Explain why the area of $S$ is equal to $int_C x,dsigma$ ( line integral )












1












$begingroup$



Let $S$ be a surface in $mathbb{R^3}
$ with the parametrization $g(s, t) = (t, t^2
, st)$
where $g : [0, 1] × [0, 10] → mathbb{R^3}
$
. Explain why the area of $S$ is equal
to $int_C xdσ$
, where $C$ is the curve in $mathbb{R^2}$
parameterized by $h(t) = (t, t^2
)$,
$h : [0, 10] → mathbb{R^2}$
and Find the area of $S$.




i didn't understand the question at all how can i explain that . Area $S$ =
$int_C x,dsigma$ .



i know that the surface area is given by :
$ int f(x(t),y(t)|r'(t)| { dt}$



but the curve siting in the $[xy]$ plane is $h(t)$ so my guess (might be wrong) that $|r'(t)|$ = $sqrt{4t^2 + 1}$ and $f(x(t),y(t)) = st = sx$ .






integration multivariable-calculus curvature line-integrals







share|cite|improve this question











$endgroup$

















    1












    $begingroup$



    Let $S$ be a surface in $mathbb{R^3}
    $     with the parametrization $g(s, t) = (t, t^2
    , st)$
    where $g : [0, 1] × [0, 10] → mathbb{R^3}
    $
    . Explain why the area of $S$ is equal
    to $int_C xdσ$
    , where $C$ is the curve in $mathbb{R^2}$
    parameterized by $h(t) = (t, t^2
    )$    ,
    $h : [0, 10] → mathbb{R^2}$
    and Find the area of $S$.




    i didn't understand the question at all how can i explain that . Area $S$ =
    $int_C x,dsigma$ .



    i know that the surface area is given by :
    $ int f(x(t),y(t)|r'(t)| { dt}$



    but the curve siting in the $[xy]$ plane is $h(t)$ so my guess (might be wrong) that $|r'(t)|$ = $sqrt{4t^2 + 1}$ and $f(x(t),y(t)) = st = sx$ .






    integration multivariable-calculus curvature line-integrals







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$



      Let $S$ be a surface in $mathbb{R^3}
      $       with the parametrization $g(s, t) = (t, t^2
      , st)$
      where $g : [0, 1] × [0, 10] → mathbb{R^3}
      $
      . Explain why the area of $S$ is equal
      to $int_C xdσ$
      , where $C$ is the curve in $mathbb{R^2}$
      parameterized by $h(t) = (t, t^2
      )$      ,
      $h : [0, 10] → mathbb{R^2}$
      and Find the area of $S$.




      i didn't understand the question at all how can i explain that . Area $S$ =
      $int_C x,dsigma$ .



      i know that the surface area is given by :
      $ int f(x(t),y(t)|r'(t)| { dt}$



      but the curve siting in the $[xy]$ plane is $h(t)$ so my guess (might be wrong) that $|r'(t)|$ = $sqrt{4t^2 + 1}$ and $f(x(t),y(t)) = st = sx$ .






      integration multivariable-calculus curvature line-integrals







      share|cite|improve this question











      $endgroup$





      Let $S$ be a surface in $mathbb{R^3}
      $       with the parametrization $g(s, t) = (t, t^2
      , st)$
      where $g : [0, 1] × [0, 10] → mathbb{R^3}
      $
      . Explain why the area of $S$ is equal
      to $int_C xdσ$
      , where $C$ is the curve in $mathbb{R^2}$
      parameterized by $h(t) = (t, t^2
      )$      ,
      $h : [0, 10] → mathbb{R^2}$
      and Find the area of $S$.




      i didn't understand the question at all how can i explain that . Area $S$ =
      $int_C x,dsigma$ .



      i know that the surface area is given by :
      $ int f(x(t),y(t)|r'(t)| { dt}$



      but the curve siting in the $[xy]$ plane is $h(t)$ so my guess (might be wrong) that $|r'(t)|$ = $sqrt{4t^2 + 1}$ and $f(x(t),y(t)) = st = sx$ .






      integration multivariable-calculus curvature line-integrals




      integration multivariable-calculus curvature line-integrals






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 29 '18 at 12:25









      mechanodroid

      28.9k62648




      28.9k62648










      asked Dec 29 '18 at 10:56









      Mather Mather

      4028




      4028






















          2 Answers
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          $begingroup$

          The question is only valid for this particular case, not in general.



          Let me introduce a notation $mathbf{g}(s,t)$ to denote the vector pointing from origin $(0,0,0)$ to the point $(t,t^2,st)$. Hence, $mathbf{g}(s,t)=t,mathbf{i}+t^2,mathbf{j}+st,mathbf{k}$. The surface area is (refer any book on Advanced Engineering Mathematics)
          $$A=int_0^{10}int_0^1 leftlVert frac{partialmathbf{g}}{partial s}timesfrac{partialmathbf{g}}{partial t} rightrVert,text{d}s,text{d}t = int_0^{10}int_0^1 sqrt{4t^4+t^2},text{d}s,text{d}t=int_0^{10}tsqrt{4t^2+1},text{d}t.$$
          Incidentally, in this case,
          $$int_C x,text{d}sigma=int_0^{10}tsqrt{4t^2+1},text{d}t,$$
          since $C$ is parametrised by $(t,t^2)$ and, as you already pointed out, the infinitesimal arc length is $text{d}sigma=sqrt{4t^2+1},text{d}t$.






          share|cite|improve this answer











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            1












            $begingroup$

            Notice that your surface $S$ is the area between the curves $tmapsto (t,t^2,0)$ and $t mapsto (t,t^2,t)$.



            Informally, if you consider the curve $C$ given by $tmapsto (t,t^2)$ as the $x$-axis, you are calculating the area between the $x$-axis and the curve $y = x$.



            Hence $$text{Area}(S) = int_{C} x,dsigma$$






            share|cite|improve this answer









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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              0












              $begingroup$

              The question is only valid for this particular case, not in general.



              Let me introduce a notation $mathbf{g}(s,t)$ to denote the vector pointing from origin $(0,0,0)$ to the point $(t,t^2,st)$. Hence, $mathbf{g}(s,t)=t,mathbf{i}+t^2,mathbf{j}+st,mathbf{k}$. The surface area is (refer any book on Advanced Engineering Mathematics)
              $$A=int_0^{10}int_0^1 leftlVert frac{partialmathbf{g}}{partial s}timesfrac{partialmathbf{g}}{partial t} rightrVert,text{d}s,text{d}t = int_0^{10}int_0^1 sqrt{4t^4+t^2},text{d}s,text{d}t=int_0^{10}tsqrt{4t^2+1},text{d}t.$$
              Incidentally, in this case,
              $$int_C x,text{d}sigma=int_0^{10}tsqrt{4t^2+1},text{d}t,$$
              since $C$ is parametrised by $(t,t^2)$ and, as you already pointed out, the infinitesimal arc length is $text{d}sigma=sqrt{4t^2+1},text{d}t$.






              share|cite|improve this answer











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                0












                $begingroup$

                The question is only valid for this particular case, not in general.



                Let me introduce a notation $mathbf{g}(s,t)$ to denote the vector pointing from origin $(0,0,0)$ to the point $(t,t^2,st)$. Hence, $mathbf{g}(s,t)=t,mathbf{i}+t^2,mathbf{j}+st,mathbf{k}$. The surface area is (refer any book on Advanced Engineering Mathematics)
                $$A=int_0^{10}int_0^1 leftlVert frac{partialmathbf{g}}{partial s}timesfrac{partialmathbf{g}}{partial t} rightrVert,text{d}s,text{d}t = int_0^{10}int_0^1 sqrt{4t^4+t^2},text{d}s,text{d}t=int_0^{10}tsqrt{4t^2+1},text{d}t.$$
                Incidentally, in this case,
                $$int_C x,text{d}sigma=int_0^{10}tsqrt{4t^2+1},text{d}t,$$
                since $C$ is parametrised by $(t,t^2)$ and, as you already pointed out, the infinitesimal arc length is $text{d}sigma=sqrt{4t^2+1},text{d}t$.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  The question is only valid for this particular case, not in general.



                  Let me introduce a notation $mathbf{g}(s,t)$ to denote the vector pointing from origin $(0,0,0)$ to the point $(t,t^2,st)$. Hence, $mathbf{g}(s,t)=t,mathbf{i}+t^2,mathbf{j}+st,mathbf{k}$. The surface area is (refer any book on Advanced Engineering Mathematics)
                  $$A=int_0^{10}int_0^1 leftlVert frac{partialmathbf{g}}{partial s}timesfrac{partialmathbf{g}}{partial t} rightrVert,text{d}s,text{d}t = int_0^{10}int_0^1 sqrt{4t^4+t^2},text{d}s,text{d}t=int_0^{10}tsqrt{4t^2+1},text{d}t.$$
                  Incidentally, in this case,
                  $$int_C x,text{d}sigma=int_0^{10}tsqrt{4t^2+1},text{d}t,$$
                  since $C$ is parametrised by $(t,t^2)$ and, as you already pointed out, the infinitesimal arc length is $text{d}sigma=sqrt{4t^2+1},text{d}t$.






                  share|cite|improve this answer











                  $endgroup$



                  The question is only valid for this particular case, not in general.



                  Let me introduce a notation $mathbf{g}(s,t)$ to denote the vector pointing from origin $(0,0,0)$ to the point $(t,t^2,st)$. Hence, $mathbf{g}(s,t)=t,mathbf{i}+t^2,mathbf{j}+st,mathbf{k}$. The surface area is (refer any book on Advanced Engineering Mathematics)
                  $$A=int_0^{10}int_0^1 leftlVert frac{partialmathbf{g}}{partial s}timesfrac{partialmathbf{g}}{partial t} rightrVert,text{d}s,text{d}t = int_0^{10}int_0^1 sqrt{4t^4+t^2},text{d}s,text{d}t=int_0^{10}tsqrt{4t^2+1},text{d}t.$$
                  Incidentally, in this case,
                  $$int_C x,text{d}sigma=int_0^{10}tsqrt{4t^2+1},text{d}t,$$
                  since $C$ is parametrised by $(t,t^2)$ and, as you already pointed out, the infinitesimal arc length is $text{d}sigma=sqrt{4t^2+1},text{d}t$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 29 '18 at 12:24

























                  answered Dec 29 '18 at 12:17









                  ZxcvasdfZxcvasdf

                  835




                  835























                      1












                      $begingroup$

                      Notice that your surface $S$ is the area between the curves $tmapsto (t,t^2,0)$ and $t mapsto (t,t^2,t)$.



                      Informally, if you consider the curve $C$ given by $tmapsto (t,t^2)$ as the $x$-axis, you are calculating the area between the $x$-axis and the curve $y = x$.



                      Hence $$text{Area}(S) = int_{C} x,dsigma$$






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        Notice that your surface $S$ is the area between the curves $tmapsto (t,t^2,0)$ and $t mapsto (t,t^2,t)$.



                        Informally, if you consider the curve $C$ given by $tmapsto (t,t^2)$ as the $x$-axis, you are calculating the area between the $x$-axis and the curve $y = x$.



                        Hence $$text{Area}(S) = int_{C} x,dsigma$$






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Notice that your surface $S$ is the area between the curves $tmapsto (t,t^2,0)$ and $t mapsto (t,t^2,t)$.



                          Informally, if you consider the curve $C$ given by $tmapsto (t,t^2)$ as the $x$-axis, you are calculating the area between the $x$-axis and the curve $y = x$.



                          Hence $$text{Area}(S) = int_{C} x,dsigma$$






                          share|cite|improve this answer









                          $endgroup$



                          Notice that your surface $S$ is the area between the curves $tmapsto (t,t^2,0)$ and $t mapsto (t,t^2,t)$.



                          Informally, if you consider the curve $C$ given by $tmapsto (t,t^2)$ as the $x$-axis, you are calculating the area between the $x$-axis and the curve $y = x$.



                          Hence $$text{Area}(S) = int_{C} x,dsigma$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 29 '18 at 12:23









                          mechanodroidmechanodroid

                          28.9k62648




                          28.9k62648






























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