Prove: $omega$ Is Exact

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Let $omega=omega_1dx_1+...+omega_ndx_n$ be a continuous form on an open and connected set $Omega subset mathbbR^n$.

Let assume that for all $x,yin Omega$ there is $cin mathbbR$ such that $int_gammaomega=c$ for every $gamma$ that connects $x$ to $y$.

Prove: $omega$ is exact in $Omega$

$int_x^yomega=c$ for any $gamma$ so $-int_y^xomega=-c$ for any $gamma$ summing them both we get $oint omega=0$ and therefore $omega$ is exact

Am I missing out something here? I have used the theorem:

$omega$ is exact $iffoint omega=0 iff$ any $gamma_1,gamma_2$ with same start and end $int_gamma_1omega=int_gamma_2omega$

multivariable-calculus differential-forms

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asked Jul 31 at 12:50

newhere

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up vote
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Let $omega=omega_1dx_1+...+omega_ndx_n$ be a continuous form on an open and connected set $Omega subset mathbbR^n$.

Let assume that for all $x,yin Omega$ there is $cin mathbbR$ such that $int_gammaomega=c$ for every $gamma$ that connects $x$ to $y$.

Prove: $omega$ is exact in $Omega$

$int_x^yomega=c$ for any $gamma$ so $-int_y^xomega=-c$ for any $gamma$ summing them both we get $oint omega=0$ and therefore $omega$ is exact

Am I missing out something here? I have used the theorem:

$omega$ is exact $iffoint omega=0 iff$ any $gamma_1,gamma_2$ with same start and end $int_gamma_1omega=int_gamma_2omega$

multivariable-calculus differential-forms

share|cite|improve this question

asked Jul 31 at 12:50

newhere

742310

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let $omega=omega_1dx_1+...+omega_ndx_n$ be a continuous form on an open and connected set $Omega subset mathbbR^n$.

Let assume that for all $x,yin Omega$ there is $cin mathbbR$ such that $int_gammaomega=c$ for every $gamma$ that connects $x$ to $y$.

Prove: $omega$ is exact in $Omega$

$int_x^yomega=c$ for any $gamma$ so $-int_y^xomega=-c$ for any $gamma$ summing them both we get $oint omega=0$ and therefore $omega$ is exact

Am I missing out something here? I have used the theorem:

$omega$ is exact $iffoint omega=0 iff$ any $gamma_1,gamma_2$ with same start and end $int_gamma_1omega=int_gamma_2omega$

multivariable-calculus differential-forms

share|cite|improve this question

asked Jul 31 at 12:50

newhere

742310

Let $omega=omega_1dx_1+...+omega_ndx_n$ be a continuous form on an open and connected set $Omega subset mathbbR^n$.

Let assume that for all $x,yin Omega$ there is $cin mathbbR$ such that $int_gammaomega=c$ for every $gamma$ that connects $x$ to $y$.

Prove: $omega$ is exact in $Omega$

$int_x^yomega=c$ for any $gamma$ so $-int_y^xomega=-c$ for any $gamma$ summing them both we get $oint omega=0$ and therefore $omega$ is exact

Am I missing out something here? I have used the theorem:

$omega$ is exact $iffoint omega=0 iff$ any $gamma_1,gamma_2$ with same start and end $int_gamma_1omega=int_gamma_2omega$

multivariable-calculus differential-forms

share|cite|improve this question

asked Jul 31 at 12:50

newhere

742310

share|cite|improve this question

share|cite|improve this question

asked Jul 31 at 12:50

newhere

742310

asked Jul 31 at 12:50

newhere

742310

asked Jul 31 at 12:50

newhere

742310

742310

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

You don't need the middle part of your theorem, or to think about closed curves at all. The right part

any $gamma_1,gamma_2$ with same start and end $int_gamma_1omega=int_gamma_2omega$

is exactly the property you've been given, only worded differently (this is the only thing I can see that is of any merit to "show"). The left part

$omega$ is exact

is what you are after. Connect the two using your theorem, and you're done.

share|cite|improve this answer

edited Jul 31 at 13:11

answered Jul 31 at 13:06

Arthur

98.3k793174

• 
  
  
  

  
  

  
  
  
  
  
  

  
  any $gamma_1,gamma_2$ with same start and end $int_gamma_1omega=int_gamma_2omega$ is path independent if $omegain C^1$ right?
  – newhere
  Jul 31 at 13:14
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @newhere That's what that line is saying, yes.
  – Arthur
  Jul 31 at 13:14
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

You don't need the middle part of your theorem, or to think about closed curves at all. The right part

any $gamma_1,gamma_2$ with same start and end $int_gamma_1omega=int_gamma_2omega$

is exactly the property you've been given, only worded differently (this is the only thing I can see that is of any merit to "show"). The left part

$omega$ is exact

is what you are after. Connect the two using your theorem, and you're done.

share|cite|improve this answer

edited Jul 31 at 13:11

answered Jul 31 at 13:06

Arthur

98.3k793174

• 
  
  
  

  
  

  
  
  
  
  
  

  
  any $gamma_1,gamma_2$ with same start and end $int_gamma_1omega=int_gamma_2omega$ is path independent if $omegain C^1$ right?
  – newhere
  Jul 31 at 13:14
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @newhere That's what that line is saying, yes.
  – Arthur
  Jul 31 at 13:14
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

You don't need the middle part of your theorem, or to think about closed curves at all. The right part

any $gamma_1,gamma_2$ with same start and end $int_gamma_1omega=int_gamma_2omega$

is exactly the property you've been given, only worded differently (this is the only thing I can see that is of any merit to "show"). The left part

$omega$ is exact

is what you are after. Connect the two using your theorem, and you're done.

share|cite|improve this answer

edited Jul 31 at 13:11

answered Jul 31 at 13:06

Arthur

98.3k793174

• 
  
  
  

  
  

  
  
  
  
  
  

  
  any $gamma_1,gamma_2$ with same start and end $int_gamma_1omega=int_gamma_2omega$ is path independent if $omegain C^1$ right?
  – newhere
  Jul 31 at 13:14
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @newhere That's what that line is saying, yes.
  – Arthur
  Jul 31 at 13:14
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

You don't need the middle part of your theorem, or to think about closed curves at all. The right part

any $gamma_1,gamma_2$ with same start and end $int_gamma_1omega=int_gamma_2omega$

is exactly the property you've been given, only worded differently (this is the only thing I can see that is of any merit to "show"). The left part

$omega$ is exact

is what you are after. Connect the two using your theorem, and you're done.

share|cite|improve this answer

edited Jul 31 at 13:11

answered Jul 31 at 13:06

Arthur

98.3k793174

You don't need the middle part of your theorem, or to think about closed curves at all. The right part

any $gamma_1,gamma_2$ with same start and end $int_gamma_1omega=int_gamma_2omega$

is exactly the property you've been given, only worded differently (this is the only thing I can see that is of any merit to "show"). The left part

$omega$ is exact

is what you are after. Connect the two using your theorem, and you're done.

share|cite|improve this answer

edited Jul 31 at 13:11

answered Jul 31 at 13:06

Arthur

98.3k793174

share|cite|improve this answer

share|cite|improve this answer

edited Jul 31 at 13:11

edited Jul 31 at 13:11

edited Jul 31 at 13:11

answered Jul 31 at 13:06

Arthur

98.3k793174

answered Jul 31 at 13:06

Arthur

98.3k793174

answered Jul 31 at 13:06

Arthur

98.3k793174

98.3k793174

• 
  
  
  

  
  

  
  
  
  
  
  

  
  any $gamma_1,gamma_2$ with same start and end $int_gamma_1omega=int_gamma_2omega$ is path independent if $omegain C^1$ right?
  – newhere
  Jul 31 at 13:14
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @newhere That's what that line is saying, yes.
  – Arthur
  Jul 31 at 13:14
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  any $gamma_1,gamma_2$ with same start and end $int_gamma_1omega=int_gamma_2omega$ is path independent if $omegain C^1$ right?
  – newhere
  Jul 31 at 13:14
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @newhere That's what that line is saying, yes.
  – Arthur
  Jul 31 at 13:14
  

  
  
  
  

any $gamma_1,gamma_2$ with same start and end $int_gamma_1omega=int_gamma_2omega$ is path independent if $omegain C^1$ right?
– newhere
Jul 31 at 13:14

any $gamma_1,gamma_2$ with same start and end $int_gamma_1omega=int_gamma_2omega$ is path independent if $omegain C^1$ right?
– newhere
Jul 31 at 13:14

1

1

@newhere That's what that line is saying, yes.
– Arthur
Jul 31 at 13:14

@newhere That's what that line is saying, yes.
– Arthur
Jul 31 at 13:14

add a comment | 

 

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