_*)=lim_{epsilon rightarrow 0} frac{1}{pi epsilon ^2} int_{partial B_epsilon (x_*)}f(x)dx$
I know that
$$
lim_{epsilon rightarrow 0} frac{1}{pi epsilon ^2} int_{partial B_epsilon (x_*)}f(x)dx =lim_{epsilon rightarrow 0} frac{1}{pi epsilon ^2} int_{B_epsilon (x_*)}nabla times f(x)dx
$$
But how can I continue from here? As I obviously can't calculate explicitly the right side of the equation...
calculus integration multivariable-calculus surface-integrals
calculus integration multivariable-calculus surface-integrals
edited Nov 17 at 17:11
rafa11111
893317
asked Nov 17 at 16:56
ChikChak
769418
edited Nov 17 at 17:11
rafa11111
893317
asked Nov 17 at 16:56
ChikChak
769418
edited Nov 17 at 17:11
rafa11111
893317
edited Nov 17 at 17:11
rafa11111
893317
edited Nov 17 at 17:11
rafa11111
893317
893317
asked Nov 17 at 16:56
ChikChak
769418
asked Nov 17 at 16:56
ChikChak
769418
asked Nov 17 at 16:56
ChikChak
769418
769418
Shouldn't it be something like $int_{B_epsilon (x_*)} nabla times f cdot mathbf{n} dS$ in the integral on RHS?
– rafa11111
Nov 17 at 17:15
@rafa11111 Why? And also if so, how to continue?
– ChikChak
Nov 17 at 17:17
I suppose you used Kelvin-Stokes theorem to get the second expression. However, it states that the integral of $f$ along a closed path $C$ is equal to the integral of $nabla times f$ in the surface enclosed by $C$. Therefore, the integral on RHS must be evaluated in a surface. I think that you can get the original equation from the second one by using the fact that the area of $B_epsilon(x_*)$ is proportional to $epsilon^2$, which is also dividing the RHS. It's just a guess.
– rafa11111
Nov 17 at 17:23
1
For the 3D case I have recently posted this answer to the question "How do I use Stokes' theorem to find curl?.
– Daniele Tampieri
Nov 17 at 17:27
add a comment |
Shouldn't it be something like $int_{B_epsilon (x_*)} nabla times f cdot mathbf{n} dS$ in the integral on RHS?
– rafa11111
Nov 17 at 17:15
@rafa11111 Why? And also if so, how to continue?
– ChikChak
Nov 17 at 17:17
I suppose you used Kelvin-Stokes theorem to get the second expression. However, it states that the integral of $f$ along a closed path $C$ is equal to the integral of $nabla times f$ in the surface enclosed by $C$. Therefore, the integral on RHS must be evaluated in a surface. I think that you can get the original equation from the second one by using the fact that the area of $B_epsilon(x_*)$ is proportional to $epsilon^2$, which is also dividing the RHS. It's just a guess.
– rafa11111
Nov 17 at 17:23
1
For the 3D case I have recently posted this answer to the question "How do I use Stokes' theorem to find curl?.
– Daniele Tampieri
Nov 17 at 17:27
Shouldn't it be something like $int_{B_epsilon (x_*)} nabla times f cdot mathbf{n} dS$ in the integral on RHS?
– rafa11111
Nov 17 at 17:15
Shouldn't it be something like $int_{B_epsilon (x_*)} nabla times f cdot mathbf{n} dS$ in the integral on RHS?
– rafa11111
Nov 17 at 17:15
@rafa11111 Why? And also if so, how to continue?
– ChikChak
Nov 17 at 17:17
@rafa11111 Why? And also if so, how to continue?
– ChikChak
Nov 17 at 17:17
I suppose you used Kelvin-Stokes theorem to get the second expression. However, it states that the integral of $f$ along a closed path $C$ is equal to the integral of $nabla times f$ in the surface enclosed by $C$. Therefore, the integral on RHS must be evaluated in a surface. I think that you can get the original equation from the second one by using the fact that the area of $B_epsilon(x_*)$ is proportional to $epsilon^2$, which is also dividing the RHS. It's just a guess.
– rafa11111
Nov 17 at 17:23
I suppose you used Kelvin-Stokes theorem to get the second expression. However, it states that the integral of $f$ along a closed path $C$ is equal to the integral of $nabla times f$ in the surface enclosed by $C$. Therefore, the integral on RHS must be evaluated in a surface. I think that you can get the original equation from the second one by using the fact that the area of $B_epsilon(x_*)$ is proportional to $epsilon^2$, which is also dividing the RHS. It's just a guess.
– rafa11111
Nov 17 at 17:23
1
1
For the 3D case I have recently posted this answer to the question "How do I use Stokes' theorem to find curl?.
– Daniele Tampieri
Nov 17 at 17:27
For the 3D case I have recently posted this answer to the question "How do I use Stokes' theorem to find curl?.
– Daniele Tampieri
Nov 17 at 17:27
add a comment |
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Shouldn't it be something like $int_{B_epsilon (x_*)} nabla times f cdot mathbf{n} dS$ in the integral on RHS?
– rafa11111
Nov 17 at 17:15
@rafa11111 Why? And also if so, how to continue?
– ChikChak
Nov 17 at 17:17
I suppose you used Kelvin-Stokes theorem to get the second expression. However, it states that the integral of $f$ along a closed path $C$ is equal to the integral of $nabla times f$ in the surface enclosed by $C$. Therefore, the integral on RHS must be evaluated in a surface. I think that you can get the original equation from the second one by using the fact that the area of $B_epsilon(x_*)$ is proportional to $epsilon^2$, which is also dividing the RHS. It's just a guess.
– rafa11111
Nov 17 at 17:23
1
For the 3D case I have recently posted this answer to the question "How do I use Stokes' theorem to find curl?.
– Daniele Tampieri
Nov 17 at 17:27