Calculate a Line Integral of differentiable form of degree $1$.integral of a 2-form over an oriented...

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Calculate a Line Integral of differentiable form of degree $1$.


integral of a 2-form over an oriented manifoldUsing Stokes's theorem to calculate a value of integralDifferential forms and determinantsAn Integral InequalityProving a formula for the exterior derivative of a specific $k$-form, given in base representationCalculate the integral using Green's TheoremCalculate $domega$ of an $n-1$ forms $omega$$d(Iomega) + I(domega) = omega$ for differential formsCalculus with Differential Forms.$f^{*}omega = det(df)dx_{1}wedge cdots wedge dx_{n}$ for $f(x_{1},…,x_{n}) = (y_{1},…,y_{n})$ and $omega = dy_{1}wedge cdots dy_{n}$













1












$begingroup$



Show that the form $omega = 2xy^3dx + 3x^2y^2dy$ is closed and calculate $int_{c}omega$, where $c$ is given by $y=x^2$, $(0,0)$ to $(x,y)$.




My attempt.



$begin{eqnarray*}
domega &=& d(2xy^3)wedge dx + d(3x^2y^2)wedge dy\
&=& (2y^3dx + 6xy^2dy)wedge dx + (6xy^2dx + 6xy^2dy)dy\
& = & -6xy^2dxwedge dy + 6xy^2dxwedge dy\
&=& 0.
end{eqnarray*}$



Now, by definition



$$int_{c}omega = int_{a}^{b}left(sum_{i}a_{i}(t)frac{dx_{i}}{dt}right)dt$$



where $omega = sum_{i}a_{i}dx_{i}$. Then



$$sum_{i}a_{i}(t)frac{dx_{i}}{dt} = 2x(t)y(t)^3frac{dx}{dt} + 3x(t)^2y(t)^2frac{dy}{dt}.$$



We can write $c: [0,x] to {(x,x^2)}$ given by $c(t) = (t,t^2)$, by taking $x(t) = t$ and $y(t) = t^2$. So, we have



$begin{eqnarray*}
int_{c}omega &=& int_{0}^{x}left(2x(t)y(t)^3x'(t) + 3x(t)^2y(t)^2y'(t)right)dt\
&=& int_{0}^{x}2t^{7}dt + int_{0}^{x}6t^{7}dt\
& = & frac{t^{8}}{4}Bigg|_{0}^{x} + frac{3t^{8}}{4}Bigg|_{0}^{x}\
& = & t^{8}Bigg|_{0}^{x}\
&=& x^{8}
end{eqnarray*}$



Is correct? This is my first problem about Line Integrals on $1$-forms.






real-analysis integration differential-forms







share|cite|improve this question









$endgroup$

















    1












    $begingroup$



    Show that the form $omega = 2xy^3dx + 3x^2y^2dy$ is closed and calculate $int_{c}omega$, where $c$ is given by $y=x^2$, $(0,0)$ to $(x,y)$.




    My attempt.



    $begin{eqnarray*}
    domega &=& d(2xy^3)wedge dx + d(3x^2y^2)wedge dy\
    &=& (2y^3dx + 6xy^2dy)wedge dx + (6xy^2dx + 6xy^2dy)dy\
    & = & -6xy^2dxwedge dy + 6xy^2dxwedge dy\
    &=& 0.
    end{eqnarray*}$



    Now, by definition



    $$int_{c}omega = int_{a}^{b}left(sum_{i}a_{i}(t)frac{dx_{i}}{dt}right)dt$$



    where $omega = sum_{i}a_{i}dx_{i}$. Then



    $$sum_{i}a_{i}(t)frac{dx_{i}}{dt} = 2x(t)y(t)^3frac{dx}{dt} + 3x(t)^2y(t)^2frac{dy}{dt}.$$



    We can write $c: [0,x] to {(x,x^2)}$ given by $c(t) = (t,t^2)$, by taking $x(t) = t$ and $y(t) = t^2$. So, we have



    $begin{eqnarray*}
    int_{c}omega &=& int_{0}^{x}left(2x(t)y(t)^3x'(t) + 3x(t)^2y(t)^2y'(t)right)dt\
    &=& int_{0}^{x}2t^{7}dt + int_{0}^{x}6t^{7}dt\
    & = & frac{t^{8}}{4}Bigg|_{0}^{x} + frac{3t^{8}}{4}Bigg|_{0}^{x}\
    & = & t^{8}Bigg|_{0}^{x}\
    &=& x^{8}
    end{eqnarray*}$



    Is correct? This is my first problem about Line Integrals on $1$-forms.






    real-analysis integration differential-forms







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$



      Show that the form $omega = 2xy^3dx + 3x^2y^2dy$ is closed and calculate $int_{c}omega$, where $c$ is given by $y=x^2$, $(0,0)$ to $(x,y)$.




      My attempt.



      $begin{eqnarray*}
      domega &=& d(2xy^3)wedge dx + d(3x^2y^2)wedge dy\
      &=& (2y^3dx + 6xy^2dy)wedge dx + (6xy^2dx + 6xy^2dy)dy\
      & = & -6xy^2dxwedge dy + 6xy^2dxwedge dy\
      &=& 0.
      end{eqnarray*}$



      Now, by definition



      $$int_{c}omega = int_{a}^{b}left(sum_{i}a_{i}(t)frac{dx_{i}}{dt}right)dt$$



      where $omega = sum_{i}a_{i}dx_{i}$. Then



      $$sum_{i}a_{i}(t)frac{dx_{i}}{dt} = 2x(t)y(t)^3frac{dx}{dt} + 3x(t)^2y(t)^2frac{dy}{dt}.$$



      We can write $c: [0,x] to {(x,x^2)}$ given by $c(t) = (t,t^2)$, by taking $x(t) = t$ and $y(t) = t^2$. So, we have



      $begin{eqnarray*}
      int_{c}omega &=& int_{0}^{x}left(2x(t)y(t)^3x'(t) + 3x(t)^2y(t)^2y'(t)right)dt\
      &=& int_{0}^{x}2t^{7}dt + int_{0}^{x}6t^{7}dt\
      & = & frac{t^{8}}{4}Bigg|_{0}^{x} + frac{3t^{8}}{4}Bigg|_{0}^{x}\
      & = & t^{8}Bigg|_{0}^{x}\
      &=& x^{8}
      end{eqnarray*}$



      Is correct? This is my first problem about Line Integrals on $1$-forms.






      real-analysis integration differential-forms







      share|cite|improve this question









      $endgroup$





      Show that the form $omega = 2xy^3dx + 3x^2y^2dy$ is closed and calculate $int_{c}omega$, where $c$ is given by $y=x^2$, $(0,0)$ to $(x,y)$.




      My attempt.



      $begin{eqnarray*}
      domega &=& d(2xy^3)wedge dx + d(3x^2y^2)wedge dy\
      &=& (2y^3dx + 6xy^2dy)wedge dx + (6xy^2dx + 6xy^2dy)dy\
      & = & -6xy^2dxwedge dy + 6xy^2dxwedge dy\
      &=& 0.
      end{eqnarray*}$



      Now, by definition



      $$int_{c}omega = int_{a}^{b}left(sum_{i}a_{i}(t)frac{dx_{i}}{dt}right)dt$$



      where $omega = sum_{i}a_{i}dx_{i}$. Then



      $$sum_{i}a_{i}(t)frac{dx_{i}}{dt} = 2x(t)y(t)^3frac{dx}{dt} + 3x(t)^2y(t)^2frac{dy}{dt}.$$



      We can write $c: [0,x] to {(x,x^2)}$ given by $c(t) = (t,t^2)$, by taking $x(t) = t$ and $y(t) = t^2$. So, we have



      $begin{eqnarray*}
      int_{c}omega &=& int_{0}^{x}left(2x(t)y(t)^3x'(t) + 3x(t)^2y(t)^2y'(t)right)dt\
      &=& int_{0}^{x}2t^{7}dt + int_{0}^{x}6t^{7}dt\
      & = & frac{t^{8}}{4}Bigg|_{0}^{x} + frac{3t^{8}}{4}Bigg|_{0}^{x}\
      & = & t^{8}Bigg|_{0}^{x}\
      &=& x^{8}
      end{eqnarray*}$



      Is correct? This is my first problem about Line Integrals on $1$-forms.






      real-analysis integration differential-forms




      real-analysis integration differential-forms






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked yesterday









      Lucas CorrêaLucas Corrêa

      1,6321321




      1,6321321






















          1 Answer
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          1












          $begingroup$

          The differential form is even exact:
          $$
          omega = 2xy^3,dx + 3x^2y^2,dy
          = frac{partial(x^2y^3)}{partial x} dx + frac{partial(x^2y^3)}{partial y} dy
          = d(x^2y^3).
          $$
          From this follows that $omega$ is closed, since $d^2omega = 0$ for any differential form $omega.$



          Therefore we can calculate the integral as
          $$
          int_c omega
          = int_{(0,0)}^{(x,x^2)} d(x^2y^3)
          = x^2y^3 {Bigg|}_{(0,0)}^{(x,x^2)}
          = x^2(x^2)^3 - 0^2 cdot 0^3
          = x^8.
          $$



          Thus, your result is correct.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you! Your approach is very clever!
            $endgroup$
            – Lucas Corrêa
            yesterday











          Your Answer





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          active

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          active

          oldest

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          1












          $begingroup$

          The differential form is even exact:
          $$
          omega = 2xy^3,dx + 3x^2y^2,dy
          = frac{partial(x^2y^3)}{partial x} dx + frac{partial(x^2y^3)}{partial y} dy
          = d(x^2y^3).
          $$
          From this follows that $omega$ is closed, since $d^2omega = 0$ for any differential form $omega.$



          Therefore we can calculate the integral as
          $$
          int_c omega
          = int_{(0,0)}^{(x,x^2)} d(x^2y^3)
          = x^2y^3 {Bigg|}_{(0,0)}^{(x,x^2)}
          = x^2(x^2)^3 - 0^2 cdot 0^3
          = x^8.
          $$



          Thus, your result is correct.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you! Your approach is very clever!
            $endgroup$
            – Lucas Corrêa
            yesterday
















          1












          $begingroup$

          The differential form is even exact:
          $$
          omega = 2xy^3,dx + 3x^2y^2,dy
          = frac{partial(x^2y^3)}{partial x} dx + frac{partial(x^2y^3)}{partial y} dy
          = d(x^2y^3).
          $$
          From this follows that $omega$ is closed, since $d^2omega = 0$ for any differential form $omega.$



          Therefore we can calculate the integral as
          $$
          int_c omega
          = int_{(0,0)}^{(x,x^2)} d(x^2y^3)
          = x^2y^3 {Bigg|}_{(0,0)}^{(x,x^2)}
          = x^2(x^2)^3 - 0^2 cdot 0^3
          = x^8.
          $$



          Thus, your result is correct.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you! Your approach is very clever!
            $endgroup$
            – Lucas Corrêa
            yesterday














          1












          1








          1





          $begingroup$

          The differential form is even exact:
          $$
          omega = 2xy^3,dx + 3x^2y^2,dy
          = frac{partial(x^2y^3)}{partial x} dx + frac{partial(x^2y^3)}{partial y} dy
          = d(x^2y^3).
          $$
          From this follows that $omega$ is closed, since $d^2omega = 0$ for any differential form $omega.$



          Therefore we can calculate the integral as
          $$
          int_c omega
          = int_{(0,0)}^{(x,x^2)} d(x^2y^3)
          = x^2y^3 {Bigg|}_{(0,0)}^{(x,x^2)}
          = x^2(x^2)^3 - 0^2 cdot 0^3
          = x^8.
          $$



          Thus, your result is correct.






          share|cite|improve this answer









          $endgroup$



          The differential form is even exact:
          $$
          omega = 2xy^3,dx + 3x^2y^2,dy
          = frac{partial(x^2y^3)}{partial x} dx + frac{partial(x^2y^3)}{partial y} dy
          = d(x^2y^3).
          $$
          From this follows that $omega$ is closed, since $d^2omega = 0$ for any differential form $omega.$



          Therefore we can calculate the integral as
          $$
          int_c omega
          = int_{(0,0)}^{(x,x^2)} d(x^2y^3)
          = x^2y^3 {Bigg|}_{(0,0)}^{(x,x^2)}
          = x^2(x^2)^3 - 0^2 cdot 0^3
          = x^8.
          $$



          Thus, your result is correct.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          md2perpemd2perpe

          8,21111028




          8,21111028












          • $begingroup$
            Thank you! Your approach is very clever!
            $endgroup$
            – Lucas Corrêa
            yesterday


















          • $begingroup$
            Thank you! Your approach is very clever!
            $endgroup$
            – Lucas Corrêa
            yesterday
















          $begingroup$
          Thank you! Your approach is very clever!
          $endgroup$
          – Lucas Corrêa
          yesterday




          $begingroup$
          Thank you! Your approach is very clever!
          $endgroup$
          – Lucas Corrêa
          yesterday


















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