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

Having the player face themselves after the mid-game

Was this cameo in Captain Marvel computer generated?

Is it appropriate to ask a former professor to order a library book for me through ILL?

Precision notation for voltmeters

Did Amazon pay $0 in taxes last year?

Giving a career talk in my old university, how prominently should I tell students my salary?

What should I do when a paper is published similar to my PhD thesis without citation?

Short story about cities being connected by a conveyor belt

Does the US political system, in principle, allow for a no-party system?

How spaceships determine each other's mass in space?

How do you use environments that have the same name within a single latex document?

When Central Limit Theorem breaks down

Ultrafilters as a double dual

Why aren't there more Gauls like Obelix?

How can I portion out frozen cookie dough?

Should I apply for my boss's promotion?

Insult for someone who "doesn't know anything"

How to write a chaotic neutral protagonist and prevent my readers from thinking they are evil?

How to make sure I'm assertive enough in contact with subordinates?

How does learning spells work when leveling a multiclass character?

Can I challenge the interviewer to give me a proper technical feedback?

If nine coins are tossed, what is the probability that the number of heads is even?

Help! My Character is too much for her story!

Why does a car's steering wheel get lighter with increasing speed



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
          1






          active

          oldest

          votes


















          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





          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%2f3139355%2fcalculate-a-line-integral-of-differentiable-form-of-degree-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 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


















          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%2f3139355%2fcalculate-a-line-integral-of-differentiable-form-of-degree-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

          Nidaros erkebispedøme

          Image
          For det lutherske bispedømet, sjå Nidaros bispedøme. Nidaros erkebispedøme var eit katolsk erkebispedøme med sete i Nidaros. Det vart oppretta av kong Olav Kyrre mellom 1070 og 1080 som bispedøme. Bispedømet vart opphøgd til erkebispedøme i 1153. Erkebispedømet vart innstifta av pave Anastasius IV i 1154, og var i funksjon fram til reformasjonen i 1537. Erkebiskopen sitt hovudsete var Erkebispegarden, rett ved domkyrkja Nidarosdomen. Frå 1152 til 1154 var Nicholas Breakspear i Skandinavia som paveleg utsending, og organiserte opprettinga av erkesetet. Innhaldsliste 1 Utbreiing og jurisdiksjon 2 Overhyrdar 2.1 Lydbiskopar 3 Biskopar 3.1 Biskopar 3.2 Erkebiskopar Utbreiing og jurisdiksjon | Kart over den nordiske kyrkjeprovinsen (1153-1387) Erkebispedømet omfatta i tillegg til Noreg øyane i vest i Noregsveldet. Nidaros var eit metropolitanerkebispedøme med fylgjande underliggjande bispedøme suffraganar: Bj...
          Read more

          Birsay

          Image
          Tettstaden Birsay på Orknøyane. Ruinane av jarleplasset om lag midt i biletet. (Fotografi frå 2012) Utsikt mot Birsay og Brough of Brisay (fotografi frå 2009) Birsay er ein tettstad i eit sokn (parish) med same namn, nordvest på øya Mainland [Note 1] på Orknøyane, om lag 31 km frå Kirkwall. Staden er omtala fleire stader i Orknøyingasoga [1] , der han er kalla «Byrgesherad» (norrønt: Birgisherað ). Namnet skriv seg truleg frå elvenamnet Birgisá [2] . Innhaldsliste 1 Den gamle hovudstaden 2 Bispesetet 3 Jarlepalasset 4 Kjelder og utfyllande opplysningar 4.1 Notar 4.2 Fotnotar 4.3 Bakgrunnsstoff Den gamle hovudstaden | Hovudartikkelen for dette emnet er Torfinn Sigurdsson . Soknet er i dag i hovudsak ei jordbrukskommune men er mest kjend for dei mange kulturminna i området. Tettstaden Birsay blir gjerne omtala som «den gamle hovudstaden på Orknøyane» [3] , og ein viser d...
          Read more

          Was Woodrow Wilson really a Liberal?Was World War I a war of liberals against authoritarians?Founding Fathers...

          Image
          Do VLANs within a subnet need to have their own subnet for router on a stick? What are the differences between the usage of 'it' and 'they'? Dragon forelimb placement How do we improve the relationship with a client software team that performs poorly and is becoming less collaborative? What typically incentivizes a professor to change jobs to a lower ranking university? How to write a macro that is braces sensitive? Modeling an IPv4 Address Writing rule stating superpower from different root cause is bad writing Can I ask the recruiters in my resume to put the reason why I am rejected? Approximately how much travel t...
          Read more