Why does Green's theorem fail for this exact differential form $int Mdx + Ndy$
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Why does Green's theorem fail for this exact differential form $int Mdx + Ndy = 0$ since it's an exact differential and M and N both are not functions of z ( the case for which i want to ask the question). Green says this $$int int partial N/partial X - partial M/partial YdA$$ which is equal to zero as this is an exact differential. If i do the line integral in some path to some point to some point then it is equal to the change of function between the final and initial points which may not be zero. So my question is, why does not Gauss' theorem become true for this?
differential-equations differential-forms gaussian-integral
asked Aug 18 at 0:37
user187604
2286
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up vote
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down vote
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Why does Green's theorem fail for this exact differential form $int Mdx + Ndy = 0$ since it's an exact differential and M and N both are not functions of z ( the case for which i want to ask the question). Green says this $$int int partial N/partial X - partial M/partial YdA$$ which is equal to zero as this is an exact differential. If i do the line integral in some path to some point to some point then it is equal to the change of function between the final and initial points which may not be zero. So my question is, why does not Gauss' theorem become true for this?
differential-equations differential-forms gaussian-integral
asked Aug 18 at 0:37
user187604
2286
First, your $ds$ needs to be $dA$ (ordinarily $ds$ stands for an arclength integral). But if the form is exact and it's continuously differentiable, that integrand is precisely $0$. I.e., $partial N/partial x - partial M/partial y = 0$ (assuming the functions are defined and continuously differentiable everywhere inside your curve). What's your issue?
– Ted Shifrin
Aug 18 at 0:47
@TedShifrin is it okay now?
– user187604
Aug 18 at 0:51
1
The integral of an exact form is path-independent. What you wrote in the first line applies only to line integrals around closed paths.
– Ted Shifrin
Aug 18 at 0:57
@TedShifrin so when im moving from a point to another which is not a closed integral that also doesn't enclose a surface. So we can't apply gauss theorem here. Is it the answer?
– user187604
Aug 18 at 1:21
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Why does Green's theorem fail for this exact differential form $int Mdx + Ndy = 0$ since it's an exact differential and M and N both are not functions of z ( the case for which i want to ask the question). Green says this $$int int partial N/partial X - partial M/partial YdA$$ which is equal to zero as this is an exact differential. If i do the line integral in some path to some point to some point then it is equal to the change of function between the final and initial points which may not be zero. So my question is, why does not Gauss' theorem become true for this?
differential-equations differential-forms gaussian-integral
asked Aug 18 at 0:37
user187604
2286
Why does Green's theorem fail for this exact differential form $int Mdx + Ndy = 0$ since it's an exact differential and M and N both are not functions of z ( the case for which i want to ask the question). Green says this $$int int partial N/partial X - partial M/partial YdA$$ which is equal to zero as this is an exact differential. If i do the line integral in some path to some point to some point then it is equal to the change of function between the final and initial points which may not be zero. So my question is, why does not Gauss' theorem become true for this?
differential-equations differential-forms gaussian-integral
asked Aug 18 at 0:37
user187604
2286
edited Aug 18 at 1:52
asked Aug 18 at 0:37
user187604
2286
asked Aug 18 at 0:37
user187604
2286
asked Aug 18 at 0:37
user187604
2286
2286
First, your $ds$ needs to be $dA$ (ordinarily $ds$ stands for an arclength integral). But if the form is exact and it's continuously differentiable, that integrand is precisely $0$. I.e., $partial N/partial x - partial M/partial y = 0$ (assuming the functions are defined and continuously differentiable everywhere inside your curve). What's your issue?
– Ted Shifrin
Aug 18 at 0:47
@TedShifrin is it okay now?
– user187604
Aug 18 at 0:51
1
The integral of an exact form is path-independent. What you wrote in the first line applies only to line integrals around closed paths.
– Ted Shifrin
Aug 18 at 0:57
@TedShifrin so when im moving from a point to another which is not a closed integral that also doesn't enclose a surface. So we can't apply gauss theorem here. Is it the answer?
– user187604
Aug 18 at 1:21
add a comment |Â
First, your $ds$ needs to be $dA$ (ordinarily $ds$ stands for an arclength integral). But if the form is exact and it's continuously differentiable, that integrand is precisely $0$. I.e., $partial N/partial x - partial M/partial y = 0$ (assuming the functions are defined and continuously differentiable everywhere inside your curve). What's your issue?
– Ted Shifrin
Aug 18 at 0:47
@TedShifrin is it okay now?
– user187604
Aug 18 at 0:51
1
The integral of an exact form is path-independent. What you wrote in the first line applies only to line integrals around closed paths.
– Ted Shifrin
Aug 18 at 0:57
@TedShifrin so when im moving from a point to another which is not a closed integral that also doesn't enclose a surface. So we can't apply gauss theorem here. Is it the answer?
– user187604
Aug 18 at 1:21
First, your $ds$ needs to be $dA$ (ordinarily $ds$ stands for an arclength integral). But if the form is exact and it's continuously differentiable, that integrand is precisely $0$. I.e., $partial N/partial x - partial M/partial y = 0$ (assuming the functions are defined and continuously differentiable everywhere inside your curve). What's your issue?
– Ted Shifrin
Aug 18 at 0:47
First, your $ds$ needs to be $dA$ (ordinarily $ds$ stands for an arclength integral). But if the form is exact and it's continuously differentiable, that integrand is precisely $0$. I.e., $partial N/partial x - partial M/partial y = 0$ (assuming the functions are defined and continuously differentiable everywhere inside your curve). What's your issue?
– Ted Shifrin
Aug 18 at 0:47
@TedShifrin is it okay now?
– user187604
Aug 18 at 0:51
@TedShifrin is it okay now?
– user187604
Aug 18 at 0:51
1
1
The integral of an exact form is path-independent. What you wrote in the first line applies only to line integrals around closed paths.
– Ted Shifrin
Aug 18 at 0:57
The integral of an exact form is path-independent. What you wrote in the first line applies only to line integrals around closed paths.
– Ted Shifrin
Aug 18 at 0:57
@TedShifrin so when im moving from a point to another which is not a closed integral that also doesn't enclose a surface. So we can't apply gauss theorem here. Is it the answer?
– user187604
Aug 18 at 1:21
@TedShifrin so when im moving from a point to another which is not a closed integral that also doesn't enclose a surface. So we can't apply gauss theorem here. Is it the answer?
– user187604
Aug 18 at 1:21
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
A note on terminology: The theorem you refer to as Gauss's theorem is more commonly called Green's theorem. Gauss's theorem often refers to the divergence theorem (which is a generalazation of green's theorem, and thus both may be reffered to as Gauss's theorem).
Now, let us state Green's theorem (by wikipedia):
Let $C$ be a positively oriented, piecewise smooth, simple closed curve in a plane, and let $D$ be the region bounded by $C$. If $M$ and $N$ are functions of $(x, y)$ defined on an open region containing $D$ and have continuous partial derivatives there, then
$$ int_C M,dx + N, dy = iint_D left(fracpartial Npartial x - fracpartial Mpartial yright) , dx, dy $$
where the path of integration along $C$ is anticlockwise.
Note that the theorem requires the curve to be closed. Thus, you may not apply the theorem for curves that have different end points.
answered Aug 18 at 1:29
yakobyd
579111
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
A note on terminology: The theorem you refer to as Gauss's theorem is more commonly called Green's theorem. Gauss's theorem often refers to the divergence theorem (which is a generalazation of green's theorem, and thus both may be reffered to as Gauss's theorem).
Now, let us state Green's theorem (by wikipedia):
Let $C$ be a positively oriented, piecewise smooth, simple closed curve in a plane, and let $D$ be the region bounded by $C$. If $M$ and $N$ are functions of $(x, y)$ defined on an open region containing $D$ and have continuous partial derivatives there, then
$$ int_C M,dx + N, dy = iint_D left(fracpartial Npartial x - fracpartial Mpartial yright) , dx, dy $$
where the path of integration along $C$ is anticlockwise.
Note that the theorem requires the curve to be closed. Thus, you may not apply the theorem for curves that have different end points.
answered Aug 18 at 1:29
yakobyd
579111
add a comment |Â
up vote
2
down vote
accepted
A note on terminology: The theorem you refer to as Gauss's theorem is more commonly called Green's theorem. Gauss's theorem often refers to the divergence theorem (which is a generalazation of green's theorem, and thus both may be reffered to as Gauss's theorem).
Now, let us state Green's theorem (by wikipedia):
Let $C$ be a positively oriented, piecewise smooth, simple closed curve in a plane, and let $D$ be the region bounded by $C$. If $M$ and $N$ are functions of $(x, y)$ defined on an open region containing $D$ and have continuous partial derivatives there, then
$$ int_C M,dx + N, dy = iint_D left(fracpartial Npartial x - fracpartial Mpartial yright) , dx, dy $$
where the path of integration along $C$ is anticlockwise.
Note that the theorem requires the curve to be closed. Thus, you may not apply the theorem for curves that have different end points.
answered Aug 18 at 1:29
yakobyd
579111
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
A note on terminology: The theorem you refer to as Gauss's theorem is more commonly called Green's theorem. Gauss's theorem often refers to the divergence theorem (which is a generalazation of green's theorem, and thus both may be reffered to as Gauss's theorem).
Now, let us state Green's theorem (by wikipedia):
Let $C$ be a positively oriented, piecewise smooth, simple closed curve in a plane, and let $D$ be the region bounded by $C$. If $M$ and $N$ are functions of $(x, y)$ defined on an open region containing $D$ and have continuous partial derivatives there, then
$$ int_C M,dx + N, dy = iint_D left(fracpartial Npartial x - fracpartial Mpartial yright) , dx, dy $$
where the path of integration along $C$ is anticlockwise.
Note that the theorem requires the curve to be closed. Thus, you may not apply the theorem for curves that have different end points.
answered Aug 18 at 1:29
yakobyd
579111
A note on terminology: The theorem you refer to as Gauss's theorem is more commonly called Green's theorem. Gauss's theorem often refers to the divergence theorem (which is a generalazation of green's theorem, and thus both may be reffered to as Gauss's theorem).
Now, let us state Green's theorem (by wikipedia):
Let $C$ be a positively oriented, piecewise smooth, simple closed curve in a plane, and let $D$ be the region bounded by $C$. If $M$ and $N$ are functions of $(x, y)$ defined on an open region containing $D$ and have continuous partial derivatives there, then
$$ int_C M,dx + N, dy = iint_D left(fracpartial Npartial x - fracpartial Mpartial yright) , dx, dy $$
where the path of integration along $C$ is anticlockwise.
Note that the theorem requires the curve to be closed. Thus, you may not apply the theorem for curves that have different end points.
answered Aug 18 at 1:29
yakobyd
579111
edited Aug 18 at 1:49
answered Aug 18 at 1:29
yakobyd
579111
answered Aug 18 at 1:29
yakobyd
579111
answered Aug 18 at 1:29
yakobyd
579111
579111
add a comment |Â
add a comment |Â
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First, your $ds$ needs to be $dA$ (ordinarily $ds$ stands for an arclength integral). But if the form is exact and it's continuously differentiable, that integrand is precisely $0$. I.e., $partial N/partial x - partial M/partial y = 0$ (assuming the functions are defined and continuously differentiable everywhere inside your curve). What's your issue?
– Ted Shifrin
Aug 18 at 0:47
@TedShifrin is it okay now?
– user187604
Aug 18 at 0:51
1
The integral of an exact form is path-independent. What you wrote in the first line applies only to line integrals around closed paths.
– Ted Shifrin
Aug 18 at 0:57
@TedShifrin so when im moving from a point to another which is not a closed integral that also doesn't enclose a surface. So we can't apply gauss theorem here. Is it the answer?
– user187604
Aug 18 at 1:21