Surface integral and parametrization











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I'm struggling with surface integrals, and I still do not have much confidence with the parameterization of functions. This is the exercise I would like to solve:




Calculate the surface integral of the function



$$f = (x-1)^2 + (y-2)^2$$



extended to the surface $P equiv (1+rho cos vartheta, 2 + rho sin vartheta, 4 - rho^2)$ with $vartheta in [0, 2pi], rho in [sqrt{2}, sqrt{3}]$




without going into detail in the steps and in the final result, how should I proceed? The first thing I should do is find the parameterization of the surface, right? And can I retrieve it?



Thanks in advances






surface-integrals







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    up vote
    0
    down vote

    favorite












    I'm struggling with surface integrals, and I still do not have much confidence with the parameterization of functions. This is the exercise I would like to solve:




    Calculate the surface integral of the function



    $$f = (x-1)^2 + (y-2)^2$$



    extended to the surface $P equiv (1+rho cos vartheta, 2 + rho sin vartheta, 4 - rho^2)$ with $vartheta in [0, 2pi], rho in [sqrt{2}, sqrt{3}]$




    without going into detail in the steps and in the final result, how should I proceed? The first thing I should do is find the parameterization of the surface, right? And can I retrieve it?



    Thanks in advances






    surface-integrals







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I'm struggling with surface integrals, and I still do not have much confidence with the parameterization of functions. This is the exercise I would like to solve:




      Calculate the surface integral of the function



      $$f = (x-1)^2 + (y-2)^2$$



      extended to the surface $P equiv (1+rho cos vartheta, 2 + rho sin vartheta, 4 - rho^2)$ with $vartheta in [0, 2pi], rho in [sqrt{2}, sqrt{3}]$




      without going into detail in the steps and in the final result, how should I proceed? The first thing I should do is find the parameterization of the surface, right? And can I retrieve it?



      Thanks in advances






      surface-integrals







      share|cite|improve this question













      I'm struggling with surface integrals, and I still do not have much confidence with the parameterization of functions. This is the exercise I would like to solve:




      Calculate the surface integral of the function



      $$f = (x-1)^2 + (y-2)^2$$



      extended to the surface $P equiv (1+rho cos vartheta, 2 + rho sin vartheta, 4 - rho^2)$ with $vartheta in [0, 2pi], rho in [sqrt{2}, sqrt{3}]$




      without going into detail in the steps and in the final result, how should I proceed? The first thing I should do is find the parameterization of the surface, right? And can I retrieve it?



      Thanks in advances






      surface-integrals




      surface-integrals






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      asked 2 days ago









      user3204810

      1806




      1806






















          1 Answer
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          Your surface $S$ is already given in a parametrized form, i.e. your parametrization is
          $$ P(u,v) = (1+ucos v, 2+usin v, 4-v^2), quad uin [sqrt 2, sqrt 3], quad v in [0,2pi]. $$
          Now you should compute your surface area element $mathrm dA$, which is defined as
          $$ mathrm dA = |partial_u P times partial_v P| , mathrm du , mathrm dv, $$
          and the sought integral then becomes
          $$ int_S f , mathrm dA = int_{0}^{2pi} int_{sqrt 2}^{sqrt 3} f(P(u,v)) |partial_u P times partial_v P| , mathrm du ,mathrm dv. $$
          Can you proceed?






          share|cite|improve this answer

















          • 1




            Tomorrow I try calmly. In any case, thanks for the tip!
            – user3204810
            2 days ago











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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          Your surface $S$ is already given in a parametrized form, i.e. your parametrization is
          $$ P(u,v) = (1+ucos v, 2+usin v, 4-v^2), quad uin [sqrt 2, sqrt 3], quad v in [0,2pi]. $$
          Now you should compute your surface area element $mathrm dA$, which is defined as
          $$ mathrm dA = |partial_u P times partial_v P| , mathrm du , mathrm dv, $$
          and the sought integral then becomes
          $$ int_S f , mathrm dA = int_{0}^{2pi} int_{sqrt 2}^{sqrt 3} f(P(u,v)) |partial_u P times partial_v P| , mathrm du ,mathrm dv. $$
          Can you proceed?






          share|cite|improve this answer

















          • 1




            Tomorrow I try calmly. In any case, thanks for the tip!
            – user3204810
            2 days ago















          up vote
          1
          down vote



          accepted










          Your surface $S$ is already given in a parametrized form, i.e. your parametrization is
          $$ P(u,v) = (1+ucos v, 2+usin v, 4-v^2), quad uin [sqrt 2, sqrt 3], quad v in [0,2pi]. $$
          Now you should compute your surface area element $mathrm dA$, which is defined as
          $$ mathrm dA = |partial_u P times partial_v P| , mathrm du , mathrm dv, $$
          and the sought integral then becomes
          $$ int_S f , mathrm dA = int_{0}^{2pi} int_{sqrt 2}^{sqrt 3} f(P(u,v)) |partial_u P times partial_v P| , mathrm du ,mathrm dv. $$
          Can you proceed?






          share|cite|improve this answer

















          • 1




            Tomorrow I try calmly. In any case, thanks for the tip!
            – user3204810
            2 days ago













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Your surface $S$ is already given in a parametrized form, i.e. your parametrization is
          $$ P(u,v) = (1+ucos v, 2+usin v, 4-v^2), quad uin [sqrt 2, sqrt 3], quad v in [0,2pi]. $$
          Now you should compute your surface area element $mathrm dA$, which is defined as
          $$ mathrm dA = |partial_u P times partial_v P| , mathrm du , mathrm dv, $$
          and the sought integral then becomes
          $$ int_S f , mathrm dA = int_{0}^{2pi} int_{sqrt 2}^{sqrt 3} f(P(u,v)) |partial_u P times partial_v P| , mathrm du ,mathrm dv. $$
          Can you proceed?






          share|cite|improve this answer












          Your surface $S$ is already given in a parametrized form, i.e. your parametrization is
          $$ P(u,v) = (1+ucos v, 2+usin v, 4-v^2), quad uin [sqrt 2, sqrt 3], quad v in [0,2pi]. $$
          Now you should compute your surface area element $mathrm dA$, which is defined as
          $$ mathrm dA = |partial_u P times partial_v P| , mathrm du , mathrm dv, $$
          and the sought integral then becomes
          $$ int_S f , mathrm dA = int_{0}^{2pi} int_{sqrt 2}^{sqrt 3} f(P(u,v)) |partial_u P times partial_v P| , mathrm du ,mathrm dv. $$
          Can you proceed?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          MisterRiemann

          4,4751622




          4,4751622








          • 1




            Tomorrow I try calmly. In any case, thanks for the tip!
            – user3204810
            2 days ago














          • 1




            Tomorrow I try calmly. In any case, thanks for the tip!
            – user3204810
            2 days ago








          1




          1




          Tomorrow I try calmly. In any case, thanks for the tip!
          – user3204810
          2 days ago




          Tomorrow I try calmly. In any case, thanks for the tip!
          – user3204810
          2 days ago


















           

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