Simple Curve Integral using Parametrization












1












$begingroup$


I know the answer but cannot for the love of everything figure out how the book got the answer and the solution is only the answer nothing step-by step at all.
$$
F(D) = F*dr
$$calculate curve integral using the parametrisation
where F is
$$
F(x,y,z) = yz, xz, xy
$$
over domain D, where D is
$$
(x,y,z) = cos(t), sin(t), t
$$
where t is between (0, pi/4)



So im supposed to use the parametrization that is already there



I get this:



$$
F(x,y,z) =tsin(t), tcos(t), sin(t)cos(t) $$
$$
r'(t) =-sin(t), cos(t), 1 $$



I know im supposed to multiply F with r'(t) but I can't simplify in any way that get the answer pi/8 that should be the final answer for the integral
$$
int_0^{pi/4}-tsin^2(t) + tcos^2(t) + cos(t)sin(t)$$



Any idea how to get from there to pi/8?






integration algebraic-curves parametrization







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I know the answer but cannot for the love of everything figure out how the book got the answer and the solution is only the answer nothing step-by step at all.
    $$
    F(D) = F*dr
    $$    calculate curve integral using the parametrisation
    where F is
    $$
    F(x,y,z) = yz, xz, xy
    $$
    over domain D, where D is
    $$
    (x,y,z) = cos(t), sin(t), t
    $$
    where t is between (0, pi/4)



    So im supposed to use the parametrization that is already there



    I get this:



    $$
    F(x,y,z) =tsin(t), tcos(t), sin(t)cos(t) $$
    $$
    r'(t) =-sin(t), cos(t), 1 $$



    I know im supposed to multiply F with r'(t) but I can't simplify in any way that get the answer pi/8 that should be the final answer for the integral
    $$
    int_0^{pi/4}-tsin^2(t) + tcos^2(t) + cos(t)sin(t)$$



    Any idea how to get from there to pi/8?






    integration algebraic-curves parametrization







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I know the answer but cannot for the love of everything figure out how the book got the answer and the solution is only the answer nothing step-by step at all.
      $$
      F(D) = F*dr
      $$      calculate curve integral using the parametrisation
      where F is
      $$
      F(x,y,z) = yz, xz, xy
      $$
      over domain D, where D is
      $$
      (x,y,z) = cos(t), sin(t), t
      $$
      where t is between (0, pi/4)



      So im supposed to use the parametrization that is already there



      I get this:



      $$
      F(x,y,z) =tsin(t), tcos(t), sin(t)cos(t) $$
      $$
      r'(t) =-sin(t), cos(t), 1 $$



      I know im supposed to multiply F with r'(t) but I can't simplify in any way that get the answer pi/8 that should be the final answer for the integral
      $$
      int_0^{pi/4}-tsin^2(t) + tcos^2(t) + cos(t)sin(t)$$



      Any idea how to get from there to pi/8?






      integration algebraic-curves parametrization







      share|cite|improve this question









      $endgroup$




      I know the answer but cannot for the love of everything figure out how the book got the answer and the solution is only the answer nothing step-by step at all.
      $$
      F(D) = F*dr
      $$      calculate curve integral using the parametrisation
      where F is
      $$
      F(x,y,z) = yz, xz, xy
      $$
      over domain D, where D is
      $$
      (x,y,z) = cos(t), sin(t), t
      $$
      where t is between (0, pi/4)



      So im supposed to use the parametrization that is already there



      I get this:



      $$
      F(x,y,z) =tsin(t), tcos(t), sin(t)cos(t) $$
      $$
      r'(t) =-sin(t), cos(t), 1 $$



      I know im supposed to multiply F with r'(t) but I can't simplify in any way that get the answer pi/8 that should be the final answer for the integral
      $$
      int_0^{pi/4}-tsin^2(t) + tcos^2(t) + cos(t)sin(t)$$



      Any idea how to get from there to pi/8?






      integration algebraic-curves parametrization




      integration algebraic-curves parametrization






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 18 '18 at 15:38









      user3514461user3514461

      82




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          1 Answer
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          $begingroup$

          You are given the vector field ${bf F}(x,y,z):=(yz,xz,xy)$ and the path $$gamma:quad {bf r}(t):=(cos t,sin t, t)qquadleft(0leq tleq{piover4}right) .$$ The line integral of thie field ${bf F}$ along $gamma$ is given by
          $$int_gamma{bf F}cdot d{bf r}=int_0^{piover4}{bf F}bigl({bf r}(t)bigr)cdot{bf r}'(t)>dt=int_0^{piover4}bigl(-tsin^2 t+tcos^2 t+sin tcos t)>dt={piover8} .$$
          For doing the integral you can use $t(cos^2 t-sin^2 t)=tcos(2t)$ and proceed by partial integration; furthermore $sin tcos t={1over2}sin(2t)$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Still get pi^2/16 + 1/4...
            $endgroup$
            – user3514461
            Dec 18 '18 at 19:35










          • $begingroup$
            Worked, had a lot of problem integrating the tcos2t but in the end only (1/2)tsin2t is left and setting t=pi/4
            $endgroup$
            – user3514461
            Dec 18 '18 at 21:32











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          1 Answer
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          oldest

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          1 Answer
          1






          active

          oldest

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          oldest

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          active

          oldest

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          1












          $begingroup$

          You are given the vector field ${bf F}(x,y,z):=(yz,xz,xy)$ and the path $$gamma:quad {bf r}(t):=(cos t,sin t, t)qquadleft(0leq tleq{piover4}right) .$$ The line integral of thie field ${bf F}$ along $gamma$ is given by
          $$int_gamma{bf F}cdot d{bf r}=int_0^{piover4}{bf F}bigl({bf r}(t)bigr)cdot{bf r}'(t)>dt=int_0^{piover4}bigl(-tsin^2 t+tcos^2 t+sin tcos t)>dt={piover8} .$$
          For doing the integral you can use $t(cos^2 t-sin^2 t)=tcos(2t)$ and proceed by partial integration; furthermore $sin tcos t={1over2}sin(2t)$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Still get pi^2/16 + 1/4...
            $endgroup$
            – user3514461
            Dec 18 '18 at 19:35










          • $begingroup$
            Worked, had a lot of problem integrating the tcos2t but in the end only (1/2)tsin2t is left and setting t=pi/4
            $endgroup$
            – user3514461
            Dec 18 '18 at 21:32
















          1












          $begingroup$

          You are given the vector field ${bf F}(x,y,z):=(yz,xz,xy)$ and the path $$gamma:quad {bf r}(t):=(cos t,sin t, t)qquadleft(0leq tleq{piover4}right) .$$ The line integral of thie field ${bf F}$ along $gamma$ is given by
          $$int_gamma{bf F}cdot d{bf r}=int_0^{piover4}{bf F}bigl({bf r}(t)bigr)cdot{bf r}'(t)>dt=int_0^{piover4}bigl(-tsin^2 t+tcos^2 t+sin tcos t)>dt={piover8} .$$
          For doing the integral you can use $t(cos^2 t-sin^2 t)=tcos(2t)$ and proceed by partial integration; furthermore $sin tcos t={1over2}sin(2t)$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Still get pi^2/16 + 1/4...
            $endgroup$
            – user3514461
            Dec 18 '18 at 19:35










          • $begingroup$
            Worked, had a lot of problem integrating the tcos2t but in the end only (1/2)tsin2t is left and setting t=pi/4
            $endgroup$
            – user3514461
            Dec 18 '18 at 21:32














          1












          1








          1





          $begingroup$

          You are given the vector field ${bf F}(x,y,z):=(yz,xz,xy)$ and the path $$gamma:quad {bf r}(t):=(cos t,sin t, t)qquadleft(0leq tleq{piover4}right) .$$ The line integral of thie field ${bf F}$ along $gamma$ is given by
          $$int_gamma{bf F}cdot d{bf r}=int_0^{piover4}{bf F}bigl({bf r}(t)bigr)cdot{bf r}'(t)>dt=int_0^{piover4}bigl(-tsin^2 t+tcos^2 t+sin tcos t)>dt={piover8} .$$
          For doing the integral you can use $t(cos^2 t-sin^2 t)=tcos(2t)$ and proceed by partial integration; furthermore $sin tcos t={1over2}sin(2t)$.






          share|cite|improve this answer











          $endgroup$



          You are given the vector field ${bf F}(x,y,z):=(yz,xz,xy)$ and the path $$gamma:quad {bf r}(t):=(cos t,sin t, t)qquadleft(0leq tleq{piover4}right) .$$ The line integral of thie field ${bf F}$ along $gamma$ is given by
          $$int_gamma{bf F}cdot d{bf r}=int_0^{piover4}{bf F}bigl({bf r}(t)bigr)cdot{bf r}'(t)>dt=int_0^{piover4}bigl(-tsin^2 t+tcos^2 t+sin tcos t)>dt={piover8} .$$
          For doing the integral you can use $t(cos^2 t-sin^2 t)=tcos(2t)$ and proceed by partial integration; furthermore $sin tcos t={1over2}sin(2t)$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 18 '18 at 20:00

























          answered Dec 18 '18 at 17:10









          Christian BlatterChristian Blatter

          175k8115327




          175k8115327












          • $begingroup$
            Still get pi^2/16 + 1/4...
            $endgroup$
            – user3514461
            Dec 18 '18 at 19:35










          • $begingroup$
            Worked, had a lot of problem integrating the tcos2t but in the end only (1/2)tsin2t is left and setting t=pi/4
            $endgroup$
            – user3514461
            Dec 18 '18 at 21:32


















          • $begingroup$
            Still get pi^2/16 + 1/4...
            $endgroup$
            – user3514461
            Dec 18 '18 at 19:35










          • $begingroup$
            Worked, had a lot of problem integrating the tcos2t but in the end only (1/2)tsin2t is left and setting t=pi/4
            $endgroup$
            – user3514461
            Dec 18 '18 at 21:32
















          $begingroup$
          Still get pi^2/16 + 1/4...
          $endgroup$
          – user3514461
          Dec 18 '18 at 19:35




          $begingroup$
          Still get pi^2/16 + 1/4...
          $endgroup$
          – user3514461
          Dec 18 '18 at 19:35












          $begingroup$
          Worked, had a lot of problem integrating the tcos2t but in the end only (1/2)tsin2t is left and setting t=pi/4
          $endgroup$
          – user3514461
          Dec 18 '18 at 21:32




          $begingroup$
          Worked, had a lot of problem integrating the tcos2t but in the end only (1/2)tsin2t is left and setting t=pi/4
          $endgroup$
          – user3514461
          Dec 18 '18 at 21:32


















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