Solve the integral $ int{y^2dl}$ where L:$x^2+y^2+z^2=a^2;$ $x+y+z=0$












0












$begingroup$


Solve the integral $ int{y^2dl}$ where L:$x^2+y^2+z^2=a^2;$ $x+y+z=0$.



I tried to apply a 2 parametrizations:



1)
$x= sqrt{frac{2t^2}{-3t-frac{a^2}{2}}}$
$y= sqrt{-frac{3}{2}t-frac{a^2}{4}}$
$z= sqrt{-6t-a^2}$



From this (using $int{f sqrt{(frac{dx}{dt})^2+(frac{dy}{dt})^2+(frac{dz}{dt})^2}}$ i got something like $int f(t^2) sqrt{frac{p(t^2)}{g^3(t)}}dt$
which I don't know how to solve.



2)
Moving to spherical coordinates (r, fi, theta):



$r=a$
and there is an equation with fi and theta (from $x+y+z=0)$ which seems too complex do deal with.



Now I hope to find the parametrization that will be more fitting.






integration parametrization







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Solve the integral $ int{y^2dl}$ where L:$x^2+y^2+z^2=a^2;$ $x+y+z=0$.



    I tried to apply a 2 parametrizations:



    1)
    $x= sqrt{frac{2t^2}{-3t-frac{a^2}{2}}}$
    $y= sqrt{-frac{3}{2}t-frac{a^2}{4}}$
    $z= sqrt{-6t-a^2}$



    From this (using $int{f sqrt{(frac{dx}{dt})^2+(frac{dy}{dt})^2+(frac{dz}{dt})^2}}$ i got something like $int f(t^2) sqrt{frac{p(t^2)}{g^3(t)}}dt$
    which I don't know how to solve.



    2)
    Moving to spherical coordinates (r, fi, theta):



    $r=a$
    and there is an equation with fi and theta (from $x+y+z=0)$ which seems too complex do deal with.



    Now I hope to find the parametrization that will be more fitting.






    integration parametrization







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Solve the integral $ int{y^2dl}$ where L:$x^2+y^2+z^2=a^2;$ $x+y+z=0$.



      I tried to apply a 2 parametrizations:



      1)
      $x= sqrt{frac{2t^2}{-3t-frac{a^2}{2}}}$
      $y= sqrt{-frac{3}{2}t-frac{a^2}{4}}$
      $z= sqrt{-6t-a^2}$



      From this (using $int{f sqrt{(frac{dx}{dt})^2+(frac{dy}{dt})^2+(frac{dz}{dt})^2}}$ i got something like $int f(t^2) sqrt{frac{p(t^2)}{g^3(t)}}dt$
      which I don't know how to solve.



      2)
      Moving to spherical coordinates (r, fi, theta):



      $r=a$
      and there is an equation with fi and theta (from $x+y+z=0)$ which seems too complex do deal with.



      Now I hope to find the parametrization that will be more fitting.






      integration parametrization







      share|cite|improve this question











      $endgroup$




      Solve the integral $ int{y^2dl}$ where L:$x^2+y^2+z^2=a^2;$ $x+y+z=0$.



      I tried to apply a 2 parametrizations:



      1)
      $x= sqrt{frac{2t^2}{-3t-frac{a^2}{2}}}$
      $y= sqrt{-frac{3}{2}t-frac{a^2}{4}}$
      $z= sqrt{-6t-a^2}$



      From this (using $int{f sqrt{(frac{dx}{dt})^2+(frac{dy}{dt})^2+(frac{dz}{dt})^2}}$ i got something like $int f(t^2) sqrt{frac{p(t^2)}{g^3(t)}}dt$
      which I don't know how to solve.



      2)
      Moving to spherical coordinates (r, fi, theta):



      $r=a$
      and there is an equation with fi and theta (from $x+y+z=0)$ which seems too complex do deal with.



      Now I hope to find the parametrization that will be more fitting.






      integration parametrization




      integration parametrization






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 6 '18 at 12:09







      user9393410

















      asked Dec 6 '18 at 12:00









      user9393410user9393410

      32




      32






















          1 Answer
          1






          active

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          2












          $begingroup$

          I think $int{x^2}d l$=$int{y^2}d l$=$int{z^2}d l$ by the symmetry of $L$，so
          $$int{y^2}d l=frac{1}{3}int{(x^2+y^2+z^2)}d l=frac{1}{3}int{a^2}d l$$
          And
          $$x=frac{sqrt{2}}{2}acos{t}-frac{sqrt{6}}{6}asin{t},,,y=frac{sqrt{6}}{3}asin{t},,,z=-frac{sqrt{2}}{2}acos{t}-frac{sqrt{6}}{6}asin{t}$$
          where $tin[0,2pi]$ will be a good parametrization.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Damn, that's a badass solution! How did you come up with this parametrization?
            $endgroup$
            – user9393410
            Dec 6 '18 at 13:12










          • $begingroup$
            Consider the projection of $L$ on $xOy$: $x^2+xy+y^2=frac{a^2}{2}$, which is also $(x+frac{y}{2})^2+frac{3}{4}y^2=frac{a^2}{2}$.
            $endgroup$
            – Lau
            Dec 6 '18 at 13:28










          • $begingroup$
            Oh, I see. Thank you!
            $endgroup$
            – user9393410
            Dec 6 '18 at 14:03











          Your Answer





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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          I think $int{x^2}d l$=$int{y^2}d l$=$int{z^2}d l$ by the symmetry of $L$，so
          $$int{y^2}d l=frac{1}{3}int{(x^2+y^2+z^2)}d l=frac{1}{3}int{a^2}d l$$
          And
          $$x=frac{sqrt{2}}{2}acos{t}-frac{sqrt{6}}{6}asin{t},,,y=frac{sqrt{6}}{3}asin{t},,,z=-frac{sqrt{2}}{2}acos{t}-frac{sqrt{6}}{6}asin{t}$$
          where $tin[0,2pi]$ will be a good parametrization.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Damn, that's a badass solution! How did you come up with this parametrization?
            $endgroup$
            – user9393410
            Dec 6 '18 at 13:12










          • $begingroup$
            Consider the projection of $L$ on $xOy$: $x^2+xy+y^2=frac{a^2}{2}$, which is also $(x+frac{y}{2})^2+frac{3}{4}y^2=frac{a^2}{2}$.
            $endgroup$
            – Lau
            Dec 6 '18 at 13:28










          • $begingroup$
            Oh, I see. Thank you!
            $endgroup$
            – user9393410
            Dec 6 '18 at 14:03
















          2












          $begingroup$

          I think $int{x^2}d l$=$int{y^2}d l$=$int{z^2}d l$ by the symmetry of $L$，so
          $$int{y^2}d l=frac{1}{3}int{(x^2+y^2+z^2)}d l=frac{1}{3}int{a^2}d l$$
          And
          $$x=frac{sqrt{2}}{2}acos{t}-frac{sqrt{6}}{6}asin{t},,,y=frac{sqrt{6}}{3}asin{t},,,z=-frac{sqrt{2}}{2}acos{t}-frac{sqrt{6}}{6}asin{t}$$
          where $tin[0,2pi]$ will be a good parametrization.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Damn, that's a badass solution! How did you come up with this parametrization?
            $endgroup$
            – user9393410
            Dec 6 '18 at 13:12










          • $begingroup$
            Consider the projection of $L$ on $xOy$: $x^2+xy+y^2=frac{a^2}{2}$, which is also $(x+frac{y}{2})^2+frac{3}{4}y^2=frac{a^2}{2}$.
            $endgroup$
            – Lau
            Dec 6 '18 at 13:28










          • $begingroup$
            Oh, I see. Thank you!
            $endgroup$
            – user9393410
            Dec 6 '18 at 14:03














          2












          2








          2





          $begingroup$

          I think $int{x^2}d l$=$int{y^2}d l$=$int{z^2}d l$ by the symmetry of $L$，so
          $$int{y^2}d l=frac{1}{3}int{(x^2+y^2+z^2)}d l=frac{1}{3}int{a^2}d l$$
          And
          $$x=frac{sqrt{2}}{2}acos{t}-frac{sqrt{6}}{6}asin{t},,,y=frac{sqrt{6}}{3}asin{t},,,z=-frac{sqrt{2}}{2}acos{t}-frac{sqrt{6}}{6}asin{t}$$
          where $tin[0,2pi]$ will be a good parametrization.






          share|cite|improve this answer











          $endgroup$



          I think $int{x^2}d l$=$int{y^2}d l$=$int{z^2}d l$ by the symmetry of $L$，so
          $$int{y^2}d l=frac{1}{3}int{(x^2+y^2+z^2)}d l=frac{1}{3}int{a^2}d l$$
          And
          $$x=frac{sqrt{2}}{2}acos{t}-frac{sqrt{6}}{6}asin{t},,,y=frac{sqrt{6}}{3}asin{t},,,z=-frac{sqrt{2}}{2}acos{t}-frac{sqrt{6}}{6}asin{t}$$
          where $tin[0,2pi]$ will be a good parametrization.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 29 '18 at 4:46

























          answered Dec 6 '18 at 12:17









          LauLau

          527315




          527315












          • $begingroup$
            Damn, that's a badass solution! How did you come up with this parametrization?
            $endgroup$
            – user9393410
            Dec 6 '18 at 13:12










          • $begingroup$
            Consider the projection of $L$ on $xOy$: $x^2+xy+y^2=frac{a^2}{2}$, which is also $(x+frac{y}{2})^2+frac{3}{4}y^2=frac{a^2}{2}$.
            $endgroup$
            – Lau
            Dec 6 '18 at 13:28










          • $begingroup$
            Oh, I see. Thank you!
            $endgroup$
            – user9393410
            Dec 6 '18 at 14:03


















          • $begingroup$
            Damn, that's a badass solution! How did you come up with this parametrization?
            $endgroup$
            – user9393410
            Dec 6 '18 at 13:12










          • $begingroup$
            Consider the projection of $L$ on $xOy$: $x^2+xy+y^2=frac{a^2}{2}$, which is also $(x+frac{y}{2})^2+frac{3}{4}y^2=frac{a^2}{2}$.
            $endgroup$
            – Lau
            Dec 6 '18 at 13:28










          • $begingroup$
            Oh, I see. Thank you!
            $endgroup$
            – user9393410
            Dec 6 '18 at 14:03
















          $begingroup$
          Damn, that's a badass solution! How did you come up with this parametrization?
          $endgroup$
          – user9393410
          Dec 6 '18 at 13:12




          $begingroup$
          Damn, that's a badass solution! How did you come up with this parametrization?
          $endgroup$
          – user9393410
          Dec 6 '18 at 13:12












          $begingroup$
          Consider the projection of $L$ on $xOy$: $x^2+xy+y^2=frac{a^2}{2}$, which is also $(x+frac{y}{2})^2+frac{3}{4}y^2=frac{a^2}{2}$.
          $endgroup$
          – Lau
          Dec 6 '18 at 13:28




          $begingroup$
          Consider the projection of $L$ on $xOy$: $x^2+xy+y^2=frac{a^2}{2}$, which is also $(x+frac{y}{2})^2+frac{3}{4}y^2=frac{a^2}{2}$.
          $endgroup$
          – Lau
          Dec 6 '18 at 13:28












          $begingroup$
          Oh, I see. Thank you!
          $endgroup$
          – user9393410
          Dec 6 '18 at 14:03




          $begingroup$
          Oh, I see. Thank you!
          $endgroup$
          – user9393410
          Dec 6 '18 at 14:03


















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