Determine $int_C Fcdot dx$ where $F(x,y,z)=(z,-y,x)$ and $C$ is the line segment from $(5,0,2)$ to $(5,3,4)$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Use Stokes' Theorem to evaluate integralEvaluate the line integral $int_C x^2 dx+(x+y)dy $Line integrals and path independenceFind a plane that passes through a given point and contains a given lineParametric line segment in 3-spaceEvaluate the line integral $int_C (x^2-y+3z);ds$Maximizing the line integral over a line segmentWork on a line segment - Final stepFinding the distance between a point and line (3-D)Formula of curvature not defined in arc length

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Determine $int_C Fcdot dx$ where $F(x,y,z)=(z,-y,x)$ and $C$ is the line segment from $(5,0,2)$ to $(5,3,4)$



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Use Stokes' Theorem to evaluate integralEvaluate the line integral $int_C x^2 dx+(x+y)dy $Line integrals and path independenceFind a plane that passes through a given point and contains a given lineParametric line segment in 3-spaceEvaluate the line integral $int_C (x^2-y+3z);ds$Maximizing the line integral over a line segmentWork on a line segment - Final stepFinding the distance between a point and line (3-D)Formula of curvature not defined in arc length










0












$begingroup$


So first thing I did was parametrize the line. I used $gamma(t)=(5,t,2+frac23t)$ For $tin [0,3]$



I then have $F(gamma(t))=(2+frac23t,-t,5)$



and $dotgamma(t)=(0,1,frac23)$



Then $int_C Fdx=int_0^3 langle F,Trangle=int_0^3 -t+frac103dt=-frac23$



I'm pretty sure my parametrization is incorrect as I dont think I should get this negative.










share|cite|improve this question











$endgroup$
















    0












    $begingroup$


    So first thing I did was parametrize the line. I used $gamma(t)=(5,t,2+frac23t)$ For $tin [0,3]$



    I then have $F(gamma(t))=(2+frac23t,-t,5)$



    and $dotgamma(t)=(0,1,frac23)$



    Then $int_C Fdx=int_0^3 langle F,Trangle=int_0^3 -t+frac103dt=-frac23$



    I'm pretty sure my parametrization is incorrect as I dont think I should get this negative.










    share|cite|improve this question











    $endgroup$














      0












      0








      0





      $begingroup$


      So first thing I did was parametrize the line. I used $gamma(t)=(5,t,2+frac23t)$ For $tin [0,3]$



      I then have $F(gamma(t))=(2+frac23t,-t,5)$



      and $dotgamma(t)=(0,1,frac23)$



      Then $int_C Fdx=int_0^3 langle F,Trangle=int_0^3 -t+frac103dt=-frac23$



      I'm pretty sure my parametrization is incorrect as I dont think I should get this negative.










      share|cite|improve this question











      $endgroup$




      So first thing I did was parametrize the line. I used $gamma(t)=(5,t,2+frac23t)$ For $tin [0,3]$



      I then have $F(gamma(t))=(2+frac23t,-t,5)$



      and $dotgamma(t)=(0,1,frac23)$



      Then $int_C Fdx=int_0^3 langle F,Trangle=int_0^3 -t+frac103dt=-frac23$



      I'm pretty sure my parametrization is incorrect as I dont think I should get this negative.







      multivariable-calculus






      share|cite|improve this question















      share|cite|improve this question













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      edited Mar 27 at 17:12







      AColoredReptile

















      asked Mar 27 at 17:06









      AColoredReptileAColoredReptile

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      401210




















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          $begingroup$

          Your calculations are correct; except the last step of the integration, which is $$int_0^3left(frac103-tright)mathrm dt=frac112.$$






          share|cite|improve this answer









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            $begingroup$

            Your calculations are correct; except the last step of the integration, which is $$int_0^3left(frac103-tright)mathrm dt=frac112.$$






            share|cite|improve this answer









            $endgroup$

















              2












              $begingroup$

              Your calculations are correct; except the last step of the integration, which is $$int_0^3left(frac103-tright)mathrm dt=frac112.$$






              share|cite|improve this answer









              $endgroup$















                2












                2








                2





                $begingroup$

                Your calculations are correct; except the last step of the integration, which is $$int_0^3left(frac103-tright)mathrm dt=frac112.$$






                share|cite|improve this answer









                $endgroup$



                Your calculations are correct; except the last step of the integration, which is $$int_0^3left(frac103-tright)mathrm dt=frac112.$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 27 at 17:11









                st.mathst.math

                1,268115




                1,268115



























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