How to prove that $ | Phi_u times Phi_v | = sqrt{ det g} $?












2














Let be $ Phi $ a parametrization of a surface $ in mathbb{R}^3 $
and $g$ the metric tensor.



As the title says...how to show that How to prove that $ | Phi_u times Phi_v | = sqrt{det g} $ ?



I started like this:



$$| Phi_u times Phi_v | = left| begin{pmatrix} frac{delta phi_1}{ delta u} \ frac{delta phi_2}{ delta u} \ frac{delta phi_2}{ delta u} end{pmatrix} times begin{pmatrix} frac{delta phi_1}{ delta v} \ frac{delta phi_2}{ delta v} \ frac{delta phi_2}{ delta v} end{pmatrix} right| = left| begin{pmatrix} frac{delta phi_2}{ delta u} frac{delta phi_3}{ delta v}- frac{delta phi_3}{ delta u} frac{delta phi_2}{ delta v}\ frac{delta phi_1}{ delta u} frac{delta phi_3}{ delta v}- frac{delta phi_3}{ delta u} frac{delta phi_1}{ delta v}\ frac{delta phi_1}{ delta u} frac{delta phi_2}{ delta v}- frac{delta phi_2}{ delta u} frac{delta phi_1}{ delta v}end{pmatrix} right|$$



I am not sure how to proceed..If I continue I am coming to a dead end..hope you can help me out a bit :) !






integration multivariable-calculus parametrization







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    2














    Let be $ Phi $ a parametrization of a surface $ in mathbb{R}^3 $
    and $g$ the metric tensor.



    As the title says...how to show that How to prove that $ | Phi_u times Phi_v | = sqrt{det g} $ ?



    I started like this:



    $$| Phi_u times Phi_v | = left| begin{pmatrix} frac{delta phi_1}{ delta u} \ frac{delta phi_2}{ delta u} \ frac{delta phi_2}{ delta u} end{pmatrix} times begin{pmatrix} frac{delta phi_1}{ delta v} \ frac{delta phi_2}{ delta v} \ frac{delta phi_2}{ delta v} end{pmatrix} right| = left| begin{pmatrix} frac{delta phi_2}{ delta u} frac{delta phi_3}{ delta v}- frac{delta phi_3}{ delta u} frac{delta phi_2}{ delta v}\ frac{delta phi_1}{ delta u} frac{delta phi_3}{ delta v}- frac{delta phi_3}{ delta u} frac{delta phi_1}{ delta v}\ frac{delta phi_1}{ delta u} frac{delta phi_2}{ delta v}- frac{delta phi_2}{ delta u} frac{delta phi_1}{ delta v}end{pmatrix} right|$$



    I am not sure how to proceed..If I continue I am coming to a dead end..hope you can help me out a bit :) !






    integration multivariable-calculus parametrization







    share|cite|improve this question



























      2












      2








      2







      Let be $ Phi $ a parametrization of a surface $ in mathbb{R}^3 $
      and $g$ the metric tensor.



      As the title says...how to show that How to prove that $ | Phi_u times Phi_v | = sqrt{det g} $ ?



      I started like this:



      $$| Phi_u times Phi_v | = left| begin{pmatrix} frac{delta phi_1}{ delta u} \ frac{delta phi_2}{ delta u} \ frac{delta phi_2}{ delta u} end{pmatrix} times begin{pmatrix} frac{delta phi_1}{ delta v} \ frac{delta phi_2}{ delta v} \ frac{delta phi_2}{ delta v} end{pmatrix} right| = left| begin{pmatrix} frac{delta phi_2}{ delta u} frac{delta phi_3}{ delta v}- frac{delta phi_3}{ delta u} frac{delta phi_2}{ delta v}\ frac{delta phi_1}{ delta u} frac{delta phi_3}{ delta v}- frac{delta phi_3}{ delta u} frac{delta phi_1}{ delta v}\ frac{delta phi_1}{ delta u} frac{delta phi_2}{ delta v}- frac{delta phi_2}{ delta u} frac{delta phi_1}{ delta v}end{pmatrix} right|$$



      I am not sure how to proceed..If I continue I am coming to a dead end..hope you can help me out a bit :) !






      integration multivariable-calculus parametrization







      share|cite|improve this question















      Let be $ Phi $ a parametrization of a surface $ in mathbb{R}^3 $
      and $g$ the metric tensor.



      As the title says...how to show that How to prove that $ | Phi_u times Phi_v | = sqrt{det g} $ ?



      I started like this:



      $$| Phi_u times Phi_v | = left| begin{pmatrix} frac{delta phi_1}{ delta u} \ frac{delta phi_2}{ delta u} \ frac{delta phi_2}{ delta u} end{pmatrix} times begin{pmatrix} frac{delta phi_1}{ delta v} \ frac{delta phi_2}{ delta v} \ frac{delta phi_2}{ delta v} end{pmatrix} right| = left| begin{pmatrix} frac{delta phi_2}{ delta u} frac{delta phi_3}{ delta v}- frac{delta phi_3}{ delta u} frac{delta phi_2}{ delta v}\ frac{delta phi_1}{ delta u} frac{delta phi_3}{ delta v}- frac{delta phi_3}{ delta u} frac{delta phi_1}{ delta v}\ frac{delta phi_1}{ delta u} frac{delta phi_2}{ delta v}- frac{delta phi_2}{ delta u} frac{delta phi_1}{ delta v}end{pmatrix} right|$$



      I am not sure how to proceed..If I continue I am coming to a dead end..hope you can help me out a bit :) !






      integration multivariable-calculus parametrization




      integration multivariable-calculus parametrization






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      share|cite|improve this question













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      edited Dec 12 '18 at 22:02









      mechanodroid

      27k62446




      27k62446










      asked Dec 12 '18 at 21:39









      constant94constant94

      1128




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          2 Answers
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          2














          Recall Lagrange's identity for cross product:
          $$|a times b|^2 = |a|^2|b|^2 - langle a,brangle^2, qquad a,b in mathbb{R}^3$$



          Therefore
          $$|Phi_u times Phi_v|^2 = |Phi_u|^2|Phi_v|^2 - langle Phi_u, Phi_vrangle^2 = begin{vmatrix} |Phi_u|^2 & langle Phi_u, Phi_vrangle \ langle Phi_u, Phi_vrangle & |Phi_v|^2end{vmatrix} = det g$$



          where $g = begin{bmatrix} langle Phi_u, Phi_urangle & langle Phi_u, Phi_vrangle \ langle Phi_u, Phi_vrangle & langlePhi_v, Phi_vrangleend{bmatrix}$ is the metric tensor.



          Since $det g > 0$ we conclude $|Phi_u times Phi_v| = sqrt{det g}$.






          share|cite|improve this answer





























            0














            Call ${bf r} = Phi(u, v)$, and recall the unitary vector along the direction $u$ is



            $$
            h_u hat e_u = frac{partial {bf r}}{partial u} = Phi_u
            $$



            with $h_u$ the scale factor. You have then



            $$
            |Phi_u times Phi_v| = |h_u h_v| |hat{e}_utimes hat{e}_v| = |h_u h_v| = sqrt{g}
            $$






            share|cite|improve this answer





















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              2 Answers
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              active

              oldest

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              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2














              Recall Lagrange's identity for cross product:
              $$|a times b|^2 = |a|^2|b|^2 - langle a,brangle^2, qquad a,b in mathbb{R}^3$$



              Therefore
              $$|Phi_u times Phi_v|^2 = |Phi_u|^2|Phi_v|^2 - langle Phi_u, Phi_vrangle^2 = begin{vmatrix} |Phi_u|^2 & langle Phi_u, Phi_vrangle \ langle Phi_u, Phi_vrangle & |Phi_v|^2end{vmatrix} = det g$$



              where $g = begin{bmatrix} langle Phi_u, Phi_urangle & langle Phi_u, Phi_vrangle \ langle Phi_u, Phi_vrangle & langlePhi_v, Phi_vrangleend{bmatrix}$ is the metric tensor.



              Since $det g > 0$ we conclude $|Phi_u times Phi_v| = sqrt{det g}$.






              share|cite|improve this answer


























                2














                Recall Lagrange's identity for cross product:
                $$|a times b|^2 = |a|^2|b|^2 - langle a,brangle^2, qquad a,b in mathbb{R}^3$$



                Therefore
                $$|Phi_u times Phi_v|^2 = |Phi_u|^2|Phi_v|^2 - langle Phi_u, Phi_vrangle^2 = begin{vmatrix} |Phi_u|^2 & langle Phi_u, Phi_vrangle \ langle Phi_u, Phi_vrangle & |Phi_v|^2end{vmatrix} = det g$$



                where $g = begin{bmatrix} langle Phi_u, Phi_urangle & langle Phi_u, Phi_vrangle \ langle Phi_u, Phi_vrangle & langlePhi_v, Phi_vrangleend{bmatrix}$ is the metric tensor.



                Since $det g > 0$ we conclude $|Phi_u times Phi_v| = sqrt{det g}$.






                share|cite|improve this answer
























                  2












                  2








                  2






                  Recall Lagrange's identity for cross product:
                  $$|a times b|^2 = |a|^2|b|^2 - langle a,brangle^2, qquad a,b in mathbb{R}^3$$



                  Therefore
                  $$|Phi_u times Phi_v|^2 = |Phi_u|^2|Phi_v|^2 - langle Phi_u, Phi_vrangle^2 = begin{vmatrix} |Phi_u|^2 & langle Phi_u, Phi_vrangle \ langle Phi_u, Phi_vrangle & |Phi_v|^2end{vmatrix} = det g$$



                  where $g = begin{bmatrix} langle Phi_u, Phi_urangle & langle Phi_u, Phi_vrangle \ langle Phi_u, Phi_vrangle & langlePhi_v, Phi_vrangleend{bmatrix}$ is the metric tensor.



                  Since $det g > 0$ we conclude $|Phi_u times Phi_v| = sqrt{det g}$.






                  share|cite|improve this answer












                  Recall Lagrange's identity for cross product:
                  $$|a times b|^2 = |a|^2|b|^2 - langle a,brangle^2, qquad a,b in mathbb{R}^3$$



                  Therefore
                  $$|Phi_u times Phi_v|^2 = |Phi_u|^2|Phi_v|^2 - langle Phi_u, Phi_vrangle^2 = begin{vmatrix} |Phi_u|^2 & langle Phi_u, Phi_vrangle \ langle Phi_u, Phi_vrangle & |Phi_v|^2end{vmatrix} = det g$$



                  where $g = begin{bmatrix} langle Phi_u, Phi_urangle & langle Phi_u, Phi_vrangle \ langle Phi_u, Phi_vrangle & langlePhi_v, Phi_vrangleend{bmatrix}$ is the metric tensor.



                  Since $det g > 0$ we conclude $|Phi_u times Phi_v| = sqrt{det g}$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 12 '18 at 22:01









                  mechanodroidmechanodroid

                  27k62446




                  27k62446























                      0














                      Call ${bf r} = Phi(u, v)$, and recall the unitary vector along the direction $u$ is



                      $$
                      h_u hat e_u = frac{partial {bf r}}{partial u} = Phi_u
                      $$



                      with $h_u$ the scale factor. You have then



                      $$
                      |Phi_u times Phi_v| = |h_u h_v| |hat{e}_utimes hat{e}_v| = |h_u h_v| = sqrt{g}
                      $$






                      share|cite|improve this answer


























                        0














                        Call ${bf r} = Phi(u, v)$, and recall the unitary vector along the direction $u$ is



                        $$
                        h_u hat e_u = frac{partial {bf r}}{partial u} = Phi_u
                        $$



                        with $h_u$ the scale factor. You have then



                        $$
                        |Phi_u times Phi_v| = |h_u h_v| |hat{e}_utimes hat{e}_v| = |h_u h_v| = sqrt{g}
                        $$






                        share|cite|improve this answer
























                          0












                          0








                          0






                          Call ${bf r} = Phi(u, v)$, and recall the unitary vector along the direction $u$ is



                          $$
                          h_u hat e_u = frac{partial {bf r}}{partial u} = Phi_u
                          $$



                          with $h_u$ the scale factor. You have then



                          $$
                          |Phi_u times Phi_v| = |h_u h_v| |hat{e}_utimes hat{e}_v| = |h_u h_v| = sqrt{g}
                          $$






                          share|cite|improve this answer












                          Call ${bf r} = Phi(u, v)$, and recall the unitary vector along the direction $u$ is



                          $$
                          h_u hat e_u = frac{partial {bf r}}{partial u} = Phi_u
                          $$



                          with $h_u$ the scale factor. You have then



                          $$
                          |Phi_u times Phi_v| = |h_u h_v| |hat{e}_utimes hat{e}_v| = |h_u h_v| = sqrt{g}
                          $$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 12 '18 at 21:55









                          caveraccaverac

                          14k21130




                          14k21130






























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