integrate a differential form over the indicated smooth cube

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I'm supposed to integrate the differential form over the smooth cube and I'm not sure if I'm on the right track.



So suppose,



$ int_c x dy+ y dz$ where $c: [-1,1] rightarrow mathbb{R^{3}}$ is $c(t) = (sin(t), cos(t), t) = (x,y,z)$



I took the derivative of $c(t)$ so I have



$c'(t) = (cos(t), -sin(t), 1) = (dx,dy,dz)$



and then I substituted back into the equation which is



$ int_{[-1,1]} sin(t) (-sin(t))+ cos(t)) (1)$



$ int_{[-1,1]} -(sin(t))^{2}+ cos(t) $



So do I now take the antiderivative of this and evaluate from $-1$ to $1$?






differential-geometry vector-analysis differential-forms







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    0














    I'm supposed to integrate the differential form over the smooth cube and I'm not sure if I'm on the right track.



    So suppose,



    $ int_c x dy+ y dz$ where $c: [-1,1] rightarrow mathbb{R^{3}}$ is $c(t) = (sin(t), cos(t), t) = (x,y,z)$



    I took the derivative of $c(t)$ so I have



    $c'(t) = (cos(t), -sin(t), 1) = (dx,dy,dz)$



    and then I substituted back into the equation which is



    $ int_{[-1,1]} sin(t) (-sin(t))+ cos(t)) (1)$



    $ int_{[-1,1]} -(sin(t))^{2}+ cos(t) $



    So do I now take the antiderivative of this and evaluate from $-1$ to $1$?






    differential-geometry vector-analysis differential-forms







    share|cite|improve this question



























      0












      0








      0







      I'm supposed to integrate the differential form over the smooth cube and I'm not sure if I'm on the right track.



      So suppose,



      $ int_c x dy+ y dz$ where $c: [-1,1] rightarrow mathbb{R^{3}}$ is $c(t) = (sin(t), cos(t), t) = (x,y,z)$



      I took the derivative of $c(t)$ so I have



      $c'(t) = (cos(t), -sin(t), 1) = (dx,dy,dz)$



      and then I substituted back into the equation which is



      $ int_{[-1,1]} sin(t) (-sin(t))+ cos(t)) (1)$



      $ int_{[-1,1]} -(sin(t))^{2}+ cos(t) $



      So do I now take the antiderivative of this and evaluate from $-1$ to $1$?






      differential-geometry vector-analysis differential-forms







      share|cite|improve this question















      I'm supposed to integrate the differential form over the smooth cube and I'm not sure if I'm on the right track.



      So suppose,



      $ int_c x dy+ y dz$ where $c: [-1,1] rightarrow mathbb{R^{3}}$ is $c(t) = (sin(t), cos(t), t) = (x,y,z)$



      I took the derivative of $c(t)$ so I have



      $c'(t) = (cos(t), -sin(t), 1) = (dx,dy,dz)$



      and then I substituted back into the equation which is



      $ int_{[-1,1]} sin(t) (-sin(t))+ cos(t)) (1)$



      $ int_{[-1,1]} -(sin(t))^{2}+ cos(t) $



      So do I now take the antiderivative of this and evaluate from $-1$ to $1$?






      differential-geometry vector-analysis differential-forms




      differential-geometry vector-analysis differential-forms






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 1 at 0:30









      amWhy

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      191k28224439










      asked Dec 1 at 0:25









      usukidoll

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