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







share|cite|improve this question











$endgroup$

















    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













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 17 '18 at 23:18







      Bidon

















      asked Dec 17 '18 at 20:39









      BidonBidon

      967




      967






















          1 Answer
          1






          active

          oldest

          votes


















          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











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044419%2fdivergence-of-a-parametrized-vector-field%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          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












          $begingroup$

          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

          14.8k31130




          14.8k31130












          • $begingroup$
            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


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044419%2fdivergence-of-a-parametrized-vector-field%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Le Mesnil-Réaume

          Image
          Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
          Read more

          Ida-Boy-Ed-Garten

          Image
          Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
          Read more

          web3.py web3.isConnected() returns false always

          Image
          1 I am trying an tutorial on web3.py so i created a profile in infura and used its project id.But it shows false always when i try web3.isConnected() PS C:UsersHP> python Python 3.6.3 (v3.6.3:2c5fed8, Oct 3 2017, 18:11:49) [MSC v.1900 64 bit (AMD64)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> from web3 import Web3 >>> infura_url = "mainnet.infura.io/v3/******************************" >>> web3 = Web3(Web3.HTTPProvider(infura_url)) >>> web3.isConnected() False web3.py share | improve this question
          Read more