Integral of 1-form along any path between two points

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Let $K$ be an oriented curve starting at $a = (pi, -1)$ and ending at $b = (3pi, 1)$.
$$
I = int_K (e^y+e^-y)cos xdx + (e^y-e^-y)sin xdy
$$
Does the value of the integral depend on the choice of the path between $a$ and $b$? Calculate the value of $I$.




Let $H = (e^y+e^-y)sin x$ then $grad H = [(e^y+e^-y)cos x, (e^y-e^-y)sin x]^T$ so I suppose that the value of integral does not depend on the choice of the path between $a$ and $b$. Is it



But how to calculate the value? I cannot use Green Theorem because it is not the closed path and therefore is not the edge of any area.






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    Let $K$ be an oriented curve starting at $a = (pi, -1)$ and ending at $b = (3pi, 1)$.
    $$
    I = int_K (e^y+e^-y)cos xdx + (e^y-e^-y)sin xdy
    $$
    Does the value of the integral depend on the choice of the path between $a$ and $b$? Calculate the value of $I$.




    Let $H = (e^y+e^-y)sin x$ then $grad H = [(e^y+e^-y)cos x, (e^y-e^-y)sin x]^T$ so I suppose that the value of integral does not depend on the choice of the path between $a$ and $b$. Is it



    But how to calculate the value? I cannot use Green Theorem because it is not the closed path and therefore is not the edge of any area.






    lebesgue-integral differential-forms







    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite












      Let $K$ be an oriented curve starting at $a = (pi, -1)$ and ending at $b = (3pi, 1)$.
      $$
      I = int_K (e^y+e^-y)cos xdx + (e^y-e^-y)sin xdy
      $$
      Does the value of the integral depend on the choice of the path between $a$ and $b$? Calculate the value of $I$.




      Let $H = (e^y+e^-y)sin x$ then $grad H = [(e^y+e^-y)cos x, (e^y-e^-y)sin x]^T$ so I suppose that the value of integral does not depend on the choice of the path between $a$ and $b$. Is it



      But how to calculate the value? I cannot use Green Theorem because it is not the closed path and therefore is not the edge of any area.






      lebesgue-integral differential-forms







      share|cite|improve this question














      Let $K$ be an oriented curve starting at $a = (pi, -1)$ and ending at $b = (3pi, 1)$.
      $$
      I = int_K (e^y+e^-y)cos xdx + (e^y-e^-y)sin xdy
      $$
      Does the value of the integral depend on the choice of the path between $a$ and $b$? Calculate the value of $I$.




      Let $H = (e^y+e^-y)sin x$ then $grad H = [(e^y+e^-y)cos x, (e^y-e^-y)sin x]^T$ so I suppose that the value of integral does not depend on the choice of the path between $a$ and $b$. Is it



      But how to calculate the value? I cannot use Green Theorem because it is not the closed path and therefore is not the edge of any area.






      lebesgue-integral differential-forms




      lebesgue-integral differential-forms






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      asked Aug 31 at 7:42









      latonimus

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          You are correct that the integral is path-independent, because the 1-form you are integrating is exact (by which I mean, it is the gradient of a function).



          This hugely simplifies the evaluation of the integral, because we can simply use the fundamental theorem of calculus!



          $$ int_a^b ,nabla H cdot mathrmdx = int_a^b ,mathrmdH = H|_b - H|_a ,.$$






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            1 Answer
            1






            active

            oldest

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            oldest

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            active

            oldest

            votes








            up vote
            1
            down vote



            accepted










            You are correct that the integral is path-independent, because the 1-form you are integrating is exact (by which I mean, it is the gradient of a function).



            This hugely simplifies the evaluation of the integral, because we can simply use the fundamental theorem of calculus!



            $$ int_a^b ,nabla H cdot mathrmdx = int_a^b ,mathrmdH = H|_b - H|_a ,.$$






            share|cite|improve this answer
























              up vote
              1
              down vote



              accepted










              You are correct that the integral is path-independent, because the 1-form you are integrating is exact (by which I mean, it is the gradient of a function).



              This hugely simplifies the evaluation of the integral, because we can simply use the fundamental theorem of calculus!



              $$ int_a^b ,nabla H cdot mathrmdx = int_a^b ,mathrmdH = H|_b - H|_a ,.$$






              share|cite|improve this answer






















                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                You are correct that the integral is path-independent, because the 1-form you are integrating is exact (by which I mean, it is the gradient of a function).



                This hugely simplifies the evaluation of the integral, because we can simply use the fundamental theorem of calculus!



                $$ int_a^b ,nabla H cdot mathrmdx = int_a^b ,mathrmdH = H|_b - H|_a ,.$$






                share|cite|improve this answer












                You are correct that the integral is path-independent, because the 1-form you are integrating is exact (by which I mean, it is the gradient of a function).



                This hugely simplifies the evaluation of the integral, because we can simply use the fundamental theorem of calculus!



                $$ int_a^b ,nabla H cdot mathrmdx = int_a^b ,mathrmdH = H|_b - H|_a ,.$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 31 at 7:48









                gj255

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