orientation problem in computing a 2-form integral












0












$begingroup$


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







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    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







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      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







      share|cite|improve this question









      $endgroup$




      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 30 '18 at 3:39









      PerelManPerelMan

      524211




      524211






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          This is a question of orientation, you have to choose a normal vector which points outward.



          Inducing orientations on boundary manifolds






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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











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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          This is a question of orientation, you have to choose a normal vector which points outward.



          Inducing orientations on boundary manifolds






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          1












          $begingroup$

          This is a question of orientation, you have to choose a normal vector which points outward.



          Inducing orientations on boundary manifolds






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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














          1












          1








          1





          $begingroup$

          This is a question of orientation, you have to choose a normal vector which points outward.



          Inducing orientations on boundary manifolds






          share|cite|improve this answer









          $endgroup$



          This is a question of orientation, you have to choose a normal vector which points outward.



          Inducing orientations on boundary manifolds







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • $begingroup$
            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


















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