How to calculate the line integral with respect to the circle in counterclockwise direction












1














Consider the vector field $F=<y,-x>$.



Compute the line integral $int_cFcdot dr$ where $C$ is the circle of radius $3$ centered at the origin counterclockwise.



My Try:



The circle is $x^2+y^2=9$



$x=3cos t$



$y=3sin t$ for $0le tle2pi$



Now how do I calculate $int_cFcdot dr$?



Can anyone explain how to solve this.






calculus integration multivariable-calculus line-integrals







share|cite|improve this question






















  • Just follow the definition. Can you look up the definition?
    – Zachary Selk
    Nov 26 at 2:09










  • @ZacharySelk Here $F=<y,-x>$. Then what about $r(t)$?
    – user982787
    Nov 26 at 2:16
















1














Consider the vector field $F=<y,-x>$.



Compute the line integral $int_cFcdot dr$ where $C$ is the circle of radius $3$ centered at the origin counterclockwise.



My Try:



The circle is $x^2+y^2=9$



$x=3cos t$



$y=3sin t$ for $0le tle2pi$



Now how do I calculate $int_cFcdot dr$?



Can anyone explain how to solve this.






calculus integration multivariable-calculus line-integrals







share|cite|improve this question






















  • Just follow the definition. Can you look up the definition?
    – Zachary Selk
    Nov 26 at 2:09










  • @ZacharySelk Here $F=<y,-x>$. Then what about $r(t)$?
    – user982787
    Nov 26 at 2:16














1












1








1


2





Consider the vector field $F=<y,-x>$.



Compute the line integral $int_cFcdot dr$ where $C$ is the circle of radius $3$ centered at the origin counterclockwise.



My Try:



The circle is $x^2+y^2=9$



$x=3cos t$



$y=3sin t$ for $0le tle2pi$



Now how do I calculate $int_cFcdot dr$?



Can anyone explain how to solve this.






calculus integration multivariable-calculus line-integrals







share|cite|improve this question













Consider the vector field $F=<y,-x>$.



Compute the line integral $int_cFcdot dr$ where $C$ is the circle of radius $3$ centered at the origin counterclockwise.



My Try:



The circle is $x^2+y^2=9$



$x=3cos t$



$y=3sin t$ for $0le tle2pi$



Now how do I calculate $int_cFcdot dr$?



Can anyone explain how to solve this.






calculus integration multivariable-calculus line-integrals




calculus integration multivariable-calculus line-integrals






share|cite|improve this question













share|cite|improve this question











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asked Nov 26 at 2:02









user982787

1117




1117












  • Just follow the definition. Can you look up the definition?
    – Zachary Selk
    Nov 26 at 2:09










  • @ZacharySelk Here $F=<y,-x>$. Then what about $r(t)$?
    – user982787
    Nov 26 at 2:16


















  • Just follow the definition. Can you look up the definition?
    – Zachary Selk
    Nov 26 at 2:09










  • @ZacharySelk Here $F=<y,-x>$. Then what about $r(t)$?
    – user982787
    Nov 26 at 2:16
















Just follow the definition. Can you look up the definition?
– Zachary Selk
Nov 26 at 2:09




Just follow the definition. Can you look up the definition?
– Zachary Selk
Nov 26 at 2:09












@ZacharySelk Here $F=<y,-x>$. Then what about $r(t)$?
– user982787
Nov 26 at 2:16




@ZacharySelk Here $F=<y,-x>$. Then what about $r(t)$?
– user982787
Nov 26 at 2:16










1 Answer
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The radial vector is



$vec r = begin{pmatrix} x \ y end{pmatrix}; tag 1$



also,



$F = begin{pmatrix} y \ - x end{pmatrix}; tag 2$



with



$x = 3 cos t, tag 3$



$y = 3 sin t, tag 4$



we have



$vec r = begin{pmatrix} 3 cos t \ 3sin t end{pmatrix}, tag 5$



$d vec r = dfrac{d vec r}{dt} dt = begin{pmatrix} -3sin t \ 3cos t end{pmatrix} dt, tag 5$



and



$F = begin{pmatrix} 3sin t \ -3cos t end{pmatrix}; tag 6$



then



$F cdot d vec r = begin{pmatrix} 3sin t \ -3cos t end{pmatrix} cdot begin{pmatrix} -3sin t \ 3cos t end{pmatrix} dt$
$= (-9 sin^2 t - 9 cos^2 t)dt = -9(sin^2 t + cos^2 t)dt = -9(1)dt = -9dt; tag 7$



finally,



$displaystyle int_C F cdot d vec r = int_C -9 dt = -9 int_C dt = -9(2pi) = -18pi. tag 8$






share|cite|improve this answer





















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    1 Answer
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    The radial vector is



    $vec r = begin{pmatrix} x \ y end{pmatrix}; tag 1$



    also,



    $F = begin{pmatrix} y \ - x end{pmatrix}; tag 2$



    with



    $x = 3 cos t, tag 3$



    $y = 3 sin t, tag 4$



    we have



    $vec r = begin{pmatrix} 3 cos t \ 3sin t end{pmatrix}, tag 5$



    $d vec r = dfrac{d vec r}{dt} dt = begin{pmatrix} -3sin t \ 3cos t end{pmatrix} dt, tag 5$



    and



    $F = begin{pmatrix} 3sin t \ -3cos t end{pmatrix}; tag 6$



    then



    $F cdot d vec r = begin{pmatrix} 3sin t \ -3cos t end{pmatrix} cdot begin{pmatrix} -3sin t \ 3cos t end{pmatrix} dt$
    $= (-9 sin^2 t - 9 cos^2 t)dt = -9(sin^2 t + cos^2 t)dt = -9(1)dt = -9dt; tag 7$



    finally,



    $displaystyle int_C F cdot d vec r = int_C -9 dt = -9 int_C dt = -9(2pi) = -18pi. tag 8$






    share|cite|improve this answer


























      1














      The radial vector is



      $vec r = begin{pmatrix} x \ y end{pmatrix}; tag 1$



      also,



      $F = begin{pmatrix} y \ - x end{pmatrix}; tag 2$



      with



      $x = 3 cos t, tag 3$



      $y = 3 sin t, tag 4$



      we have



      $vec r = begin{pmatrix} 3 cos t \ 3sin t end{pmatrix}, tag 5$



      $d vec r = dfrac{d vec r}{dt} dt = begin{pmatrix} -3sin t \ 3cos t end{pmatrix} dt, tag 5$



      and



      $F = begin{pmatrix} 3sin t \ -3cos t end{pmatrix}; tag 6$



      then



      $F cdot d vec r = begin{pmatrix} 3sin t \ -3cos t end{pmatrix} cdot begin{pmatrix} -3sin t \ 3cos t end{pmatrix} dt$
      $= (-9 sin^2 t - 9 cos^2 t)dt = -9(sin^2 t + cos^2 t)dt = -9(1)dt = -9dt; tag 7$



      finally,



      $displaystyle int_C F cdot d vec r = int_C -9 dt = -9 int_C dt = -9(2pi) = -18pi. tag 8$






      share|cite|improve this answer
























        1












        1








        1






        The radial vector is



        $vec r = begin{pmatrix} x \ y end{pmatrix}; tag 1$



        also,



        $F = begin{pmatrix} y \ - x end{pmatrix}; tag 2$



        with



        $x = 3 cos t, tag 3$



        $y = 3 sin t, tag 4$



        we have



        $vec r = begin{pmatrix} 3 cos t \ 3sin t end{pmatrix}, tag 5$



        $d vec r = dfrac{d vec r}{dt} dt = begin{pmatrix} -3sin t \ 3cos t end{pmatrix} dt, tag 5$



        and



        $F = begin{pmatrix} 3sin t \ -3cos t end{pmatrix}; tag 6$



        then



        $F cdot d vec r = begin{pmatrix} 3sin t \ -3cos t end{pmatrix} cdot begin{pmatrix} -3sin t \ 3cos t end{pmatrix} dt$
        $= (-9 sin^2 t - 9 cos^2 t)dt = -9(sin^2 t + cos^2 t)dt = -9(1)dt = -9dt; tag 7$



        finally,



        $displaystyle int_C F cdot d vec r = int_C -9 dt = -9 int_C dt = -9(2pi) = -18pi. tag 8$






        share|cite|improve this answer












        The radial vector is



        $vec r = begin{pmatrix} x \ y end{pmatrix}; tag 1$



        also,



        $F = begin{pmatrix} y \ - x end{pmatrix}; tag 2$



        with



        $x = 3 cos t, tag 3$



        $y = 3 sin t, tag 4$



        we have



        $vec r = begin{pmatrix} 3 cos t \ 3sin t end{pmatrix}, tag 5$



        $d vec r = dfrac{d vec r}{dt} dt = begin{pmatrix} -3sin t \ 3cos t end{pmatrix} dt, tag 5$



        and



        $F = begin{pmatrix} 3sin t \ -3cos t end{pmatrix}; tag 6$



        then



        $F cdot d vec r = begin{pmatrix} 3sin t \ -3cos t end{pmatrix} cdot begin{pmatrix} -3sin t \ 3cos t end{pmatrix} dt$
        $= (-9 sin^2 t - 9 cos^2 t)dt = -9(sin^2 t + cos^2 t)dt = -9(1)dt = -9dt; tag 7$



        finally,



        $displaystyle int_C F cdot d vec r = int_C -9 dt = -9 int_C dt = -9(2pi) = -18pi. tag 8$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 26 at 2:26









        Robert Lewis

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        43.5k22863






























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