Divergence of a parametrized vector field












2












$begingroup$


Here's the problem:
Use Gauss's divergence theorem to calculate the total flux through the solid,$V$, enclosed by the set $M={(x,y,z)in mathbb{R^3}: z=x^2+y^2, 0leq zleq 1}$ and the vector field $X:mathbb{R^3} to mathbb{R^{3times 1}}, (x,y,z)to begin{bmatrix} xz \ z \ -frac{z^2}{2}end{bmatrix}$.



My attempt:
$$Phi_X(M)=int_Vdiv(X)dV$$



A parametrization for M is:
$$psi :]-pi;pi[times]0;1[tomathbb{R^3} \ (theta,v)to (sqrt{v}costheta,sqrt{v}sin(theta),v)$$



The divergence of $X$ is (And it I think it is here where I got it wrong):
$$div(X)=frac{1}{sqrt{det(G(psi;(theta,v))}}trbigg(Jbig(sqrt{det(G(psi;(theta,v)))}Xcircpsi;(theta,v)big)bigg)$$



Where $G(psi;(theta,v))$ is the Gram matrix of $psi$ at $(theta,v)$ and $J$ denotes the Jacobian matrix.



But if this definition is correct then it is not clear how I can compute the trace of that Jacobian matrix because it is not a square matrix, and if the definion is incorrect how do I compute the divergent of $X$ when using the parametrization $psi$?






differential-geometry vector-fields divergence







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    2












    $begingroup$


    Here's the problem:
    Use Gauss's divergence theorem to calculate the total flux through the solid,$V$, enclosed by the set $M={(x,y,z)in mathbb{R^3}: z=x^2+y^2, 0leq zleq 1}$ and the vector field $X:mathbb{R^3} to mathbb{R^{3times 1}}, (x,y,z)to begin{bmatrix} xz \ z \ -frac{z^2}{2}end{bmatrix}$.



    My attempt:
    $$Phi_X(M)=int_Vdiv(X)dV$$



    A parametrization for M is:
    $$psi :]-pi;pi[times]0;1[tomathbb{R^3} \ (theta,v)to (sqrt{v}costheta,sqrt{v}sin(theta),v)$$



    The divergence of $X$ is (And it I think it is here where I got it wrong):
    $$div(X)=frac{1}{sqrt{det(G(psi;(theta,v))}}trbigg(Jbig(sqrt{det(G(psi;(theta,v)))}Xcircpsi;(theta,v)big)bigg)$$



    Where $G(psi;(theta,v))$ is the Gram matrix of $psi$ at $(theta,v)$ and $J$ denotes the Jacobian matrix.



    But if this definition is correct then it is not clear how I can compute the trace of that Jacobian matrix because it is not a square matrix, and if the definion is incorrect how do I compute the divergent of $X$ when using the parametrization $psi$?






    differential-geometry vector-fields divergence







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Here's the problem:
      Use Gauss's divergence theorem to calculate the total flux through the solid,$V$, enclosed by the set $M={(x,y,z)in mathbb{R^3}: z=x^2+y^2, 0leq zleq 1}$ and the vector field $X:mathbb{R^3} to mathbb{R^{3times 1}}, (x,y,z)to begin{bmatrix} xz \ z \ -frac{z^2}{2}end{bmatrix}$.



      My attempt:
      $$Phi_X(M)=int_Vdiv(X)dV$$



      A parametrization for M is:
      $$psi :]-pi;pi[times]0;1[tomathbb{R^3} \ (theta,v)to (sqrt{v}costheta,sqrt{v}sin(theta),v)$$



      The divergence of $X$ is (And it I think it is here where I got it wrong):
      $$div(X)=frac{1}{sqrt{det(G(psi;(theta,v))}}trbigg(Jbig(sqrt{det(G(psi;(theta,v)))}Xcircpsi;(theta,v)big)bigg)$$



      Where $G(psi;(theta,v))$ is the Gram matrix of $psi$ at $(theta,v)$ and $J$ denotes the Jacobian matrix.



      But if this definition is correct then it is not clear how I can compute the trace of that Jacobian matrix because it is not a square matrix, and if the definion is incorrect how do I compute the divergent of $X$ when using the parametrization $psi$?






      differential-geometry vector-fields divergence







      share|cite|improve this question











      $endgroup$




      Here's the problem:
      Use Gauss's divergence theorem to calculate the total flux through the solid,$V$, enclosed by the set $M={(x,y,z)in mathbb{R^3}: z=x^2+y^2, 0leq zleq 1}$ and the vector field $X:mathbb{R^3} to mathbb{R^{3times 1}}, (x,y,z)to begin{bmatrix} xz \ z \ -frac{z^2}{2}end{bmatrix}$.



      My attempt:
      $$Phi_X(M)=int_Vdiv(X)dV$$



      A parametrization for M is:
      $$psi :]-pi;pi[times]0;1[tomathbb{R^3} \ (theta,v)to (sqrt{v}costheta,sqrt{v}sin(theta),v)$$



      The divergence of $X$ is (And it I think it is here where I got it wrong):
      $$div(X)=frac{1}{sqrt{det(G(psi;(theta,v))}}trbigg(Jbig(sqrt{det(G(psi;(theta,v)))}Xcircpsi;(theta,v)big)bigg)$$



      Where $G(psi;(theta,v))$ is the Gram matrix of $psi$ at $(theta,v)$ and $J$ denotes the Jacobian matrix.



      But if this definition is correct then it is not clear how I can compute the trace of that Jacobian matrix because it is not a square matrix, and if the definion is incorrect how do I compute the divergent of $X$ when using the parametrization $psi$?






      differential-geometry vector-fields divergence




      differential-geometry vector-fields divergence






      share|cite|improve this question















      share|cite|improve this question













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      edited Dec 17 '18 at 23:18







      Bidon

















      asked Dec 17 '18 at 20:39









      BidonBidon

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          $begingroup$

          Calculate the divergence first



          $$
          nabla cdot F = frac{partial}{partial x}(xz) + frac{partial}{partial y}(z) - frac{partial}{partial z}left(frac{z^2}{2}right) = z - z = 0
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            And then I make a sort of representation of it in the parametrization $psi$? And is that definition wrong? and why?
            $endgroup$
            – Bidon
            Dec 17 '18 at 20:52










          • $begingroup$
            @Bidon Yes, calculate the divergence and then evaluate the parametrization. The problem I see is that $X: mathbb{R}^color{red}{3} to mathbb{R}^color{red}{3}$, that's what is giving you the trouble with the dimensions
            $endgroup$
            – caverac
            Dec 17 '18 at 20:57










          • $begingroup$
            How come? I think it's quite common to that kind of vector fields, gravitational and electric for example...
            $endgroup$
            – Bidon
            Dec 17 '18 at 23:08










          • $begingroup$
            @Bidon But you say $X(x, y) = (xz, z, -z^2/2)$, what is $z$ then?
            $endgroup$
            – caverac
            Dec 17 '18 at 23:14






          • 1




            $begingroup$
            @Bidon If you do not know how the field behaves inside the region, there's not much of a point in calculating the divergence (you need the divergence inside the region to integrate it)
            $endgroup$
            – caverac
            Dec 18 '18 at 18:37











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          $begingroup$

          Calculate the divergence first



          $$
          nabla cdot F = frac{partial}{partial x}(xz) + frac{partial}{partial y}(z) - frac{partial}{partial z}left(frac{z^2}{2}right) = z - z = 0
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            And then I make a sort of representation of it in the parametrization $psi$? And is that definition wrong? and why?
            $endgroup$
            – Bidon
            Dec 17 '18 at 20:52










          • $begingroup$
            @Bidon Yes, calculate the divergence and then evaluate the parametrization. The problem I see is that $X: mathbb{R}^color{red}{3} to mathbb{R}^color{red}{3}$, that's what is giving you the trouble with the dimensions
            $endgroup$
            – caverac
            Dec 17 '18 at 20:57










          • $begingroup$
            How come? I think it's quite common to that kind of vector fields, gravitational and electric for example...
            $endgroup$
            – Bidon
            Dec 17 '18 at 23:08










          • $begingroup$
            @Bidon But you say $X(x, y) = (xz, z, -z^2/2)$, what is $z$ then?
            $endgroup$
            – caverac
            Dec 17 '18 at 23:14






          • 1




            $begingroup$
            @Bidon If you do not know how the field behaves inside the region, there's not much of a point in calculating the divergence (you need the divergence inside the region to integrate it)
            $endgroup$
            – caverac
            Dec 18 '18 at 18:37
















          1












          $begingroup$

          Calculate the divergence first



          $$
          nabla cdot F = frac{partial}{partial x}(xz) + frac{partial}{partial y}(z) - frac{partial}{partial z}left(frac{z^2}{2}right) = z - z = 0
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            And then I make a sort of representation of it in the parametrization $psi$? And is that definition wrong? and why?
            $endgroup$
            – Bidon
            Dec 17 '18 at 20:52










          • $begingroup$
            @Bidon Yes, calculate the divergence and then evaluate the parametrization. The problem I see is that $X: mathbb{R}^color{red}{3} to mathbb{R}^color{red}{3}$, that's what is giving you the trouble with the dimensions
            $endgroup$
            – caverac
            Dec 17 '18 at 20:57










          • $begingroup$
            How come? I think it's quite common to that kind of vector fields, gravitational and electric for example...
            $endgroup$
            – Bidon
            Dec 17 '18 at 23:08










          • $begingroup$
            @Bidon But you say $X(x, y) = (xz, z, -z^2/2)$, what is $z$ then?
            $endgroup$
            – caverac
            Dec 17 '18 at 23:14






          • 1




            $begingroup$
            @Bidon If you do not know how the field behaves inside the region, there's not much of a point in calculating the divergence (you need the divergence inside the region to integrate it)
            $endgroup$
            – caverac
            Dec 18 '18 at 18:37














          1












          1








          1





          $begingroup$

          Calculate the divergence first



          $$
          nabla cdot F = frac{partial}{partial x}(xz) + frac{partial}{partial y}(z) - frac{partial}{partial z}left(frac{z^2}{2}right) = z - z = 0
          $$






          share|cite|improve this answer









          $endgroup$



          Calculate the divergence first



          $$
          nabla cdot F = frac{partial}{partial x}(xz) + frac{partial}{partial y}(z) - frac{partial}{partial z}left(frac{z^2}{2}right) = z - z = 0
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 17 '18 at 20:49









          caveraccaverac

          14.6k31130




          14.6k31130












          • $begingroup$
            And then I make a sort of representation of it in the parametrization $psi$? And is that definition wrong? and why?
            $endgroup$
            – Bidon
            Dec 17 '18 at 20:52










          • $begingroup$
            @Bidon Yes, calculate the divergence and then evaluate the parametrization. The problem I see is that $X: mathbb{R}^color{red}{3} to mathbb{R}^color{red}{3}$, that's what is giving you the trouble with the dimensions
            $endgroup$
            – caverac
            Dec 17 '18 at 20:57










          • $begingroup$
            How come? I think it's quite common to that kind of vector fields, gravitational and electric for example...
            $endgroup$
            – Bidon
            Dec 17 '18 at 23:08










          • $begingroup$
            @Bidon But you say $X(x, y) = (xz, z, -z^2/2)$, what is $z$ then?
            $endgroup$
            – caverac
            Dec 17 '18 at 23:14






          • 1




            $begingroup$
            @Bidon If you do not know how the field behaves inside the region, there's not much of a point in calculating the divergence (you need the divergence inside the region to integrate it)
            $endgroup$
            – caverac
            Dec 18 '18 at 18:37


















          • $begingroup$
            And then I make a sort of representation of it in the parametrization $psi$? And is that definition wrong? and why?
            $endgroup$
            – Bidon
            Dec 17 '18 at 20:52










          • $begingroup$
            @Bidon Yes, calculate the divergence and then evaluate the parametrization. The problem I see is that $X: mathbb{R}^color{red}{3} to mathbb{R}^color{red}{3}$, that's what is giving you the trouble with the dimensions
            $endgroup$
            – caverac
            Dec 17 '18 at 20:57










          • $begingroup$
            How come? I think it's quite common to that kind of vector fields, gravitational and electric for example...
            $endgroup$
            – Bidon
            Dec 17 '18 at 23:08










          • $begingroup$
            @Bidon But you say $X(x, y) = (xz, z, -z^2/2)$, what is $z$ then?
            $endgroup$
            – caverac
            Dec 17 '18 at 23:14






          • 1




            $begingroup$
            @Bidon If you do not know how the field behaves inside the region, there's not much of a point in calculating the divergence (you need the divergence inside the region to integrate it)
            $endgroup$
            – caverac
            Dec 18 '18 at 18:37
















          $begingroup$
          And then I make a sort of representation of it in the parametrization $psi$? And is that definition wrong? and why?
          $endgroup$
          – Bidon
          Dec 17 '18 at 20:52




          $begingroup$
          And then I make a sort of representation of it in the parametrization $psi$? And is that definition wrong? and why?
          $endgroup$
          – Bidon
          Dec 17 '18 at 20:52












          $begingroup$
          @Bidon Yes, calculate the divergence and then evaluate the parametrization. The problem I see is that $X: mathbb{R}^color{red}{3} to mathbb{R}^color{red}{3}$, that's what is giving you the trouble with the dimensions
          $endgroup$
          – caverac
          Dec 17 '18 at 20:57




          $begingroup$
          @Bidon Yes, calculate the divergence and then evaluate the parametrization. The problem I see is that $X: mathbb{R}^color{red}{3} to mathbb{R}^color{red}{3}$, that's what is giving you the trouble with the dimensions
          $endgroup$
          – caverac
          Dec 17 '18 at 20:57












          $begingroup$
          How come? I think it's quite common to that kind of vector fields, gravitational and electric for example...
          $endgroup$
          – Bidon
          Dec 17 '18 at 23:08




          $begingroup$
          How come? I think it's quite common to that kind of vector fields, gravitational and electric for example...
          $endgroup$
          – Bidon
          Dec 17 '18 at 23:08












          $begingroup$
          @Bidon But you say $X(x, y) = (xz, z, -z^2/2)$, what is $z$ then?
          $endgroup$
          – caverac
          Dec 17 '18 at 23:14




          $begingroup$
          @Bidon But you say $X(x, y) = (xz, z, -z^2/2)$, what is $z$ then?
          $endgroup$
          – caverac
          Dec 17 '18 at 23:14




          1




          1




          $begingroup$
          @Bidon If you do not know how the field behaves inside the region, there's not much of a point in calculating the divergence (you need the divergence inside the region to integrate it)
          $endgroup$
          – caverac
          Dec 18 '18 at 18:37




          $begingroup$
          @Bidon If you do not know how the field behaves inside the region, there's not much of a point in calculating the divergence (you need the divergence inside the region to integrate it)
          $endgroup$
          – caverac
          Dec 18 '18 at 18:37


















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