Evaluate the path integral of $f(x, y) = y$ over the graph of the semicircle $y = sqrt{1-x^2}, -1 leq x leq...












1












$begingroup$


Evaluate the path integral of $f(x, y) = y$ over the graph of the semicircle $y = sqrt{1-x^2}, -1 leq x leq 1$



Solution attempt:



$f(x, y) = y$, along $y = sqrt{1-x^2}, -1 leq x leq 1$



$vec{r}(t) = (t, sqrt{1-t^2}), -1 leq t leq 1$



$vec{r}'(t) = (1, frac{-2t}{2sqrt{1-t^2}}) = (1, frac{-t}{sqrt{1-t^2}})$



$||vec{r}'(t)|| = sqrt{1+frac{t^2}{1-t^2}} = frac{1}{sqrt{1-t^2}}$



Show $int f(x, y) dt = int_{-1}^{1} sqrt{1-t^2}||vec{r}'(t)|| dt = int_{-1}^{1} sqrt{1-t^2}frac{1}{sqrt{1-t^2}} dt = 1+1 = 2$



therefore the value of the required path integral is 2



is this correct?






multivariable-calculus







share|cite|improve this question









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    1












    $begingroup$


    Evaluate the path integral of $f(x, y) = y$ over the graph of the semicircle $y = sqrt{1-x^2}, -1 leq x leq 1$



    Solution attempt:



    $f(x, y) = y$, along $y = sqrt{1-x^2}, -1 leq x leq 1$



    $vec{r}(t) = (t, sqrt{1-t^2}), -1 leq t leq 1$



    $vec{r}'(t) = (1, frac{-2t}{2sqrt{1-t^2}}) = (1, frac{-t}{sqrt{1-t^2}})$



    $||vec{r}'(t)|| = sqrt{1+frac{t^2}{1-t^2}} = frac{1}{sqrt{1-t^2}}$



    Show $int f(x, y) dt = int_{-1}^{1} sqrt{1-t^2}||vec{r}'(t)|| dt = int_{-1}^{1} sqrt{1-t^2}frac{1}{sqrt{1-t^2}} dt = 1+1 = 2$



    therefore the value of the required path integral is 2



    is this correct?






    multivariable-calculus







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Evaluate the path integral of $f(x, y) = y$ over the graph of the semicircle $y = sqrt{1-x^2}, -1 leq x leq 1$



      Solution attempt:



      $f(x, y) = y$, along $y = sqrt{1-x^2}, -1 leq x leq 1$



      $vec{r}(t) = (t, sqrt{1-t^2}), -1 leq t leq 1$



      $vec{r}'(t) = (1, frac{-2t}{2sqrt{1-t^2}}) = (1, frac{-t}{sqrt{1-t^2}})$



      $||vec{r}'(t)|| = sqrt{1+frac{t^2}{1-t^2}} = frac{1}{sqrt{1-t^2}}$



      Show $int f(x, y) dt = int_{-1}^{1} sqrt{1-t^2}||vec{r}'(t)|| dt = int_{-1}^{1} sqrt{1-t^2}frac{1}{sqrt{1-t^2}} dt = 1+1 = 2$



      therefore the value of the required path integral is 2



      is this correct?






      multivariable-calculus







      share|cite|improve this question









      $endgroup$




      Evaluate the path integral of $f(x, y) = y$ over the graph of the semicircle $y = sqrt{1-x^2}, -1 leq x leq 1$



      Solution attempt:



      $f(x, y) = y$, along $y = sqrt{1-x^2}, -1 leq x leq 1$



      $vec{r}(t) = (t, sqrt{1-t^2}), -1 leq t leq 1$



      $vec{r}'(t) = (1, frac{-2t}{2sqrt{1-t^2}}) = (1, frac{-t}{sqrt{1-t^2}})$



      $||vec{r}'(t)|| = sqrt{1+frac{t^2}{1-t^2}} = frac{1}{sqrt{1-t^2}}$



      Show $int f(x, y) dt = int_{-1}^{1} sqrt{1-t^2}||vec{r}'(t)|| dt = int_{-1}^{1} sqrt{1-t^2}frac{1}{sqrt{1-t^2}} dt = 1+1 = 2$



      therefore the value of the required path integral is 2



      is this correct?






      multivariable-calculus




      multivariable-calculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked Jan 28 at 5:06









      bobsag bobbobsag bob

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          1 Answer
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          $begingroup$

          You can also parameterise the curve $x=cost$, $y=sint$



          $$r'(t) = <-sint,cost>$$



          $$||r'(t)|| = 1$$



          $$F = sin t$$



          $$I = int_{0}^{pi} sint dt = -cost|_{0}^{pi} = 2$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            (+1) for typing faster than I
            $endgroup$
            – Mark Viola
            Jan 28 at 5:48











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          1 Answer
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          1 Answer
          1






          active

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          active

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          active

          oldest

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          2












          $begingroup$

          You can also parameterise the curve $x=cost$, $y=sint$



          $$r'(t) = <-sint,cost>$$



          $$||r'(t)|| = 1$$



          $$F = sin t$$



          $$I = int_{0}^{pi} sint dt = -cost|_{0}^{pi} = 2$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            (+1) for typing faster than I
            $endgroup$
            – Mark Viola
            Jan 28 at 5:48
















          2












          $begingroup$

          You can also parameterise the curve $x=cost$, $y=sint$



          $$r'(t) = <-sint,cost>$$



          $$||r'(t)|| = 1$$



          $$F = sin t$$



          $$I = int_{0}^{pi} sint dt = -cost|_{0}^{pi} = 2$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            (+1) for typing faster than I
            $endgroup$
            – Mark Viola
            Jan 28 at 5:48














          2












          2








          2





          $begingroup$

          You can also parameterise the curve $x=cost$, $y=sint$



          $$r'(t) = <-sint,cost>$$



          $$||r'(t)|| = 1$$



          $$F = sin t$$



          $$I = int_{0}^{pi} sint dt = -cost|_{0}^{pi} = 2$$






          share|cite|improve this answer









          $endgroup$



          You can also parameterise the curve $x=cost$, $y=sint$



          $$r'(t) = <-sint,cost>$$



          $$||r'(t)|| = 1$$



          $$F = sin t$$



          $$I = int_{0}^{pi} sint dt = -cost|_{0}^{pi} = 2$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 28 at 5:48









          Satish RamanathanSatish Ramanathan

          10k31323




          10k31323












          • $begingroup$
            (+1) for typing faster than I
            $endgroup$
            – Mark Viola
            Jan 28 at 5:48


















          • $begingroup$
            (+1) for typing faster than I
            $endgroup$
            – Mark Viola
            Jan 28 at 5:48
















          $begingroup$
          (+1) for typing faster than I
          $endgroup$
          – Mark Viola
          Jan 28 at 5:48




          $begingroup$
          (+1) for typing faster than I
          $endgroup$
          – Mark Viola
          Jan 28 at 5:48


















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