About a surface integral of type $1$












1












$begingroup$



Compute $displaystyle F(t)=iintlimits_{s}f(x,y,z) ds$, where $s$ is a surface $x+y+z=t$, and $f(x,y,z)=1-x^2-y^2-z^2$ when $1-x^2-y^2-z^2 ge 0$,
else $f=0$.




Here is my attempt:
I think I can rotate the space rectangular coordinate system, let the $x+y+z=t$ be the new $xy$-plane, may be easy to compute, but I don’t know how to determine the new coordinates when changing the variables $x$, $y$ and $z$ to some new $u$, $v$ and $w$.






calculus integration multivariable-calculus







share|cite|improve this question











$endgroup$

















    1












    $begingroup$



    Compute $displaystyle F(t)=iintlimits_{s}f(x,y,z) ds$, where $s$ is a surface $x+y+z=t$, and $f(x,y,z)=1-x^2-y^2-z^2$ when $1-x^2-y^2-z^2 ge 0$,
    else $f=0$.




    Here is my attempt:
    I think I can rotate the space rectangular coordinate system, let the $x+y+z=t$ be the new $xy$-plane, may be easy to compute, but I don’t know how to determine the new coordinates when changing the variables $x$, $y$ and $z$ to some new $u$, $v$ and $w$.






    calculus integration multivariable-calculus







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$



      Compute $displaystyle F(t)=iintlimits_{s}f(x,y,z) ds$, where $s$ is a surface $x+y+z=t$, and $f(x,y,z)=1-x^2-y^2-z^2$ when $1-x^2-y^2-z^2 ge 0$,
      else $f=0$.




      Here is my attempt:
      I think I can rotate the space rectangular coordinate system, let the $x+y+z=t$ be the new $xy$-plane, may be easy to compute, but I don’t know how to determine the new coordinates when changing the variables $x$, $y$ and $z$ to some new $u$, $v$ and $w$.






      calculus integration multivariable-calculus







      share|cite|improve this question











      $endgroup$





      Compute $displaystyle F(t)=iintlimits_{s}f(x,y,z) ds$, where $s$ is a surface $x+y+z=t$, and $f(x,y,z)=1-x^2-y^2-z^2$ when $1-x^2-y^2-z^2 ge 0$,
      else $f=0$.




      Here is my attempt:
      I think I can rotate the space rectangular coordinate system, let the $x+y+z=t$ be the new $xy$-plane, may be easy to compute, but I don’t know how to determine the new coordinates when changing the variables $x$, $y$ and $z$ to some new $u$, $v$ and $w$.






      calculus integration multivariable-calculus




      calculus integration multivariable-calculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 26 '18 at 3:48









      Rócherz

      2,7762721




      2,7762721










      asked Dec 26 '18 at 2:50









      jacksonjackson

      1149




      1149






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          The integral simplifies with the coordinates you propose. It simplifies even more because the form of the integrand does not change in the new coordinates and, further! the integral in spherical coordinates is very simple.



          Following your idea, the vector normal and unitary to the plane is $n_w=dfrac{1}{sqrt{3}}(1,1,1)$ we can choose two other vectors being orthogonal to $n_w$ and then use them to determine planes to build the new coordinates. $n_u=dfrac{1}{sqrt{6}}(2,-1,-1)$ and $n_v=dfrac{1}{sqrt{2}}(0,1,-1)$ will do the job. The change of coordinates is then



          $$
          begin{matrix}
          u=dfrac{1}{sqrt{6}}(2x-y-z) \
          v=dfrac{1}{sqrt{2}}(y-z) \
          w=dfrac{1}{sqrt{3}}(x+y+z) \
          end{matrix}
          $$



          As the vectors are unitary and orthogonal, the jacobian is an unitary matrix. The surface of integration is into the plane $x+y+z=sqrt{3}w=t$ and the surface element is $mathbb ds=mathbb dumathbb dv$. For the integrand, it's easy to check that $1-x^2-y^2-z^2=1-u^2-v^2-w^2$. So,



          $$iint_S(1-u^2-v^2-t^2/3)mathbb dumathbb dv$$



          Added



          Considering the integrand and the surface symmetry, it get simpler changing to cylindrical coordinates, with $u>0$ as polar axis. The surface is a circle into the plane with $w=t/sqrt{3}$ (it's the intersection of the sphere of radius $1$ with this plane). The radius of this circle is $rho(t)=sqrt{1-w(t)^2}=sqrt{1-t^2/3}$. The surface element is $rho,mathbb drho,mathbb dtheta$ and $rho^2=u^2+v^2$:



          $$I(t)=iint_S(1-u^2-v^2-t^2/3)mathbb du,mathbb dv=$$



          $$=int_0^{2pi}int_0^{sqrt{1-t^2/3}}(1-rho^2-t^2/3)rho,mathbb drho,mathbb dtheta=$$



          $$=2pi(1-t^2/3)dfrac{left(sqrt{1-t^2/3}right)^2}{2}-2pidfrac{left(sqrt{1-t^2/3}right)^4}{4}=$$



          $$=left{begin{matrix}
          0&text{if}&t<-sqrt{3}\
          2pi(1-t^2/3)^2/4&text{if}&-sqrt{3}leq tleqsqrt{3}\
          0&text{if}&sqrt{3}<t\
          end{matrix}right.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I think the surface should change to $frac{w}{3^{1/3}}$ then I compute it it’s a wired number can can you compute it
            $endgroup$
            – jackson
            Dec 26 '18 at 10:36












          • $begingroup$
            You are right, I forgot I normalized it, missing that constant. Do you mean you need help with the integral
            $endgroup$
            – Rafa Budría
            Dec 26 '18 at 10:43












          • $begingroup$
            yes thanks a lot
            $endgroup$
            – jackson
            Dec 27 '18 at 16:29











          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%2f3052596%2fabout-a-surface-integral-of-type-1%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












          $begingroup$

          The integral simplifies with the coordinates you propose. It simplifies even more because the form of the integrand does not change in the new coordinates and, further! the integral in spherical coordinates is very simple.



          Following your idea, the vector normal and unitary to the plane is $n_w=dfrac{1}{sqrt{3}}(1,1,1)$ we can choose two other vectors being orthogonal to $n_w$ and then use them to determine planes to build the new coordinates. $n_u=dfrac{1}{sqrt{6}}(2,-1,-1)$ and $n_v=dfrac{1}{sqrt{2}}(0,1,-1)$ will do the job. The change of coordinates is then



          $$
          begin{matrix}
          u=dfrac{1}{sqrt{6}}(2x-y-z) \
          v=dfrac{1}{sqrt{2}}(y-z) \
          w=dfrac{1}{sqrt{3}}(x+y+z) \
          end{matrix}
          $$



          As the vectors are unitary and orthogonal, the jacobian is an unitary matrix. The surface of integration is into the plane $x+y+z=sqrt{3}w=t$ and the surface element is $mathbb ds=mathbb dumathbb dv$. For the integrand, it's easy to check that $1-x^2-y^2-z^2=1-u^2-v^2-w^2$. So,



          $$iint_S(1-u^2-v^2-t^2/3)mathbb dumathbb dv$$



          Added



          Considering the integrand and the surface symmetry, it get simpler changing to cylindrical coordinates, with $u>0$ as polar axis. The surface is a circle into the plane with $w=t/sqrt{3}$ (it's the intersection of the sphere of radius $1$ with this plane). The radius of this circle is $rho(t)=sqrt{1-w(t)^2}=sqrt{1-t^2/3}$. The surface element is $rho,mathbb drho,mathbb dtheta$ and $rho^2=u^2+v^2$:



          $$I(t)=iint_S(1-u^2-v^2-t^2/3)mathbb du,mathbb dv=$$



          $$=int_0^{2pi}int_0^{sqrt{1-t^2/3}}(1-rho^2-t^2/3)rho,mathbb drho,mathbb dtheta=$$



          $$=2pi(1-t^2/3)dfrac{left(sqrt{1-t^2/3}right)^2}{2}-2pidfrac{left(sqrt{1-t^2/3}right)^4}{4}=$$



          $$=left{begin{matrix}
          0&text{if}&t<-sqrt{3}\
          2pi(1-t^2/3)^2/4&text{if}&-sqrt{3}leq tleqsqrt{3}\
          0&text{if}&sqrt{3}<t\
          end{matrix}right.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I think the surface should change to $frac{w}{3^{1/3}}$ then I compute it it’s a wired number can can you compute it
            $endgroup$
            – jackson
            Dec 26 '18 at 10:36












          • $begingroup$
            You are right, I forgot I normalized it, missing that constant. Do you mean you need help with the integral
            $endgroup$
            – Rafa Budría
            Dec 26 '18 at 10:43












          • $begingroup$
            yes thanks a lot
            $endgroup$
            – jackson
            Dec 27 '18 at 16:29
















          1












          $begingroup$

          The integral simplifies with the coordinates you propose. It simplifies even more because the form of the integrand does not change in the new coordinates and, further! the integral in spherical coordinates is very simple.



          Following your idea, the vector normal and unitary to the plane is $n_w=dfrac{1}{sqrt{3}}(1,1,1)$ we can choose two other vectors being orthogonal to $n_w$ and then use them to determine planes to build the new coordinates. $n_u=dfrac{1}{sqrt{6}}(2,-1,-1)$ and $n_v=dfrac{1}{sqrt{2}}(0,1,-1)$ will do the job. The change of coordinates is then



          $$
          begin{matrix}
          u=dfrac{1}{sqrt{6}}(2x-y-z) \
          v=dfrac{1}{sqrt{2}}(y-z) \
          w=dfrac{1}{sqrt{3}}(x+y+z) \
          end{matrix}
          $$



          As the vectors are unitary and orthogonal, the jacobian is an unitary matrix. The surface of integration is into the plane $x+y+z=sqrt{3}w=t$ and the surface element is $mathbb ds=mathbb dumathbb dv$. For the integrand, it's easy to check that $1-x^2-y^2-z^2=1-u^2-v^2-w^2$. So,



          $$iint_S(1-u^2-v^2-t^2/3)mathbb dumathbb dv$$



          Added



          Considering the integrand and the surface symmetry, it get simpler changing to cylindrical coordinates, with $u>0$ as polar axis. The surface is a circle into the plane with $w=t/sqrt{3}$ (it's the intersection of the sphere of radius $1$ with this plane). The radius of this circle is $rho(t)=sqrt{1-w(t)^2}=sqrt{1-t^2/3}$. The surface element is $rho,mathbb drho,mathbb dtheta$ and $rho^2=u^2+v^2$:



          $$I(t)=iint_S(1-u^2-v^2-t^2/3)mathbb du,mathbb dv=$$



          $$=int_0^{2pi}int_0^{sqrt{1-t^2/3}}(1-rho^2-t^2/3)rho,mathbb drho,mathbb dtheta=$$



          $$=2pi(1-t^2/3)dfrac{left(sqrt{1-t^2/3}right)^2}{2}-2pidfrac{left(sqrt{1-t^2/3}right)^4}{4}=$$



          $$=left{begin{matrix}
          0&text{if}&t<-sqrt{3}\
          2pi(1-t^2/3)^2/4&text{if}&-sqrt{3}leq tleqsqrt{3}\
          0&text{if}&sqrt{3}<t\
          end{matrix}right.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I think the surface should change to $frac{w}{3^{1/3}}$ then I compute it it’s a wired number can can you compute it
            $endgroup$
            – jackson
            Dec 26 '18 at 10:36












          • $begingroup$
            You are right, I forgot I normalized it, missing that constant. Do you mean you need help with the integral
            $endgroup$
            – Rafa Budría
            Dec 26 '18 at 10:43












          • $begingroup$
            yes thanks a lot
            $endgroup$
            – jackson
            Dec 27 '18 at 16:29














          1












          1








          1





          $begingroup$

          The integral simplifies with the coordinates you propose. It simplifies even more because the form of the integrand does not change in the new coordinates and, further! the integral in spherical coordinates is very simple.



          Following your idea, the vector normal and unitary to the plane is $n_w=dfrac{1}{sqrt{3}}(1,1,1)$ we can choose two other vectors being orthogonal to $n_w$ and then use them to determine planes to build the new coordinates. $n_u=dfrac{1}{sqrt{6}}(2,-1,-1)$ and $n_v=dfrac{1}{sqrt{2}}(0,1,-1)$ will do the job. The change of coordinates is then



          $$
          begin{matrix}
          u=dfrac{1}{sqrt{6}}(2x-y-z) \
          v=dfrac{1}{sqrt{2}}(y-z) \
          w=dfrac{1}{sqrt{3}}(x+y+z) \
          end{matrix}
          $$



          As the vectors are unitary and orthogonal, the jacobian is an unitary matrix. The surface of integration is into the plane $x+y+z=sqrt{3}w=t$ and the surface element is $mathbb ds=mathbb dumathbb dv$. For the integrand, it's easy to check that $1-x^2-y^2-z^2=1-u^2-v^2-w^2$. So,



          $$iint_S(1-u^2-v^2-t^2/3)mathbb dumathbb dv$$



          Added



          Considering the integrand and the surface symmetry, it get simpler changing to cylindrical coordinates, with $u>0$ as polar axis. The surface is a circle into the plane with $w=t/sqrt{3}$ (it's the intersection of the sphere of radius $1$ with this plane). The radius of this circle is $rho(t)=sqrt{1-w(t)^2}=sqrt{1-t^2/3}$. The surface element is $rho,mathbb drho,mathbb dtheta$ and $rho^2=u^2+v^2$:



          $$I(t)=iint_S(1-u^2-v^2-t^2/3)mathbb du,mathbb dv=$$



          $$=int_0^{2pi}int_0^{sqrt{1-t^2/3}}(1-rho^2-t^2/3)rho,mathbb drho,mathbb dtheta=$$



          $$=2pi(1-t^2/3)dfrac{left(sqrt{1-t^2/3}right)^2}{2}-2pidfrac{left(sqrt{1-t^2/3}right)^4}{4}=$$



          $$=left{begin{matrix}
          0&text{if}&t<-sqrt{3}\
          2pi(1-t^2/3)^2/4&text{if}&-sqrt{3}leq tleqsqrt{3}\
          0&text{if}&sqrt{3}<t\
          end{matrix}right.$$






          share|cite|improve this answer











          $endgroup$



          The integral simplifies with the coordinates you propose. It simplifies even more because the form of the integrand does not change in the new coordinates and, further! the integral in spherical coordinates is very simple.



          Following your idea, the vector normal and unitary to the plane is $n_w=dfrac{1}{sqrt{3}}(1,1,1)$ we can choose two other vectors being orthogonal to $n_w$ and then use them to determine planes to build the new coordinates. $n_u=dfrac{1}{sqrt{6}}(2,-1,-1)$ and $n_v=dfrac{1}{sqrt{2}}(0,1,-1)$ will do the job. The change of coordinates is then



          $$
          begin{matrix}
          u=dfrac{1}{sqrt{6}}(2x-y-z) \
          v=dfrac{1}{sqrt{2}}(y-z) \
          w=dfrac{1}{sqrt{3}}(x+y+z) \
          end{matrix}
          $$



          As the vectors are unitary and orthogonal, the jacobian is an unitary matrix. The surface of integration is into the plane $x+y+z=sqrt{3}w=t$ and the surface element is $mathbb ds=mathbb dumathbb dv$. For the integrand, it's easy to check that $1-x^2-y^2-z^2=1-u^2-v^2-w^2$. So,



          $$iint_S(1-u^2-v^2-t^2/3)mathbb dumathbb dv$$



          Added



          Considering the integrand and the surface symmetry, it get simpler changing to cylindrical coordinates, with $u>0$ as polar axis. The surface is a circle into the plane with $w=t/sqrt{3}$ (it's the intersection of the sphere of radius $1$ with this plane). The radius of this circle is $rho(t)=sqrt{1-w(t)^2}=sqrt{1-t^2/3}$. The surface element is $rho,mathbb drho,mathbb dtheta$ and $rho^2=u^2+v^2$:



          $$I(t)=iint_S(1-u^2-v^2-t^2/3)mathbb du,mathbb dv=$$



          $$=int_0^{2pi}int_0^{sqrt{1-t^2/3}}(1-rho^2-t^2/3)rho,mathbb drho,mathbb dtheta=$$



          $$=2pi(1-t^2/3)dfrac{left(sqrt{1-t^2/3}right)^2}{2}-2pidfrac{left(sqrt{1-t^2/3}right)^4}{4}=$$



          $$=left{begin{matrix}
          0&text{if}&t<-sqrt{3}\
          2pi(1-t^2/3)^2/4&text{if}&-sqrt{3}leq tleqsqrt{3}\
          0&text{if}&sqrt{3}<t\
          end{matrix}right.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 27 '18 at 21:51

























          answered Dec 26 '18 at 9:06









          Rafa BudríaRafa Budría

          5,7151825




          5,7151825












          • $begingroup$
            I think the surface should change to $frac{w}{3^{1/3}}$ then I compute it it’s a wired number can can you compute it
            $endgroup$
            – jackson
            Dec 26 '18 at 10:36












          • $begingroup$
            You are right, I forgot I normalized it, missing that constant. Do you mean you need help with the integral
            $endgroup$
            – Rafa Budría
            Dec 26 '18 at 10:43












          • $begingroup$
            yes thanks a lot
            $endgroup$
            – jackson
            Dec 27 '18 at 16:29


















          • $begingroup$
            I think the surface should change to $frac{w}{3^{1/3}}$ then I compute it it’s a wired number can can you compute it
            $endgroup$
            – jackson
            Dec 26 '18 at 10:36












          • $begingroup$
            You are right, I forgot I normalized it, missing that constant. Do you mean you need help with the integral
            $endgroup$
            – Rafa Budría
            Dec 26 '18 at 10:43












          • $begingroup$
            yes thanks a lot
            $endgroup$
            – jackson
            Dec 27 '18 at 16:29
















          $begingroup$
          I think the surface should change to $frac{w}{3^{1/3}}$ then I compute it it’s a wired number can can you compute it
          $endgroup$
          – jackson
          Dec 26 '18 at 10:36






          $begingroup$
          I think the surface should change to $frac{w}{3^{1/3}}$ then I compute it it’s a wired number can can you compute it
          $endgroup$
          – jackson
          Dec 26 '18 at 10:36














          $begingroup$
          You are right, I forgot I normalized it, missing that constant. Do you mean you need help with the integral
          $endgroup$
          – Rafa Budría
          Dec 26 '18 at 10:43






          $begingroup$
          You are right, I forgot I normalized it, missing that constant. Do you mean you need help with the integral
          $endgroup$
          – Rafa Budría
          Dec 26 '18 at 10:43














          $begingroup$
          yes thanks a lot
          $endgroup$
          – jackson
          Dec 27 '18 at 16:29




          $begingroup$
          yes thanks a lot
          $endgroup$
          – jackson
          Dec 27 '18 at 16:29


















          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%2f3052596%2fabout-a-surface-integral-of-type-1%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

          Bressuire

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
          Read more

          Cabo Verde

          Image
            Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
          Read more

          Gyllenstierna

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
          Read more