Integration of a 1-form

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I have to calculate the integral $int_gammadomega$
$$omega=(y+z)dx+(z+x) dy+(x-y) dz$$
$gamma$ is the intersection between the sferic surface $x^2+y^2+z^2=1$ and che plane $y=z$

I want to solve it computing $domega$ directly with wedge products, so I found $domega=-2dywedge dz$. Now I establish a parametrization for $gamma$ which is $$begincases
x=cost\
y=frac1sqrt2sint\
z=frac1sqrt2sint
endcases $$
Where $tin[0,2pi]$, so I have to evaluate this integral:
$$int_[0,2pi] -2dywedge dz(-sint, frac1sqrt2cost, frac1sqrt2cost) $$

But I don't know how to evaluate that, usually I use this method when integrating a 2-form on a 2-surface. What should I do?

differential-forms bilinear-form line-integrals

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edited Jul 17 at 11:07

asked Jul 17 at 10:50

pter26

17811

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I have to calculate the integral $int_gammadomega$
$$omega=(y+z)dx+(z+x) dy+(x-y) dz$$
$gamma$ is the intersection between the sferic surface $x^2+y^2+z^2=1$ and che plane $y=z$

I want to solve it computing $domega$ directly with wedge products, so I found $domega=-2dywedge dz$. Now I establish a parametrization for $gamma$ which is $$begincases
x=cost\
y=frac1sqrt2sint\
z=frac1sqrt2sint
endcases $$
Where $tin[0,2pi]$, so I have to evaluate this integral:
$$int_[0,2pi] -2dywedge dz(-sint, frac1sqrt2cost, frac1sqrt2cost) $$

But I don't know how to evaluate that, usually I use this method when integrating a 2-form on a 2-surface. What should I do?

differential-forms bilinear-form line-integrals

share|cite|improve this question

edited Jul 17 at 11:07

asked Jul 17 at 10:50

pter26

17811

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I have to calculate the integral $int_gammadomega$
$$omega=(y+z)dx+(z+x) dy+(x-y) dz$$
$gamma$ is the intersection between the sferic surface $x^2+y^2+z^2=1$ and che plane $y=z$

I want to solve it computing $domega$ directly with wedge products, so I found $domega=-2dywedge dz$. Now I establish a parametrization for $gamma$ which is $$begincases
x=cost\
y=frac1sqrt2sint\
z=frac1sqrt2sint
endcases $$
Where $tin[0,2pi]$, so I have to evaluate this integral:
$$int_[0,2pi] -2dywedge dz(-sint, frac1sqrt2cost, frac1sqrt2cost) $$

But I don't know how to evaluate that, usually I use this method when integrating a 2-form on a 2-surface. What should I do?

differential-forms bilinear-form line-integrals

share|cite|improve this question

edited Jul 17 at 11:07

asked Jul 17 at 10:50

pter26

17811

I have to calculate the integral $int_gammadomega$
$$omega=(y+z)dx+(z+x) dy+(x-y) dz$$
$gamma$ is the intersection between the sferic surface $x^2+y^2+z^2=1$ and che plane $y=z$

I want to solve it computing $domega$ directly with wedge products, so I found $domega=-2dywedge dz$. Now I establish a parametrization for $gamma$ which is $$begincases
x=cost\
y=frac1sqrt2sint\
z=frac1sqrt2sint
endcases $$
Where $tin[0,2pi]$, so I have to evaluate this integral:
$$int_[0,2pi] -2dywedge dz(-sint, frac1sqrt2cost, frac1sqrt2cost) $$

But I don't know how to evaluate that, usually I use this method when integrating a 2-form on a 2-surface. What should I do?

differential-forms bilinear-form line-integrals

share|cite|improve this question

edited Jul 17 at 11:07

asked Jul 17 at 10:50

pter26

17811

share|cite|improve this question

share|cite|improve this question

edited Jul 17 at 11:07

edited Jul 17 at 11:07

edited Jul 17 at 11:07

asked Jul 17 at 10:50

pter26

17811

asked Jul 17 at 10:50

pter26

17811

asked Jul 17 at 10:50

pter26

17811

17811

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

You wish to apply Stokes theorem

$$ int _gamma omega = int_D domega,$$

where $D$ is a surface with $partial D = gamma$. You need to find this $D$ first.

There are lots of them, the simplest one is the intersection of the plane $y=z$ with the unit ball. It has a parametrization

begincases
x= r cos t\
y=fracrsqrt2sin t\
z=fracrsqrt2sin t
endcases

with $0<r<1$ and $ 0<t<2pi$.

Or you can just use your parametrization of $gamma$ and integrate $omega$.

share|cite|improve this answer

answered Jul 17 at 11:07

John Ma

37.5k93669

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But now have I to integrate $omega$ and not $domega$?
  – pter26
  Jul 17 at 11:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ah yes, $int_gamma domega$ does not quite make sense.
  – John Ma
  Jul 17 at 11:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah ok! So I integrate $domega$ on the circle. Now everything makes sense, thanks a lot!
  – pter26
  Jul 17 at 11:11
  

  
  
  
  

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 wish to apply Stokes theorem

$$ int _gamma omega = int_D domega,$$

where $D$ is a surface with $partial D = gamma$. You need to find this $D$ first.

There are lots of them, the simplest one is the intersection of the plane $y=z$ with the unit ball. It has a parametrization

begincases
x= r cos t\
y=fracrsqrt2sin t\
z=fracrsqrt2sin t
endcases

with $0<r<1$ and $ 0<t<2pi$.

Or you can just use your parametrization of $gamma$ and integrate $omega$.

share|cite|improve this answer

answered Jul 17 at 11:07

John Ma

37.5k93669

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But now have I to integrate $omega$ and not $domega$?
  – pter26
  Jul 17 at 11:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ah yes, $int_gamma domega$ does not quite make sense.
  – John Ma
  Jul 17 at 11:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah ok! So I integrate $domega$ on the circle. Now everything makes sense, thanks a lot!
  – pter26
  Jul 17 at 11:11
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

You wish to apply Stokes theorem

$$ int _gamma omega = int_D domega,$$

where $D$ is a surface with $partial D = gamma$. You need to find this $D$ first.

There are lots of them, the simplest one is the intersection of the plane $y=z$ with the unit ball. It has a parametrization

begincases
x= r cos t\
y=fracrsqrt2sin t\
z=fracrsqrt2sin t
endcases

with $0<r<1$ and $ 0<t<2pi$.

Or you can just use your parametrization of $gamma$ and integrate $omega$.

share|cite|improve this answer

answered Jul 17 at 11:07

John Ma

37.5k93669

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But now have I to integrate $omega$ and not $domega$?
  – pter26
  Jul 17 at 11:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ah yes, $int_gamma domega$ does not quite make sense.
  – John Ma
  Jul 17 at 11:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah ok! So I integrate $domega$ on the circle. Now everything makes sense, thanks a lot!
  – pter26
  Jul 17 at 11:11
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

You wish to apply Stokes theorem

$$ int _gamma omega = int_D domega,$$

where $D$ is a surface with $partial D = gamma$. You need to find this $D$ first.

There are lots of them, the simplest one is the intersection of the plane $y=z$ with the unit ball. It has a parametrization

begincases
x= r cos t\
y=fracrsqrt2sin t\
z=fracrsqrt2sin t
endcases

with $0<r<1$ and $ 0<t<2pi$.

Or you can just use your parametrization of $gamma$ and integrate $omega$.

share|cite|improve this answer

answered Jul 17 at 11:07

John Ma

37.5k93669

You wish to apply Stokes theorem

$$ int _gamma omega = int_D domega,$$

where $D$ is a surface with $partial D = gamma$. You need to find this $D$ first.

There are lots of them, the simplest one is the intersection of the plane $y=z$ with the unit ball. It has a parametrization

begincases
x= r cos t\
y=fracrsqrt2sin t\
z=fracrsqrt2sin t
endcases

with $0<r<1$ and $ 0<t<2pi$.

Or you can just use your parametrization of $gamma$ and integrate $omega$.

share|cite|improve this answer

answered Jul 17 at 11:07

John Ma

37.5k93669

share|cite|improve this answer

share|cite|improve this answer

answered Jul 17 at 11:07

John Ma

37.5k93669

answered Jul 17 at 11:07

John Ma

37.5k93669

answered Jul 17 at 11:07

John Ma

37.5k93669

37.5k93669

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But now have I to integrate $omega$ and not $domega$?
  – pter26
  Jul 17 at 11:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ah yes, $int_gamma domega$ does not quite make sense.
  – John Ma
  Jul 17 at 11:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah ok! So I integrate $domega$ on the circle. Now everything makes sense, thanks a lot!
  – pter26
  Jul 17 at 11:11
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But now have I to integrate $omega$ and not $domega$?
  – pter26
  Jul 17 at 11:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ah yes, $int_gamma domega$ does not quite make sense.
  – John Ma
  Jul 17 at 11:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah ok! So I integrate $domega$ on the circle. Now everything makes sense, thanks a lot!
  – pter26
  Jul 17 at 11:11
  

  
  
  
  

But now have I to integrate $omega$ and not $domega$?
– pter26
Jul 17 at 11:08

But now have I to integrate $omega$ and not $domega$?
– pter26
Jul 17 at 11:08

ah yes, $int_gamma domega$ does not quite make sense.
– John Ma
Jul 17 at 11:09

ah yes, $int_gamma domega$ does not quite make sense.
– John Ma
Jul 17 at 11:09

Ah ok! So I integrate $domega$ on the circle. Now everything makes sense, thanks a lot!
– pter26
Jul 17 at 11:11

Ah ok! So I integrate $domega$ on the circle. Now everything makes sense, thanks a lot!
– pter26
Jul 17 at 11:11

add a comment | 

 

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