Calculate $int_{C} frac{x}{x^2+y^2} dx + frac{y}{x^2+y^2} dy~$ where $C$ is straight line segment connecting...












0












$begingroup$



Calculate $int_{C} frac{x}{x^2+y^2} dx + frac{y}{x^2+y^2} dy~$ where $C$ is straight line segment connecting $(1,1)$ to $(2,2)$




my question is , after calculating the integral using green theorem i got that $int_{C} frac{x}{x^2+y^2} dx frac{y}{x^2+y^2} dy= -ln(2)$



is it the right answer ? since we are connecting $(1,1)$ to $(2,2) $ AND NOT $(2,2)$ to $(1,1)$



so its question about the sign of the value.






integration multivariable-calculus greens-theorem







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$endgroup$








  • 2




    $begingroup$
    Green's theorem is for closed curves. Yours isn't. This is a conservative force field, so find the potential function.
    $endgroup$
    – B. Goddard
    Jan 9 at 14:36










  • $begingroup$
    I closed it with parametrization
    $endgroup$
    – Mather
    Jan 9 at 16:02










  • $begingroup$
    You should add that to your answer so we can see what you did wrong.
    $endgroup$
    – B. Goddard
    Jan 9 at 16:28










  • $begingroup$
    Because this is a conservative field, so the integral over a closed curve should be zero. Further, if you use Greens theorem, you get a double integral over a region of area zero, and so, for another reason, you should get zero.
    $endgroup$
    – B. Goddard
    Jan 9 at 18:30










  • $begingroup$
    but someone posted that the answer is $ln(2)$ @B.Goddard
    $endgroup$
    – Mather
    Jan 9 at 19:46
















0












$begingroup$



Calculate $int_{C} frac{x}{x^2+y^2} dx + frac{y}{x^2+y^2} dy~$ where $C$ is straight line segment connecting $(1,1)$ to $(2,2)$




my question is , after calculating the integral using green theorem i got that $int_{C} frac{x}{x^2+y^2} dx frac{y}{x^2+y^2} dy= -ln(2)$



is it the right answer ? since we are connecting $(1,1)$ to $(2,2) $ AND NOT $(2,2)$ to $(1,1)$



so its question about the sign of the value.






integration multivariable-calculus greens-theorem







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Green's theorem is for closed curves. Yours isn't. This is a conservative force field, so find the potential function.
    $endgroup$
    – B. Goddard
    Jan 9 at 14:36










  • $begingroup$
    I closed it with parametrization
    $endgroup$
    – Mather
    Jan 9 at 16:02










  • $begingroup$
    You should add that to your answer so we can see what you did wrong.
    $endgroup$
    – B. Goddard
    Jan 9 at 16:28










  • $begingroup$
    Because this is a conservative field, so the integral over a closed curve should be zero. Further, if you use Greens theorem, you get a double integral over a region of area zero, and so, for another reason, you should get zero.
    $endgroup$
    – B. Goddard
    Jan 9 at 18:30










  • $begingroup$
    but someone posted that the answer is $ln(2)$ @B.Goddard
    $endgroup$
    – Mather
    Jan 9 at 19:46














0












0








0





$begingroup$



Calculate $int_{C} frac{x}{x^2+y^2} dx + frac{y}{x^2+y^2} dy~$ where $C$ is straight line segment connecting $(1,1)$ to $(2,2)$




my question is , after calculating the integral using green theorem i got that $int_{C} frac{x}{x^2+y^2} dx frac{y}{x^2+y^2} dy= -ln(2)$



is it the right answer ? since we are connecting $(1,1)$ to $(2,2) $ AND NOT $(2,2)$ to $(1,1)$



so its question about the sign of the value.






integration multivariable-calculus greens-theorem







share|cite|improve this question









$endgroup$





Calculate $int_{C} frac{x}{x^2+y^2} dx + frac{y}{x^2+y^2} dy~$ where $C$ is straight line segment connecting $(1,1)$ to $(2,2)$




my question is , after calculating the integral using green theorem i got that $int_{C} frac{x}{x^2+y^2} dx frac{y}{x^2+y^2} dy= -ln(2)$



is it the right answer ? since we are connecting $(1,1)$ to $(2,2) $ AND NOT $(2,2)$ to $(1,1)$



so its question about the sign of the value.






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










asked Jan 9 at 14:30









Mather Mather

1947




1947








  • 2




    $begingroup$
    Green's theorem is for closed curves. Yours isn't. This is a conservative force field, so find the potential function.
    $endgroup$
    – B. Goddard
    Jan 9 at 14:36










  • $begingroup$
    I closed it with parametrization
    $endgroup$
    – Mather
    Jan 9 at 16:02










  • $begingroup$
    You should add that to your answer so we can see what you did wrong.
    $endgroup$
    – B. Goddard
    Jan 9 at 16:28










  • $begingroup$
    Because this is a conservative field, so the integral over a closed curve should be zero. Further, if you use Greens theorem, you get a double integral over a region of area zero, and so, for another reason, you should get zero.
    $endgroup$
    – B. Goddard
    Jan 9 at 18:30










  • $begingroup$
    but someone posted that the answer is $ln(2)$ @B.Goddard
    $endgroup$
    – Mather
    Jan 9 at 19:46














  • 2




    $begingroup$
    Green's theorem is for closed curves. Yours isn't. This is a conservative force field, so find the potential function.
    $endgroup$
    – B. Goddard
    Jan 9 at 14:36










  • $begingroup$
    I closed it with parametrization
    $endgroup$
    – Mather
    Jan 9 at 16:02










  • $begingroup$
    You should add that to your answer so we can see what you did wrong.
    $endgroup$
    – B. Goddard
    Jan 9 at 16:28










  • $begingroup$
    Because this is a conservative field, so the integral over a closed curve should be zero. Further, if you use Greens theorem, you get a double integral over a region of area zero, and so, for another reason, you should get zero.
    $endgroup$
    – B. Goddard
    Jan 9 at 18:30










  • $begingroup$
    but someone posted that the answer is $ln(2)$ @B.Goddard
    $endgroup$
    – Mather
    Jan 9 at 19:46








2




2




$begingroup$
Green's theorem is for closed curves. Yours isn't. This is a conservative force field, so find the potential function.
$endgroup$
– B. Goddard
Jan 9 at 14:36




$begingroup$
Green's theorem is for closed curves. Yours isn't. This is a conservative force field, so find the potential function.
$endgroup$
– B. Goddard
Jan 9 at 14:36












$begingroup$
I closed it with parametrization
$endgroup$
– Mather
Jan 9 at 16:02




$begingroup$
I closed it with parametrization
$endgroup$
– Mather
Jan 9 at 16:02












$begingroup$
You should add that to your answer so we can see what you did wrong.
$endgroup$
– B. Goddard
Jan 9 at 16:28




$begingroup$
You should add that to your answer so we can see what you did wrong.
$endgroup$
– B. Goddard
Jan 9 at 16:28












$begingroup$
Because this is a conservative field, so the integral over a closed curve should be zero. Further, if you use Greens theorem, you get a double integral over a region of area zero, and so, for another reason, you should get zero.
$endgroup$
– B. Goddard
Jan 9 at 18:30




$begingroup$
Because this is a conservative field, so the integral over a closed curve should be zero. Further, if you use Greens theorem, you get a double integral over a region of area zero, and so, for another reason, you should get zero.
$endgroup$
– B. Goddard
Jan 9 at 18:30












$begingroup$
but someone posted that the answer is $ln(2)$ @B.Goddard
$endgroup$
– Mather
Jan 9 at 19:46




$begingroup$
but someone posted that the answer is $ln(2)$ @B.Goddard
$endgroup$
– Mather
Jan 9 at 19:46










2 Answers
2






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The fundamental theorem of calculus tells you that if ${bf F} = nabla f$ in a simply connected region containing the curve, then $$int_C {bf F}cdot d{bf r}= f(b) - f(a)$$ where the curve $C$ begins at the point $a$ and ends at the point $b$.



Here $${bf F}(x,y) = left(frac{x}{x^2 + y^2}, frac y{x^2 + y^2} right). $$ Can you find a function $f(x,y)$ such that $${bf F}(x,y) = nabla f(x,y)?$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    $(1,1),(2,2)$ are joined by the line-segment $C:y=xin[1,2]$. The integral becomes
    $$int_Cfrac{xdx+ydy}{x^2+y^2}=int_Cfrac{2xdx}{2x^2}=int_1^2frac{dx}x=ln(2)$$



    Alternatively,
    $$int_Cfrac{xdx+ydy}{x^2+y^2}=int_Cfrac12cdotfrac{d(x^2+y^2)}{x^2+y^2}=frac12int_2^8frac{dm}m=frac12ln(m)Big|_2^8=ln(2)$$where $m=x^2+y^2$, that goes from $1^2+1^2to2^2+2^2$.






    share|cite|improve this answer









    $endgroup$













      Your 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 fundamental theorem of calculus tells you that if ${bf F} = nabla f$ in a simply connected region containing the curve, then $$int_C {bf F}cdot d{bf r}= f(b) - f(a)$$ where the curve $C$ begins at the point $a$ and ends at the point $b$.



      Here $${bf F}(x,y) = left(frac{x}{x^2 + y^2}, frac y{x^2 + y^2} right). $$ Can you find a function $f(x,y)$ such that $${bf F}(x,y) = nabla f(x,y)?$$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        The fundamental theorem of calculus tells you that if ${bf F} = nabla f$ in a simply connected region containing the curve, then $$int_C {bf F}cdot d{bf r}= f(b) - f(a)$$ where the curve $C$ begins at the point $a$ and ends at the point $b$.



        Here $${bf F}(x,y) = left(frac{x}{x^2 + y^2}, frac y{x^2 + y^2} right). $$ Can you find a function $f(x,y)$ such that $${bf F}(x,y) = nabla f(x,y)?$$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          The fundamental theorem of calculus tells you that if ${bf F} = nabla f$ in a simply connected region containing the curve, then $$int_C {bf F}cdot d{bf r}= f(b) - f(a)$$ where the curve $C$ begins at the point $a$ and ends at the point $b$.



          Here $${bf F}(x,y) = left(frac{x}{x^2 + y^2}, frac y{x^2 + y^2} right). $$ Can you find a function $f(x,y)$ such that $${bf F}(x,y) = nabla f(x,y)?$$






          share|cite|improve this answer









          $endgroup$



          The fundamental theorem of calculus tells you that if ${bf F} = nabla f$ in a simply connected region containing the curve, then $$int_C {bf F}cdot d{bf r}= f(b) - f(a)$$ where the curve $C$ begins at the point $a$ and ends at the point $b$.



          Here $${bf F}(x,y) = left(frac{x}{x^2 + y^2}, frac y{x^2 + y^2} right). $$ Can you find a function $f(x,y)$ such that $${bf F}(x,y) = nabla f(x,y)?$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 9 at 14:42









          User8128User8128

          10.7k1522




          10.7k1522























              0












              $begingroup$

              $(1,1),(2,2)$ are joined by the line-segment $C:y=xin[1,2]$. The integral becomes
              $$int_Cfrac{xdx+ydy}{x^2+y^2}=int_Cfrac{2xdx}{2x^2}=int_1^2frac{dx}x=ln(2)$$



              Alternatively,
              $$int_Cfrac{xdx+ydy}{x^2+y^2}=int_Cfrac12cdotfrac{d(x^2+y^2)}{x^2+y^2}=frac12int_2^8frac{dm}m=frac12ln(m)Big|_2^8=ln(2)$$where $m=x^2+y^2$, that goes from $1^2+1^2to2^2+2^2$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $(1,1),(2,2)$ are joined by the line-segment $C:y=xin[1,2]$. The integral becomes
                $$int_Cfrac{xdx+ydy}{x^2+y^2}=int_Cfrac{2xdx}{2x^2}=int_1^2frac{dx}x=ln(2)$$



                Alternatively,
                $$int_Cfrac{xdx+ydy}{x^2+y^2}=int_Cfrac12cdotfrac{d(x^2+y^2)}{x^2+y^2}=frac12int_2^8frac{dm}m=frac12ln(m)Big|_2^8=ln(2)$$where $m=x^2+y^2$, that goes from $1^2+1^2to2^2+2^2$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $(1,1),(2,2)$ are joined by the line-segment $C:y=xin[1,2]$. The integral becomes
                  $$int_Cfrac{xdx+ydy}{x^2+y^2}=int_Cfrac{2xdx}{2x^2}=int_1^2frac{dx}x=ln(2)$$



                  Alternatively,
                  $$int_Cfrac{xdx+ydy}{x^2+y^2}=int_Cfrac12cdotfrac{d(x^2+y^2)}{x^2+y^2}=frac12int_2^8frac{dm}m=frac12ln(m)Big|_2^8=ln(2)$$where $m=x^2+y^2$, that goes from $1^2+1^2to2^2+2^2$.






                  share|cite|improve this answer









                  $endgroup$



                  $(1,1),(2,2)$ are joined by the line-segment $C:y=xin[1,2]$. The integral becomes
                  $$int_Cfrac{xdx+ydy}{x^2+y^2}=int_Cfrac{2xdx}{2x^2}=int_1^2frac{dx}x=ln(2)$$



                  Alternatively,
                  $$int_Cfrac{xdx+ydy}{x^2+y^2}=int_Cfrac12cdotfrac{d(x^2+y^2)}{x^2+y^2}=frac12int_2^8frac{dm}m=frac12ln(m)Big|_2^8=ln(2)$$where $m=x^2+y^2$, that goes from $1^2+1^2to2^2+2^2$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 9 at 15:05









                  Shubham JohriShubham Johri

                  4,785717




                  4,785717






























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