how to calculate $int_{C}(2x^2+1)e^{x^2+y^2}dx+(2xy)e^{x^2+y^2}dy$ using Green theorem












1












$begingroup$



Compute $int_{C}(2x^2+1)e^{x^2+y^2}dx+(2xy)e^{x^2+y^2}dy$ where $C$ connects $(1,0)$ to $(0,1)$ by a straight line segment.




I tried to use green theorem since $Q_x = P_y$ so $int vec F_dot{}vec dr=0$



i need to close $C$ so i build a triangle with vertices
$(0,0)$ , $(1,0)$ ,$(0,1)$ now i got a complicated integral in the line parametrization :



$y = 0 ~~~ dy=0$



$x = t ~~~ dx=dt$



$0leq t leq 1$




$int_0^1 (2t^2 + 1)e^{t^2}dt$ , since $dy=0$ only the left side stays




any hints how i can build an easy enclosier ?






integration multivariable-calculus greens-theorem







share|cite|improve this question











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    1












    $begingroup$



    Compute $int_{C}(2x^2+1)e^{x^2+y^2}dx+(2xy)e^{x^2+y^2}dy$ where $C$ connects $(1,0)$ to $(0,1)$ by a straight line segment.




    I tried to use green theorem since $Q_x = P_y$ so $int vec F_dot{}vec dr=0$



    i need to close $C$ so i build a triangle with vertices
    $(0,0)$ , $(1,0)$ ,$(0,1)$ now i got a complicated integral in the line parametrization :



    $y = 0 ~~~ dy=0$



    $x = t ~~~ dx=dt$



    $0leq t leq 1$




    $int_0^1 (2t^2 + 1)e^{t^2}dt$ , since $dy=0$ only the left side stays




    any hints how i can build an easy enclosier ?






    integration multivariable-calculus greens-theorem







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$



      Compute $int_{C}(2x^2+1)e^{x^2+y^2}dx+(2xy)e^{x^2+y^2}dy$ where $C$ connects $(1,0)$ to $(0,1)$ by a straight line segment.




      I tried to use green theorem since $Q_x = P_y$ so $int vec F_dot{}vec dr=0$



      i need to close $C$ so i build a triangle with vertices
      $(0,0)$ , $(1,0)$ ,$(0,1)$ now i got a complicated integral in the line parametrization :



      $y = 0 ~~~ dy=0$



      $x = t ~~~ dx=dt$



      $0leq t leq 1$




      $int_0^1 (2t^2 + 1)e^{t^2}dt$ , since $dy=0$ only the left side stays




      any hints how i can build an easy enclosier ?






      integration multivariable-calculus greens-theorem







      share|cite|improve this question











      $endgroup$





      Compute $int_{C}(2x^2+1)e^{x^2+y^2}dx+(2xy)e^{x^2+y^2}dy$ where $C$ connects $(1,0)$ to $(0,1)$ by a straight line segment.




      I tried to use green theorem since $Q_x = P_y$ so $int vec F_dot{}vec dr=0$



      i need to close $C$ so i build a triangle with vertices
      $(0,0)$ , $(1,0)$ ,$(0,1)$ now i got a complicated integral in the line parametrization :



      $y = 0 ~~~ dy=0$



      $x = t ~~~ dx=dt$



      $0leq t leq 1$




      $int_0^1 (2t^2 + 1)e^{t^2}dt$ , since $dy=0$ only the left side stays




      any hints how i can build an easy enclosier ?






      integration multivariable-calculus greens-theorem




      integration multivariable-calculus greens-theorem






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 8 at 14:27







      Mather

















      asked Jan 8 at 14:21









      Mather Mather

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      3047






















          3 Answers
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          active

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          2












          $begingroup$

          Since $e^{x^2+y^2}$ is a radial function, it is better to choose $D:(cos t,sin t), 0le tle frac{pi}{2}$ and use the fact
          $$begin{eqnarray}
          int_{C}(2x^2+1)e^{x^2+y^2}dx+(2xy)e^{x^2+y^2}dy&=&int_{D}(2x^2+1)e^{x^2+y^2}dx+(2xy)e^{x^2+y^2}dy\
          &=&eint_D (2x^2+1)dx + 2xydy\
          &=&eleft[left(frac{2}{3}x^3+xright)Big|^0_1+frac{2}{3}right]\
          &=&-e.
          end{eqnarray}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            yes i got it now alone thaaaaanks ! i am glad to notice that
            $endgroup$
            – Mather
            Jan 8 at 14:40



















          1












          $begingroup$

          Seeing $x^2+y^2$ in the expression hints at circles and polar coordinates. I would suggest connecting the two given points with the quarter of the circle $x^2+y^2=1$ in the first quadrant. Then you can parametrize it as $x=cos t$, $y=sin t$, for $0le tledfrac{pi}{2}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you i noticed that right after asking the question !
            $endgroup$
            – Mather
            Jan 8 at 14:40










          • $begingroup$
            i have question though can i use implicit function theorem ?
            $endgroup$
            – Mather
            Jan 8 at 14:42










          • $begingroup$
            $ x^2 + y^2 = 1$ $2x + 2yy' = 0$ $dy = frac{-x}{y} dx$
            $endgroup$
            – Mather
            Jan 8 at 14:43












          • $begingroup$
            @Mather: Use it for what? Here you need to find both $dx$ and $dy$ with respect to $t$ from the chosen parametrization, not with respect to each other. For example, $x=cos t$, so $dx=-sin t,dt$.
            $endgroup$
            – zipirovich
            Jan 8 at 14:46



















          0












          $begingroup$

          Since $int x e^{x^2 + y^2} d(y^2) = x e^{x^2 + y^2} + C(x)$, we can see that $phi(x, y) = x e^{x^2 + y^2}$ is a potential for $mathbf F$. Then, by the gradient theorem,
          $$I = int_C (nabla phi) cdot dmathbf r = phi(0, 1) - phi(1, 0) = -e.$$






          share|cite|improve this answer











          $endgroup$













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            3 Answers
            3






            active

            oldest

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            3 Answers
            3






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

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            2












            $begingroup$

            Since $e^{x^2+y^2}$ is a radial function, it is better to choose $D:(cos t,sin t), 0le tle frac{pi}{2}$ and use the fact
            $$begin{eqnarray}
            int_{C}(2x^2+1)e^{x^2+y^2}dx+(2xy)e^{x^2+y^2}dy&=&int_{D}(2x^2+1)e^{x^2+y^2}dx+(2xy)e^{x^2+y^2}dy\
            &=&eint_D (2x^2+1)dx + 2xydy\
            &=&eleft[left(frac{2}{3}x^3+xright)Big|^0_1+frac{2}{3}right]\
            &=&-e.
            end{eqnarray}$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              yes i got it now alone thaaaaanks ! i am glad to notice that
              $endgroup$
              – Mather
              Jan 8 at 14:40
















            2












            $begingroup$

            Since $e^{x^2+y^2}$ is a radial function, it is better to choose $D:(cos t,sin t), 0le tle frac{pi}{2}$ and use the fact
            $$begin{eqnarray}
            int_{C}(2x^2+1)e^{x^2+y^2}dx+(2xy)e^{x^2+y^2}dy&=&int_{D}(2x^2+1)e^{x^2+y^2}dx+(2xy)e^{x^2+y^2}dy\
            &=&eint_D (2x^2+1)dx + 2xydy\
            &=&eleft[left(frac{2}{3}x^3+xright)Big|^0_1+frac{2}{3}right]\
            &=&-e.
            end{eqnarray}$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              yes i got it now alone thaaaaanks ! i am glad to notice that
              $endgroup$
              – Mather
              Jan 8 at 14:40














            2












            2








            2





            $begingroup$

            Since $e^{x^2+y^2}$ is a radial function, it is better to choose $D:(cos t,sin t), 0le tle frac{pi}{2}$ and use the fact
            $$begin{eqnarray}
            int_{C}(2x^2+1)e^{x^2+y^2}dx+(2xy)e^{x^2+y^2}dy&=&int_{D}(2x^2+1)e^{x^2+y^2}dx+(2xy)e^{x^2+y^2}dy\
            &=&eint_D (2x^2+1)dx + 2xydy\
            &=&eleft[left(frac{2}{3}x^3+xright)Big|^0_1+frac{2}{3}right]\
            &=&-e.
            end{eqnarray}$$






            share|cite|improve this answer











            $endgroup$



            Since $e^{x^2+y^2}$ is a radial function, it is better to choose $D:(cos t,sin t), 0le tle frac{pi}{2}$ and use the fact
            $$begin{eqnarray}
            int_{C}(2x^2+1)e^{x^2+y^2}dx+(2xy)e^{x^2+y^2}dy&=&int_{D}(2x^2+1)e^{x^2+y^2}dx+(2xy)e^{x^2+y^2}dy\
            &=&eint_D (2x^2+1)dx + 2xydy\
            &=&eleft[left(frac{2}{3}x^3+xright)Big|^0_1+frac{2}{3}right]\
            &=&-e.
            end{eqnarray}$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 8 at 14:44

























            answered Jan 8 at 14:38









            SongSong

            11.2k628




            11.2k628












            • $begingroup$
              yes i got it now alone thaaaaanks ! i am glad to notice that
              $endgroup$
              – Mather
              Jan 8 at 14:40


















            • $begingroup$
              yes i got it now alone thaaaaanks ! i am glad to notice that
              $endgroup$
              – Mather
              Jan 8 at 14:40
















            $begingroup$
            yes i got it now alone thaaaaanks ! i am glad to notice that
            $endgroup$
            – Mather
            Jan 8 at 14:40




            $begingroup$
            yes i got it now alone thaaaaanks ! i am glad to notice that
            $endgroup$
            – Mather
            Jan 8 at 14:40











            1












            $begingroup$

            Seeing $x^2+y^2$ in the expression hints at circles and polar coordinates. I would suggest connecting the two given points with the quarter of the circle $x^2+y^2=1$ in the first quadrant. Then you can parametrize it as $x=cos t$, $y=sin t$, for $0le tledfrac{pi}{2}$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              thank you i noticed that right after asking the question !
              $endgroup$
              – Mather
              Jan 8 at 14:40










            • $begingroup$
              i have question though can i use implicit function theorem ?
              $endgroup$
              – Mather
              Jan 8 at 14:42










            • $begingroup$
              $ x^2 + y^2 = 1$ $2x + 2yy' = 0$ $dy = frac{-x}{y} dx$
              $endgroup$
              – Mather
              Jan 8 at 14:43












            • $begingroup$
              @Mather: Use it for what? Here you need to find both $dx$ and $dy$ with respect to $t$ from the chosen parametrization, not with respect to each other. For example, $x=cos t$, so $dx=-sin t,dt$.
              $endgroup$
              – zipirovich
              Jan 8 at 14:46
















            1












            $begingroup$

            Seeing $x^2+y^2$ in the expression hints at circles and polar coordinates. I would suggest connecting the two given points with the quarter of the circle $x^2+y^2=1$ in the first quadrant. Then you can parametrize it as $x=cos t$, $y=sin t$, for $0le tledfrac{pi}{2}$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              thank you i noticed that right after asking the question !
              $endgroup$
              – Mather
              Jan 8 at 14:40










            • $begingroup$
              i have question though can i use implicit function theorem ?
              $endgroup$
              – Mather
              Jan 8 at 14:42










            • $begingroup$
              $ x^2 + y^2 = 1$ $2x + 2yy' = 0$ $dy = frac{-x}{y} dx$
              $endgroup$
              – Mather
              Jan 8 at 14:43












            • $begingroup$
              @Mather: Use it for what? Here you need to find both $dx$ and $dy$ with respect to $t$ from the chosen parametrization, not with respect to each other. For example, $x=cos t$, so $dx=-sin t,dt$.
              $endgroup$
              – zipirovich
              Jan 8 at 14:46














            1












            1








            1





            $begingroup$

            Seeing $x^2+y^2$ in the expression hints at circles and polar coordinates. I would suggest connecting the two given points with the quarter of the circle $x^2+y^2=1$ in the first quadrant. Then you can parametrize it as $x=cos t$, $y=sin t$, for $0le tledfrac{pi}{2}$.






            share|cite|improve this answer









            $endgroup$



            Seeing $x^2+y^2$ in the expression hints at circles and polar coordinates. I would suggest connecting the two given points with the quarter of the circle $x^2+y^2=1$ in the first quadrant. Then you can parametrize it as $x=cos t$, $y=sin t$, for $0le tledfrac{pi}{2}$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 8 at 14:38









            zipirovichzipirovich

            11.2k11631




            11.2k11631












            • $begingroup$
              thank you i noticed that right after asking the question !
              $endgroup$
              – Mather
              Jan 8 at 14:40










            • $begingroup$
              i have question though can i use implicit function theorem ?
              $endgroup$
              – Mather
              Jan 8 at 14:42










            • $begingroup$
              $ x^2 + y^2 = 1$ $2x + 2yy' = 0$ $dy = frac{-x}{y} dx$
              $endgroup$
              – Mather
              Jan 8 at 14:43












            • $begingroup$
              @Mather: Use it for what? Here you need to find both $dx$ and $dy$ with respect to $t$ from the chosen parametrization, not with respect to each other. For example, $x=cos t$, so $dx=-sin t,dt$.
              $endgroup$
              – zipirovich
              Jan 8 at 14:46


















            • $begingroup$
              thank you i noticed that right after asking the question !
              $endgroup$
              – Mather
              Jan 8 at 14:40










            • $begingroup$
              i have question though can i use implicit function theorem ?
              $endgroup$
              – Mather
              Jan 8 at 14:42










            • $begingroup$
              $ x^2 + y^2 = 1$ $2x + 2yy' = 0$ $dy = frac{-x}{y} dx$
              $endgroup$
              – Mather
              Jan 8 at 14:43












            • $begingroup$
              @Mather: Use it for what? Here you need to find both $dx$ and $dy$ with respect to $t$ from the chosen parametrization, not with respect to each other. For example, $x=cos t$, so $dx=-sin t,dt$.
              $endgroup$
              – zipirovich
              Jan 8 at 14:46
















            $begingroup$
            thank you i noticed that right after asking the question !
            $endgroup$
            – Mather
            Jan 8 at 14:40




            $begingroup$
            thank you i noticed that right after asking the question !
            $endgroup$
            – Mather
            Jan 8 at 14:40












            $begingroup$
            i have question though can i use implicit function theorem ?
            $endgroup$
            – Mather
            Jan 8 at 14:42




            $begingroup$
            i have question though can i use implicit function theorem ?
            $endgroup$
            – Mather
            Jan 8 at 14:42












            $begingroup$
            $ x^2 + y^2 = 1$ $2x + 2yy' = 0$ $dy = frac{-x}{y} dx$
            $endgroup$
            – Mather
            Jan 8 at 14:43






            $begingroup$
            $ x^2 + y^2 = 1$ $2x + 2yy' = 0$ $dy = frac{-x}{y} dx$
            $endgroup$
            – Mather
            Jan 8 at 14:43














            $begingroup$
            @Mather: Use it for what? Here you need to find both $dx$ and $dy$ with respect to $t$ from the chosen parametrization, not with respect to each other. For example, $x=cos t$, so $dx=-sin t,dt$.
            $endgroup$
            – zipirovich
            Jan 8 at 14:46




            $begingroup$
            @Mather: Use it for what? Here you need to find both $dx$ and $dy$ with respect to $t$ from the chosen parametrization, not with respect to each other. For example, $x=cos t$, so $dx=-sin t,dt$.
            $endgroup$
            – zipirovich
            Jan 8 at 14:46











            0












            $begingroup$

            Since $int x e^{x^2 + y^2} d(y^2) = x e^{x^2 + y^2} + C(x)$, we can see that $phi(x, y) = x e^{x^2 + y^2}$ is a potential for $mathbf F$. Then, by the gradient theorem,
            $$I = int_C (nabla phi) cdot dmathbf r = phi(0, 1) - phi(1, 0) = -e.$$






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              Since $int x e^{x^2 + y^2} d(y^2) = x e^{x^2 + y^2} + C(x)$, we can see that $phi(x, y) = x e^{x^2 + y^2}$ is a potential for $mathbf F$. Then, by the gradient theorem,
              $$I = int_C (nabla phi) cdot dmathbf r = phi(0, 1) - phi(1, 0) = -e.$$






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                Since $int x e^{x^2 + y^2} d(y^2) = x e^{x^2 + y^2} + C(x)$, we can see that $phi(x, y) = x e^{x^2 + y^2}$ is a potential for $mathbf F$. Then, by the gradient theorem,
                $$I = int_C (nabla phi) cdot dmathbf r = phi(0, 1) - phi(1, 0) = -e.$$






                share|cite|improve this answer











                $endgroup$



                Since $int x e^{x^2 + y^2} d(y^2) = x e^{x^2 + y^2} + C(x)$, we can see that $phi(x, y) = x e^{x^2 + y^2}$ is a potential for $mathbf F$. Then, by the gradient theorem,
                $$I = int_C (nabla phi) cdot dmathbf r = phi(0, 1) - phi(1, 0) = -e.$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 8 at 20:59

























                answered Jan 8 at 16:41









                MaximMaxim

                5,1381219




                5,1381219






























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