Line Integral in second quadrant of Unit Circle












0












$begingroup$


If I am asked to compute



$$int_c F . dr$$



Where



$$F(x,y) = <d/dx f(x, y), d/dy f(x,y)>$$
and
$$f(x,y) =sin(x^3 + y^3)$$



and C is the portion of the unit circle in the second quadrant, oriented counterclockwise, how would I go about doing that?






line-integrals greens-theorem







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  • $begingroup$
    What is $F(x,y)$ ?
    $endgroup$
    – Nosrati
    Nov 3 '18 at 18:56










  • $begingroup$
    I had trouble using MathJax. F(x, y) equals the partial derivatives with respect to x and y (in vector form) of the function sin(x^3 + y^3)
    $endgroup$
    – Joshua
    Nov 3 '18 at 18:59










  • $begingroup$
    Can you use Green theorem?
    $endgroup$
    – Nosrati
    Nov 3 '18 at 19:01










  • $begingroup$
    Technically not because Green's theorem, as far as I know, would be used for enclosed structures, not simple lines.
    $endgroup$
    – Joshua
    Nov 3 '18 at 19:03










  • $begingroup$
    So do straightforward
    $endgroup$
    – Nosrati
    Nov 3 '18 at 19:04
















0












$begingroup$


If I am asked to compute



$$int_c F . dr$$



Where



$$F(x,y) = <d/dx f(x, y), d/dy f(x,y)>$$
and
$$f(x,y) =sin(x^3 + y^3)$$



and C is the portion of the unit circle in the second quadrant, oriented counterclockwise, how would I go about doing that?






line-integrals greens-theorem







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is $F(x,y)$ ?
    $endgroup$
    – Nosrati
    Nov 3 '18 at 18:56










  • $begingroup$
    I had trouble using MathJax. F(x, y) equals the partial derivatives with respect to x and y (in vector form) of the function sin(x^3 + y^3)
    $endgroup$
    – Joshua
    Nov 3 '18 at 18:59










  • $begingroup$
    Can you use Green theorem?
    $endgroup$
    – Nosrati
    Nov 3 '18 at 19:01










  • $begingroup$
    Technically not because Green's theorem, as far as I know, would be used for enclosed structures, not simple lines.
    $endgroup$
    – Joshua
    Nov 3 '18 at 19:03










  • $begingroup$
    So do straightforward
    $endgroup$
    – Nosrati
    Nov 3 '18 at 19:04














0












0








0





$begingroup$


If I am asked to compute



$$int_c F . dr$$



Where



$$F(x,y) = <d/dx f(x, y), d/dy f(x,y)>$$
and
$$f(x,y) =sin(x^3 + y^3)$$



and C is the portion of the unit circle in the second quadrant, oriented counterclockwise, how would I go about doing that?






line-integrals greens-theorem







share|cite|improve this question











$endgroup$




If I am asked to compute



$$int_c F . dr$$



Where



$$F(x,y) = <d/dx f(x, y), d/dy f(x,y)>$$
and
$$f(x,y) =sin(x^3 + y^3)$$



and C is the portion of the unit circle in the second quadrant, oriented counterclockwise, how would I go about doing that?






line-integrals greens-theorem




line-integrals greens-theorem






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 3 '18 at 18:59









Nosrati

26.5k62354




26.5k62354










asked Nov 3 '18 at 18:54









JoshuaJoshua

172




172












  • $begingroup$
    What is $F(x,y)$ ?
    $endgroup$
    – Nosrati
    Nov 3 '18 at 18:56










  • $begingroup$
    I had trouble using MathJax. F(x, y) equals the partial derivatives with respect to x and y (in vector form) of the function sin(x^3 + y^3)
    $endgroup$
    – Joshua
    Nov 3 '18 at 18:59










  • $begingroup$
    Can you use Green theorem?
    $endgroup$
    – Nosrati
    Nov 3 '18 at 19:01










  • $begingroup$
    Technically not because Green's theorem, as far as I know, would be used for enclosed structures, not simple lines.
    $endgroup$
    – Joshua
    Nov 3 '18 at 19:03










  • $begingroup$
    So do straightforward
    $endgroup$
    – Nosrati
    Nov 3 '18 at 19:04


















  • $begingroup$
    What is $F(x,y)$ ?
    $endgroup$
    – Nosrati
    Nov 3 '18 at 18:56










  • $begingroup$
    I had trouble using MathJax. F(x, y) equals the partial derivatives with respect to x and y (in vector form) of the function sin(x^3 + y^3)
    $endgroup$
    – Joshua
    Nov 3 '18 at 18:59










  • $begingroup$
    Can you use Green theorem?
    $endgroup$
    – Nosrati
    Nov 3 '18 at 19:01










  • $begingroup$
    Technically not because Green's theorem, as far as I know, would be used for enclosed structures, not simple lines.
    $endgroup$
    – Joshua
    Nov 3 '18 at 19:03










  • $begingroup$
    So do straightforward
    $endgroup$
    – Nosrati
    Nov 3 '18 at 19:04
















$begingroup$
What is $F(x,y)$ ?
$endgroup$
– Nosrati
Nov 3 '18 at 18:56




$begingroup$
What is $F(x,y)$ ?
$endgroup$
– Nosrati
Nov 3 '18 at 18:56












$begingroup$
I had trouble using MathJax. F(x, y) equals the partial derivatives with respect to x and y (in vector form) of the function sin(x^3 + y^3)
$endgroup$
– Joshua
Nov 3 '18 at 18:59




$begingroup$
I had trouble using MathJax. F(x, y) equals the partial derivatives with respect to x and y (in vector form) of the function sin(x^3 + y^3)
$endgroup$
– Joshua
Nov 3 '18 at 18:59












$begingroup$
Can you use Green theorem?
$endgroup$
– Nosrati
Nov 3 '18 at 19:01




$begingroup$
Can you use Green theorem?
$endgroup$
– Nosrati
Nov 3 '18 at 19:01












$begingroup$
Technically not because Green's theorem, as far as I know, would be used for enclosed structures, not simple lines.
$endgroup$
– Joshua
Nov 3 '18 at 19:03




$begingroup$
Technically not because Green's theorem, as far as I know, would be used for enclosed structures, not simple lines.
$endgroup$
– Joshua
Nov 3 '18 at 19:03












$begingroup$
So do straightforward
$endgroup$
– Nosrati
Nov 3 '18 at 19:04




$begingroup$
So do straightforward
$endgroup$
– Nosrati
Nov 3 '18 at 19:04










2 Answers
2






active

oldest

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1












$begingroup$

Let $r=(cos t,sin t)$ with $tin[dfrac{pi}{2},pi]$ be the parametrization of $C$. Then
begin{align}
int_C F.dr
&= int_{frac{pi}{2}}^{pi}left(3cos^2tcos(cos^3t+sin^3t),3sin^2tcos(cos^3t+sin^3t)right)(-sin t,cos t) dt \
&= int_{frac{pi}{2}}^{pi}left(-3cos^2tsin t+3sin^2tcos tright)cos(cos^3t+sin^3t) dt \
&= sin(cos^3t+sin^3t)Big|_{frac{pi}{2}}^{pi} \
&= color{blue}{-2sin1}
end{align}






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    You need the gradient theorem:
    $$int_gamma (nabla f) cdot dmathbf r =
    f(mathbf r_2) - f(mathbf r_1) =
    f(-1, 0) - f(0, 1) =
    -2 sin 1.$$






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Let $r=(cos t,sin t)$ with $tin[dfrac{pi}{2},pi]$ be the parametrization of $C$. Then
      begin{align}
      int_C F.dr
      &= int_{frac{pi}{2}}^{pi}left(3cos^2tcos(cos^3t+sin^3t),3sin^2tcos(cos^3t+sin^3t)right)(-sin t,cos t) dt \
      &= int_{frac{pi}{2}}^{pi}left(-3cos^2tsin t+3sin^2tcos tright)cos(cos^3t+sin^3t) dt \
      &= sin(cos^3t+sin^3t)Big|_{frac{pi}{2}}^{pi} \
      &= color{blue}{-2sin1}
      end{align}






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        Let $r=(cos t,sin t)$ with $tin[dfrac{pi}{2},pi]$ be the parametrization of $C$. Then
        begin{align}
        int_C F.dr
        &= int_{frac{pi}{2}}^{pi}left(3cos^2tcos(cos^3t+sin^3t),3sin^2tcos(cos^3t+sin^3t)right)(-sin t,cos t) dt \
        &= int_{frac{pi}{2}}^{pi}left(-3cos^2tsin t+3sin^2tcos tright)cos(cos^3t+sin^3t) dt \
        &= sin(cos^3t+sin^3t)Big|_{frac{pi}{2}}^{pi} \
        &= color{blue}{-2sin1}
        end{align}






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          Let $r=(cos t,sin t)$ with $tin[dfrac{pi}{2},pi]$ be the parametrization of $C$. Then
          begin{align}
          int_C F.dr
          &= int_{frac{pi}{2}}^{pi}left(3cos^2tcos(cos^3t+sin^3t),3sin^2tcos(cos^3t+sin^3t)right)(-sin t,cos t) dt \
          &= int_{frac{pi}{2}}^{pi}left(-3cos^2tsin t+3sin^2tcos tright)cos(cos^3t+sin^3t) dt \
          &= sin(cos^3t+sin^3t)Big|_{frac{pi}{2}}^{pi} \
          &= color{blue}{-2sin1}
          end{align}






          share|cite|improve this answer











          $endgroup$



          Let $r=(cos t,sin t)$ with $tin[dfrac{pi}{2},pi]$ be the parametrization of $C$. Then
          begin{align}
          int_C F.dr
          &= int_{frac{pi}{2}}^{pi}left(3cos^2tcos(cos^3t+sin^3t),3sin^2tcos(cos^3t+sin^3t)right)(-sin t,cos t) dt \
          &= int_{frac{pi}{2}}^{pi}left(-3cos^2tsin t+3sin^2tcos tright)cos(cos^3t+sin^3t) dt \
          &= sin(cos^3t+sin^3t)Big|_{frac{pi}{2}}^{pi} \
          &= color{blue}{-2sin1}
          end{align}







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 3 '18 at 19:27

























          answered Nov 3 '18 at 19:15









          NosratiNosrati

          26.5k62354




          26.5k62354























              0












              $begingroup$

              You need the gradient theorem:
              $$int_gamma (nabla f) cdot dmathbf r =
              f(mathbf r_2) - f(mathbf r_1) =
              f(-1, 0) - f(0, 1) =
              -2 sin 1.$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                You need the gradient theorem:
                $$int_gamma (nabla f) cdot dmathbf r =
                f(mathbf r_2) - f(mathbf r_1) =
                f(-1, 0) - f(0, 1) =
                -2 sin 1.$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  You need the gradient theorem:
                  $$int_gamma (nabla f) cdot dmathbf r =
                  f(mathbf r_2) - f(mathbf r_1) =
                  f(-1, 0) - f(0, 1) =
                  -2 sin 1.$$






                  share|cite|improve this answer









                  $endgroup$



                  You need the gradient theorem:
                  $$int_gamma (nabla f) cdot dmathbf r =
                  f(mathbf r_2) - f(mathbf r_1) =
                  f(-1, 0) - f(0, 1) =
                  -2 sin 1.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 2 '18 at 9:49









                  MaximMaxim

                  5,0881219




                  5,0881219






























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