orientation problem in computing a 2-form integral
$begingroup$
Let $ {(x, y, z) ∈ mathbb{R}^3 | x^2 + y^2 = 1, 0le z le 1}$ and $α = z^2 xdy$ defined over $mathbb{R}^3$
Let $ω = dα$. Compute $int_{mathcal{C}} ω$.
If we use Stokes formulae: $int_{mathcal{C}} ω = int_{partial mathcal{C}} alpha = int_{mathbb{S}^1 times {0}} z^2x dy - int_{mathbb{S}^1 times {1}} z^2 x dy$.
1- Why does the orientation (the minus before the second integral) change between the two circles at $z=0$ and $z=1$?
2- If we use Stokes formulate differently with $V$ the volume of $mathcal{C}$:
$int_{mathcal{C}=partial V} omega = int_{V} dω == int_{V} dcirc dω =0 $.
There is an orientation problem in the 1- and a false reasonning in the 2- I can't see.
Thank you for your help.
differential-geometry
asked Dec 30 '18 at 3:39
PerelManPerelMan
524211
$endgroup$
add a comment |
$begingroup$
Let $ {(x, y, z) ∈ mathbb{R}^3 | x^2 + y^2 = 1, 0le z le 1}$ and $α = z^2 xdy$ defined over $mathbb{R}^3$
Let $ω = dα$. Compute $int_{mathcal{C}} ω$.
If we use Stokes formulae: $int_{mathcal{C}} ω = int_{partial mathcal{C}} alpha = int_{mathbb{S}^1 times {0}} z^2x dy - int_{mathbb{S}^1 times {1}} z^2 x dy$.
1- Why does the orientation (the minus before the second integral) change between the two circles at $z=0$ and $z=1$?
2- If we use Stokes formulate differently with $V$ the volume of $mathcal{C}$:
$int_{mathcal{C}=partial V} omega = int_{V} dω == int_{V} dcirc dω =0 $.
There is an orientation problem in the 1- and a false reasonning in the 2- I can't see.
Thank you for your help.
differential-geometry
asked Dec 30 '18 at 3:39
PerelManPerelMan
524211
$endgroup$
add a comment |
$begingroup$
Let $ {(x, y, z) ∈ mathbb{R}^3 | x^2 + y^2 = 1, 0le z le 1}$ and $α = z^2 xdy$ defined over $mathbb{R}^3$
Let $ω = dα$. Compute $int_{mathcal{C}} ω$.
If we use Stokes formulae: $int_{mathcal{C}} ω = int_{partial mathcal{C}} alpha = int_{mathbb{S}^1 times {0}} z^2x dy - int_{mathbb{S}^1 times {1}} z^2 x dy$.
1- Why does the orientation (the minus before the second integral) change between the two circles at $z=0$ and $z=1$?
2- If we use Stokes formulate differently with $V$ the volume of $mathcal{C}$:
$int_{mathcal{C}=partial V} omega = int_{V} dω == int_{V} dcirc dω =0 $.
There is an orientation problem in the 1- and a false reasonning in the 2- I can't see.
Thank you for your help.
differential-geometry
asked Dec 30 '18 at 3:39
PerelManPerelMan
524211
$endgroup$
Let $ {(x, y, z) ∈ mathbb{R}^3 | x^2 + y^2 = 1, 0le z le 1}$ and $α = z^2 xdy$ defined over $mathbb{R}^3$
Let $ω = dα$. Compute $int_{mathcal{C}} ω$.
If we use Stokes formulae: $int_{mathcal{C}} ω = int_{partial mathcal{C}} alpha = int_{mathbb{S}^1 times {0}} z^2x dy - int_{mathbb{S}^1 times {1}} z^2 x dy$.
1- Why does the orientation (the minus before the second integral) change between the two circles at $z=0$ and $z=1$?
2- If we use Stokes formulate differently with $V$ the volume of $mathcal{C}$:
$int_{mathcal{C}=partial V} omega = int_{V} dω == int_{V} dcirc dω =0 $.
There is an orientation problem in the 1- and a false reasonning in the 2- I can't see.
Thank you for your help.
differential-geometry
differential-geometry
asked Dec 30 '18 at 3:39
PerelManPerelMan
524211
asked Dec 30 '18 at 3:39
PerelManPerelMan
524211
asked Dec 30 '18 at 3:39
PerelManPerelMan
524211
asked Dec 30 '18 at 3:39
PerelManPerelMan
524211
asked Dec 30 '18 at 3:39
PerelManPerelMan
524211
524211
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is a question of orientation, you have to choose a normal vector which points outward.
Inducing orientations on boundary manifolds
answered Dec 30 '18 at 3:48
Tsemo AristideTsemo Aristide
56.7k11444
$endgroup$
$begingroup$
Thank you for the link. In that case, If we choose counter clockwise to be the direct direction and switch to cylindrical coordinates, I feel like the minus sign should rather go with the first integral: as $theta$ increases counter clockwise, the normal vector points inward for $mathbb{S}^1 times {0}$ and outwards for $mathbb{S}^1 times {1}$?
$endgroup$
– PerelMan
Dec 30 '18 at 21:44
1
$begingroup$
The best way tounderstand this is to consider a continuous $f:[a,b]rightarrow mathbb{R}$, $int_a^bf(t)dt=F(b)-F(a)$ where $F'=f$, the orientation of $mathbb{R}$ points towards $+infty$, the induced orientation on $a$ in the interval $[a,b]$ points towards $-infty$ because it points outwards. This is exactly the same if you consider a cube in $mathbb{R}^n$ even something more complicated by using Fubini.
$endgroup$
– Tsemo Aristide
Dec 30 '18 at 21:50
$begingroup$
Concerning Stokes, the cylinder is not a boundary.
$endgroup$
– Thomas
Dec 31 '18 at 15:14
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
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active
oldest
votes
$begingroup$
This is a question of orientation, you have to choose a normal vector which points outward.
Inducing orientations on boundary manifolds
answered Dec 30 '18 at 3:48
Tsemo AristideTsemo Aristide
56.7k11444
$endgroup$
$begingroup$
Thank you for the link. In that case, If we choose counter clockwise to be the direct direction and switch to cylindrical coordinates, I feel like the minus sign should rather go with the first integral: as $theta$ increases counter clockwise, the normal vector points inward for $mathbb{S}^1 times {0}$ and outwards for $mathbb{S}^1 times {1}$?
$endgroup$
– PerelMan
Dec 30 '18 at 21:44
1
$begingroup$
The best way tounderstand this is to consider a continuous $f:[a,b]rightarrow mathbb{R}$, $int_a^bf(t)dt=F(b)-F(a)$ where $F'=f$, the orientation of $mathbb{R}$ points towards $+infty$, the induced orientation on $a$ in the interval $[a,b]$ points towards $-infty$ because it points outwards. This is exactly the same if you consider a cube in $mathbb{R}^n$ even something more complicated by using Fubini.
$endgroup$
– Tsemo Aristide
Dec 30 '18 at 21:50
$begingroup$
Concerning Stokes, the cylinder is not a boundary.
$endgroup$
– Thomas
Dec 31 '18 at 15:14
add a comment |
$begingroup$
This is a question of orientation, you have to choose a normal vector which points outward.
Inducing orientations on boundary manifolds
answered Dec 30 '18 at 3:48
Tsemo AristideTsemo Aristide
56.7k11444
$endgroup$
$begingroup$
Thank you for the link. In that case, If we choose counter clockwise to be the direct direction and switch to cylindrical coordinates, I feel like the minus sign should rather go with the first integral: as $theta$ increases counter clockwise, the normal vector points inward for $mathbb{S}^1 times {0}$ and outwards for $mathbb{S}^1 times {1}$?
$endgroup$
– PerelMan
Dec 30 '18 at 21:44
1
$begingroup$
The best way tounderstand this is to consider a continuous $f:[a,b]rightarrow mathbb{R}$, $int_a^bf(t)dt=F(b)-F(a)$ where $F'=f$, the orientation of $mathbb{R}$ points towards $+infty$, the induced orientation on $a$ in the interval $[a,b]$ points towards $-infty$ because it points outwards. This is exactly the same if you consider a cube in $mathbb{R}^n$ even something more complicated by using Fubini.
$endgroup$
– Tsemo Aristide
Dec 30 '18 at 21:50
$begingroup$
Concerning Stokes, the cylinder is not a boundary.
$endgroup$
– Thomas
Dec 31 '18 at 15:14
add a comment |
$begingroup$
This is a question of orientation, you have to choose a normal vector which points outward.
Inducing orientations on boundary manifolds
answered Dec 30 '18 at 3:48
Tsemo AristideTsemo Aristide
56.7k11444
$endgroup$
This is a question of orientation, you have to choose a normal vector which points outward.
Inducing orientations on boundary manifolds
answered Dec 30 '18 at 3:48
Tsemo AristideTsemo Aristide
56.7k11444
answered Dec 30 '18 at 3:48
Tsemo AristideTsemo Aristide
56.7k11444
answered Dec 30 '18 at 3:48
Tsemo AristideTsemo Aristide
56.7k11444
answered Dec 30 '18 at 3:48
Tsemo AristideTsemo Aristide
56.7k11444
56.7k11444
$begingroup$
Thank you for the link. In that case, If we choose counter clockwise to be the direct direction and switch to cylindrical coordinates, I feel like the minus sign should rather go with the first integral: as $theta$ increases counter clockwise, the normal vector points inward for $mathbb{S}^1 times {0}$ and outwards for $mathbb{S}^1 times {1}$?
$endgroup$
– PerelMan
Dec 30 '18 at 21:44
1
$begingroup$
The best way tounderstand this is to consider a continuous $f:[a,b]rightarrow mathbb{R}$, $int_a^bf(t)dt=F(b)-F(a)$ where $F'=f$, the orientation of $mathbb{R}$ points towards $+infty$, the induced orientation on $a$ in the interval $[a,b]$ points towards $-infty$ because it points outwards. This is exactly the same if you consider a cube in $mathbb{R}^n$ even something more complicated by using Fubini.
$endgroup$
– Tsemo Aristide
Dec 30 '18 at 21:50
$begingroup$
Concerning Stokes, the cylinder is not a boundary.
$endgroup$
– Thomas
Dec 31 '18 at 15:14
add a comment |
$begingroup$
Thank you for the link. In that case, If we choose counter clockwise to be the direct direction and switch to cylindrical coordinates, I feel like the minus sign should rather go with the first integral: as $theta$ increases counter clockwise, the normal vector points inward for $mathbb{S}^1 times {0}$ and outwards for $mathbb{S}^1 times {1}$?
$endgroup$
– PerelMan
Dec 30 '18 at 21:44
1
$begingroup$
The best way tounderstand this is to consider a continuous $f:[a,b]rightarrow mathbb{R}$, $int_a^bf(t)dt=F(b)-F(a)$ where $F'=f$, the orientation of $mathbb{R}$ points towards $+infty$, the induced orientation on $a$ in the interval $[a,b]$ points towards $-infty$ because it points outwards. This is exactly the same if you consider a cube in $mathbb{R}^n$ even something more complicated by using Fubini.
$endgroup$
– Tsemo Aristide
Dec 30 '18 at 21:50
$begingroup$
Concerning Stokes, the cylinder is not a boundary.
$endgroup$
– Thomas
Dec 31 '18 at 15:14
$begingroup$
Thank you for the link. In that case, If we choose counter clockwise to be the direct direction and switch to cylindrical coordinates, I feel like the minus sign should rather go with the first integral: as $theta$ increases counter clockwise, the normal vector points inward for $mathbb{S}^1 times {0}$ and outwards for $mathbb{S}^1 times {1}$?
$endgroup$
– PerelMan
Dec 30 '18 at 21:44
$begingroup$
Thank you for the link. In that case, If we choose counter clockwise to be the direct direction and switch to cylindrical coordinates, I feel like the minus sign should rather go with the first integral: as $theta$ increases counter clockwise, the normal vector points inward for $mathbb{S}^1 times {0}$ and outwards for $mathbb{S}^1 times {1}$?
$endgroup$
– PerelMan
Dec 30 '18 at 21:44
1
1
$begingroup$
The best way tounderstand this is to consider a continuous $f:[a,b]rightarrow mathbb{R}$, $int_a^bf(t)dt=F(b)-F(a)$ where $F'=f$, the orientation of $mathbb{R}$ points towards $+infty$, the induced orientation on $a$ in the interval $[a,b]$ points towards $-infty$ because it points outwards. This is exactly the same if you consider a cube in $mathbb{R}^n$ even something more complicated by using Fubini.
$endgroup$
– Tsemo Aristide
Dec 30 '18 at 21:50
$begingroup$
The best way tounderstand this is to consider a continuous $f:[a,b]rightarrow mathbb{R}$, $int_a^bf(t)dt=F(b)-F(a)$ where $F'=f$, the orientation of $mathbb{R}$ points towards $+infty$, the induced orientation on $a$ in the interval $[a,b]$ points towards $-infty$ because it points outwards. This is exactly the same if you consider a cube in $mathbb{R}^n$ even something more complicated by using Fubini.
$endgroup$
– Tsemo Aristide
Dec 30 '18 at 21:50
$begingroup$
Concerning Stokes, the cylinder is not a boundary.
$endgroup$
– Thomas
Dec 31 '18 at 15:14
$begingroup$
Concerning Stokes, the cylinder is not a boundary.
$endgroup$
– Thomas
Dec 31 '18 at 15:14
add a comment |
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