Compute this integral of a form












0












$begingroup$


I need to compute the integral:
$$
int_{mathbb{S}^3}alpha,
$$
where $alphain Omega^3(mathbb{R}^4)$ is given by $alpha=-4t;dxwedge dywedge dz+4z;dxwedge dywedge dt-4y;dxwedge dzwedge dt+4x;dywedge dzwedge dt$.



My problem is that I am getting two different results. If I apply the definition:
$$
begin{array}{rccl}
varphi^+:& mathbb{S}^3cap{t geq 0}:=mathbb{S}^{3+}&longrightarrow&mathbb{D}^3\
&(x,y,z,t)&longmapsto&(x,y,z)
end{array}
$$
$$
begin{array}{rccl}
varphi^-:& mathbb{S}^{3-}cap{t leq 0}:=mathbb{S}^{3-}&longrightarrow&mathbb{D}^3\
&(x,y,z,t)&longmapsto&(x,y,z)
end{array}
$$
The inverses:
$$
begin{array}{rccl}
(varphi^+)^{-1}:& mathbb{D}^3&longrightarrow&mathbb{S}^{3+}\
&(u,v,w)&longmapsto&(u,v,w,sqrt{1-u^2-v^2-w^2})
end{array}
$$
$$
begin{array}{rccl}
(varphi^-)^{-1}:& mathbb{D}^3&longrightarrow&mathbb{S}^{3-}\
&(u,v,w)&longmapsto&(u,v,w,-sqrt{1-u^2-v^2-w^2})
end{array}
$$
If $beta=t;dxwedge dywedge dz$:
$$
int_{mathbb{S}^3}beta=int_{mathbb{S}^{3-}}beta+int_{mathbb{S}^{3+}}beta=int_{mathbb{D}^3}left[(varphi^-)^{-1}right]^*(beta)+int_{mathbb{D}^3}left[(varphi^+)^{-1}right]^*(beta)
$$
$$=-int_{mathbb{D}^3}sqrt{1-u^2-v^2-w^2};du;dv;dw+int_{mathbb{D}^3}sqrt{1-u^2-v^2-w^2};du;dv;dw=0
$$
And also $0$ for the other terms.



But using Stokes theorem I have:
$$
int_{mathbb{S}^3}alpha=int_{partial(mathbb{D}^4)}alpha=int_{mathbb{D}^4}partialalpha=int_{mathbb{D}^4}16;dxwedge dywedge dzwedge dt=16text{Vol}(mathbb{D}^4)not=0
$$
Where is the mistake? Which one is correct?






integration differential-geometry differential-forms







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    I need to compute the integral:
    $$
    int_{mathbb{S}^3}alpha,
    $$
    where $alphain Omega^3(mathbb{R}^4)$ is given by $alpha=-4t;dxwedge dywedge dz+4z;dxwedge dywedge dt-4y;dxwedge dzwedge dt+4x;dywedge dzwedge dt$.



    My problem is that I am getting two different results. If I apply the definition:
    $$
    begin{array}{rccl}
    varphi^+:& mathbb{S}^3cap{t geq 0}:=mathbb{S}^{3+}&longrightarrow&mathbb{D}^3\
    &(x,y,z,t)&longmapsto&(x,y,z)
    end{array}
    $$
    $$
    begin{array}{rccl}
    varphi^-:& mathbb{S}^{3-}cap{t leq 0}:=mathbb{S}^{3-}&longrightarrow&mathbb{D}^3\
    &(x,y,z,t)&longmapsto&(x,y,z)
    end{array}
    $$
    The inverses:
    $$
    begin{array}{rccl}
    (varphi^+)^{-1}:& mathbb{D}^3&longrightarrow&mathbb{S}^{3+}\
    &(u,v,w)&longmapsto&(u,v,w,sqrt{1-u^2-v^2-w^2})
    end{array}
    $$
    $$
    begin{array}{rccl}
    (varphi^-)^{-1}:& mathbb{D}^3&longrightarrow&mathbb{S}^{3-}\
    &(u,v,w)&longmapsto&(u,v,w,-sqrt{1-u^2-v^2-w^2})
    end{array}
    $$
    If $beta=t;dxwedge dywedge dz$:
    $$
    int_{mathbb{S}^3}beta=int_{mathbb{S}^{3-}}beta+int_{mathbb{S}^{3+}}beta=int_{mathbb{D}^3}left[(varphi^-)^{-1}right]^*(beta)+int_{mathbb{D}^3}left[(varphi^+)^{-1}right]^*(beta)
    $$
    $$=-int_{mathbb{D}^3}sqrt{1-u^2-v^2-w^2};du;dv;dw+int_{mathbb{D}^3}sqrt{1-u^2-v^2-w^2};du;dv;dw=0
    $$
    And also $0$ for the other terms.



    But using Stokes theorem I have:
    $$
    int_{mathbb{S}^3}alpha=int_{partial(mathbb{D}^4)}alpha=int_{mathbb{D}^4}partialalpha=int_{mathbb{D}^4}16;dxwedge dywedge dzwedge dt=16text{Vol}(mathbb{D}^4)not=0
    $$
    Where is the mistake? Which one is correct?






    integration differential-geometry differential-forms







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I need to compute the integral:
      $$
      int_{mathbb{S}^3}alpha,
      $$
      where $alphain Omega^3(mathbb{R}^4)$ is given by $alpha=-4t;dxwedge dywedge dz+4z;dxwedge dywedge dt-4y;dxwedge dzwedge dt+4x;dywedge dzwedge dt$.



      My problem is that I am getting two different results. If I apply the definition:
      $$
      begin{array}{rccl}
      varphi^+:& mathbb{S}^3cap{t geq 0}:=mathbb{S}^{3+}&longrightarrow&mathbb{D}^3\
      &(x,y,z,t)&longmapsto&(x,y,z)
      end{array}
      $$
      $$
      begin{array}{rccl}
      varphi^-:& mathbb{S}^{3-}cap{t leq 0}:=mathbb{S}^{3-}&longrightarrow&mathbb{D}^3\
      &(x,y,z,t)&longmapsto&(x,y,z)
      end{array}
      $$
      The inverses:
      $$
      begin{array}{rccl}
      (varphi^+)^{-1}:& mathbb{D}^3&longrightarrow&mathbb{S}^{3+}\
      &(u,v,w)&longmapsto&(u,v,w,sqrt{1-u^2-v^2-w^2})
      end{array}
      $$
      $$
      begin{array}{rccl}
      (varphi^-)^{-1}:& mathbb{D}^3&longrightarrow&mathbb{S}^{3-}\
      &(u,v,w)&longmapsto&(u,v,w,-sqrt{1-u^2-v^2-w^2})
      end{array}
      $$
      If $beta=t;dxwedge dywedge dz$:
      $$
      int_{mathbb{S}^3}beta=int_{mathbb{S}^{3-}}beta+int_{mathbb{S}^{3+}}beta=int_{mathbb{D}^3}left[(varphi^-)^{-1}right]^*(beta)+int_{mathbb{D}^3}left[(varphi^+)^{-1}right]^*(beta)
      $$
      $$=-int_{mathbb{D}^3}sqrt{1-u^2-v^2-w^2};du;dv;dw+int_{mathbb{D}^3}sqrt{1-u^2-v^2-w^2};du;dv;dw=0
      $$
      And also $0$ for the other terms.



      But using Stokes theorem I have:
      $$
      int_{mathbb{S}^3}alpha=int_{partial(mathbb{D}^4)}alpha=int_{mathbb{D}^4}partialalpha=int_{mathbb{D}^4}16;dxwedge dywedge dzwedge dt=16text{Vol}(mathbb{D}^4)not=0
      $$
      Where is the mistake? Which one is correct?






      integration differential-geometry differential-forms







      share|cite|improve this question









      $endgroup$




      I need to compute the integral:
      $$
      int_{mathbb{S}^3}alpha,
      $$
      where $alphain Omega^3(mathbb{R}^4)$ is given by $alpha=-4t;dxwedge dywedge dz+4z;dxwedge dywedge dt-4y;dxwedge dzwedge dt+4x;dywedge dzwedge dt$.



      My problem is that I am getting two different results. If I apply the definition:
      $$
      begin{array}{rccl}
      varphi^+:& mathbb{S}^3cap{t geq 0}:=mathbb{S}^{3+}&longrightarrow&mathbb{D}^3\
      &(x,y,z,t)&longmapsto&(x,y,z)
      end{array}
      $$
      $$
      begin{array}{rccl}
      varphi^-:& mathbb{S}^{3-}cap{t leq 0}:=mathbb{S}^{3-}&longrightarrow&mathbb{D}^3\
      &(x,y,z,t)&longmapsto&(x,y,z)
      end{array}
      $$
      The inverses:
      $$
      begin{array}{rccl}
      (varphi^+)^{-1}:& mathbb{D}^3&longrightarrow&mathbb{S}^{3+}\
      &(u,v,w)&longmapsto&(u,v,w,sqrt{1-u^2-v^2-w^2})
      end{array}
      $$
      $$
      begin{array}{rccl}
      (varphi^-)^{-1}:& mathbb{D}^3&longrightarrow&mathbb{S}^{3-}\
      &(u,v,w)&longmapsto&(u,v,w,-sqrt{1-u^2-v^2-w^2})
      end{array}
      $$
      If $beta=t;dxwedge dywedge dz$:
      $$
      int_{mathbb{S}^3}beta=int_{mathbb{S}^{3-}}beta+int_{mathbb{S}^{3+}}beta=int_{mathbb{D}^3}left[(varphi^-)^{-1}right]^*(beta)+int_{mathbb{D}^3}left[(varphi^+)^{-1}right]^*(beta)
      $$
      $$=-int_{mathbb{D}^3}sqrt{1-u^2-v^2-w^2};du;dv;dw+int_{mathbb{D}^3}sqrt{1-u^2-v^2-w^2};du;dv;dw=0
      $$
      And also $0$ for the other terms.



      But using Stokes theorem I have:
      $$
      int_{mathbb{S}^3}alpha=int_{partial(mathbb{D}^4)}alpha=int_{mathbb{D}^4}partialalpha=int_{mathbb{D}^4}16;dxwedge dywedge dzwedge dt=16text{Vol}(mathbb{D}^4)not=0
      $$
      Where is the mistake? Which one is correct?






      integration differential-geometry differential-forms




      integration differential-geometry differential-forms






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 7 '18 at 16:00









      davidivadfuldavidivadful

      1189




      1189






















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          $begingroup$

          The Stokes's Theorem application is correct. The mistake comes in half of the first part: With the boundary orientation on $Bbb S^3$, the chart $varphi^+$ is orientation-reversing. So you should get two positive terms for your integrals. (In your computation, in the $varphi^-$ integral, $-4tge 0$, and in the $varphi^+$ integral, you must change sign to compensate for the orientation reversal, and so $-(-4t)ge 0$ once again.)






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            $begingroup$

            The Stokes's Theorem application is correct. The mistake comes in half of the first part: With the boundary orientation on $Bbb S^3$, the chart $varphi^+$ is orientation-reversing. So you should get two positive terms for your integrals. (In your computation, in the $varphi^-$ integral, $-4tge 0$, and in the $varphi^+$ integral, you must change sign to compensate for the orientation reversal, and so $-(-4t)ge 0$ once again.)






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              The Stokes's Theorem application is correct. The mistake comes in half of the first part: With the boundary orientation on $Bbb S^3$, the chart $varphi^+$ is orientation-reversing. So you should get two positive terms for your integrals. (In your computation, in the $varphi^-$ integral, $-4tge 0$, and in the $varphi^+$ integral, you must change sign to compensate for the orientation reversal, and so $-(-4t)ge 0$ once again.)






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                The Stokes's Theorem application is correct. The mistake comes in half of the first part: With the boundary orientation on $Bbb S^3$, the chart $varphi^+$ is orientation-reversing. So you should get two positive terms for your integrals. (In your computation, in the $varphi^-$ integral, $-4tge 0$, and in the $varphi^+$ integral, you must change sign to compensate for the orientation reversal, and so $-(-4t)ge 0$ once again.)






                share|cite|improve this answer









                $endgroup$



                The Stokes's Theorem application is correct. The mistake comes in half of the first part: With the boundary orientation on $Bbb S^3$, the chart $varphi^+$ is orientation-reversing. So you should get two positive terms for your integrals. (In your computation, in the $varphi^-$ integral, $-4tge 0$, and in the $varphi^+$ integral, you must change sign to compensate for the orientation reversal, and so $-(-4t)ge 0$ once again.)







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 7 '18 at 18:21









                Ted ShifrinTed Shifrin

                63.1k44489




                63.1k44489






























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