Compute $int_C xsqrt{x^2-y^2} mathrm ds$












1












$begingroup$



Compute $$int_C xsqrt{x^2-y^2} mathrm ds$$
where $C:(x^2+y^2)^2=a^2(x^2-y^2)$ with $xgeq 0$.




My attempt:



For this I tried the parametrization: $x=rcos t,y=rsin t$, but it doesn't seem to work and I get to: $cos(2t)=dfrac 1{a^2}$, so $tin left[3dfrac {pi}2,5dfrac {pi}2 right]$?






integration curves







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$endgroup$












  • $begingroup$
    en.wikipedia.org/wiki/Lemniscate_of_Bernoulli
    $endgroup$
    – Fred
    Jan 25 at 9:17
















1












$begingroup$



Compute $$int_C xsqrt{x^2-y^2} mathrm ds$$
where $C:(x^2+y^2)^2=a^2(x^2-y^2)$ with $xgeq 0$.




My attempt:



For this I tried the parametrization: $x=rcos t,y=rsin t$, but it doesn't seem to work and I get to: $cos(2t)=dfrac 1{a^2}$, so $tin left[3dfrac {pi}2,5dfrac {pi}2 right]$?






integration curves







share|cite|improve this question











$endgroup$












  • $begingroup$
    en.wikipedia.org/wiki/Lemniscate_of_Bernoulli
    $endgroup$
    – Fred
    Jan 25 at 9:17














1












1








1





$begingroup$



Compute $$int_C xsqrt{x^2-y^2} mathrm ds$$
where $C:(x^2+y^2)^2=a^2(x^2-y^2)$ with $xgeq 0$.




My attempt:



For this I tried the parametrization: $x=rcos t,y=rsin t$, but it doesn't seem to work and I get to: $cos(2t)=dfrac 1{a^2}$, so $tin left[3dfrac {pi}2,5dfrac {pi}2 right]$?






integration curves







share|cite|improve this question











$endgroup$





Compute $$int_C xsqrt{x^2-y^2} mathrm ds$$
where $C:(x^2+y^2)^2=a^2(x^2-y^2)$ with $xgeq 0$.




My attempt:



For this I tried the parametrization: $x=rcos t,y=rsin t$, but it doesn't seem to work and I get to: $cos(2t)=dfrac 1{a^2}$, so $tin left[3dfrac {pi}2,5dfrac {pi}2 right]$?






integration curves




integration curves






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 25 at 9:57









Mohammad Zuhair Khan

1,6682625




1,6682625










asked Jan 25 at 9:07









C. CristiC. Cristi

1,634218




1,634218












  • $begingroup$
    en.wikipedia.org/wiki/Lemniscate_of_Bernoulli
    $endgroup$
    – Fred
    Jan 25 at 9:17


















  • $begingroup$
    en.wikipedia.org/wiki/Lemniscate_of_Bernoulli
    $endgroup$
    – Fred
    Jan 25 at 9:17
















$begingroup$
en.wikipedia.org/wiki/Lemniscate_of_Bernoulli
$endgroup$
– Fred
Jan 25 at 9:17




$begingroup$
en.wikipedia.org/wiki/Lemniscate_of_Bernoulli
$endgroup$
– Fred
Jan 25 at 9:17










2 Answers
2






active

oldest

votes


















0












$begingroup$

Note that the curve $(x^2+y^2)^2=a^2(x^2-y^2)$ (a lemniscate) is symmetric with respect to the $y$-axis so the line integral on the WHOLE curve $C$, $int_C xsqrt{x^2-y^2}ds$ should be zero.



With the condition $xgeq 0$, then by using polar coordinates the curve becomes $r^2(theta)=a^2cos(2theta)$. By symmetry with respect to $x$-axis, the integral can be taken along only one quarter of the whole curve:
$$int_{C^+} xsqrt{x^2-y^2}ds=2int_0^{theta_1} r(theta)cos(theta)frac{r(theta)^2}{|a|}sqrt{r(theta)^2+r'(theta)^2}dtheta.$$
where $cos(2theta_1)=0$ that is $theta_1=pi/4$.
Can you take it from here?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    uhm... How should I know that this is symettric what made you think of that? You plugged $(-x,-y)$ and it's the same expression ?, Wait there shuold be another thing, $xgeq 0$
    $endgroup$
    – C. Cristi
    Jan 25 at 9:22










  • $begingroup$
    Okay, I know the curve becaomes that but how do I continue? How do I determine the interval?
    $endgroup$
    – C. Cristi
    Jan 25 at 9:31










  • $begingroup$
    Yes, indeed, thanks
    $endgroup$
    – C. Cristi
    Jan 25 at 9:37



















0












$begingroup$

Hint:



WLOG, $a=1$, for comfort. Using the polar form, the curve has the equation



$$r^4=r^2cos2t,$$ or $$r=sqrt{cos2t}.$$



Then,



$$dot r=-frac{sin2t}{sqrt{cos 2t}}$$



The integral now reads



$$int rcos tsqrt{cos2t},ds=int rcos tsqrt{cos2t}sqrt{r^2+dot r^2},dt=int cos tsqrt{cos2t},dt$$
after simplification. It is easier to integrate in the form



$$int cos tsqrt{1-2sin^2t},dt$$



and that leads to the area of an ellipse.



The answer will be of the form $pisqrt2$ with a small integer or rational coefficient.




For $ane1$, the result just needs to be multiplied by $a^3$.






share|cite|improve this answer











$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Note that the curve $(x^2+y^2)^2=a^2(x^2-y^2)$ (a lemniscate) is symmetric with respect to the $y$-axis so the line integral on the WHOLE curve $C$, $int_C xsqrt{x^2-y^2}ds$ should be zero.



    With the condition $xgeq 0$, then by using polar coordinates the curve becomes $r^2(theta)=a^2cos(2theta)$. By symmetry with respect to $x$-axis, the integral can be taken along only one quarter of the whole curve:
    $$int_{C^+} xsqrt{x^2-y^2}ds=2int_0^{theta_1} r(theta)cos(theta)frac{r(theta)^2}{|a|}sqrt{r(theta)^2+r'(theta)^2}dtheta.$$
    where $cos(2theta_1)=0$ that is $theta_1=pi/4$.
    Can you take it from here?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      uhm... How should I know that this is symettric what made you think of that? You plugged $(-x,-y)$ and it's the same expression ?, Wait there shuold be another thing, $xgeq 0$
      $endgroup$
      – C. Cristi
      Jan 25 at 9:22










    • $begingroup$
      Okay, I know the curve becaomes that but how do I continue? How do I determine the interval?
      $endgroup$
      – C. Cristi
      Jan 25 at 9:31










    • $begingroup$
      Yes, indeed, thanks
      $endgroup$
      – C. Cristi
      Jan 25 at 9:37
















    0












    $begingroup$

    Note that the curve $(x^2+y^2)^2=a^2(x^2-y^2)$ (a lemniscate) is symmetric with respect to the $y$-axis so the line integral on the WHOLE curve $C$, $int_C xsqrt{x^2-y^2}ds$ should be zero.



    With the condition $xgeq 0$, then by using polar coordinates the curve becomes $r^2(theta)=a^2cos(2theta)$. By symmetry with respect to $x$-axis, the integral can be taken along only one quarter of the whole curve:
    $$int_{C^+} xsqrt{x^2-y^2}ds=2int_0^{theta_1} r(theta)cos(theta)frac{r(theta)^2}{|a|}sqrt{r(theta)^2+r'(theta)^2}dtheta.$$
    where $cos(2theta_1)=0$ that is $theta_1=pi/4$.
    Can you take it from here?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      uhm... How should I know that this is symettric what made you think of that? You plugged $(-x,-y)$ and it's the same expression ?, Wait there shuold be another thing, $xgeq 0$
      $endgroup$
      – C. Cristi
      Jan 25 at 9:22










    • $begingroup$
      Okay, I know the curve becaomes that but how do I continue? How do I determine the interval?
      $endgroup$
      – C. Cristi
      Jan 25 at 9:31










    • $begingroup$
      Yes, indeed, thanks
      $endgroup$
      – C. Cristi
      Jan 25 at 9:37














    0












    0








    0





    $begingroup$

    Note that the curve $(x^2+y^2)^2=a^2(x^2-y^2)$ (a lemniscate) is symmetric with respect to the $y$-axis so the line integral on the WHOLE curve $C$, $int_C xsqrt{x^2-y^2}ds$ should be zero.



    With the condition $xgeq 0$, then by using polar coordinates the curve becomes $r^2(theta)=a^2cos(2theta)$. By symmetry with respect to $x$-axis, the integral can be taken along only one quarter of the whole curve:
    $$int_{C^+} xsqrt{x^2-y^2}ds=2int_0^{theta_1} r(theta)cos(theta)frac{r(theta)^2}{|a|}sqrt{r(theta)^2+r'(theta)^2}dtheta.$$
    where $cos(2theta_1)=0$ that is $theta_1=pi/4$.
    Can you take it from here?






    share|cite|improve this answer











    $endgroup$



    Note that the curve $(x^2+y^2)^2=a^2(x^2-y^2)$ (a lemniscate) is symmetric with respect to the $y$-axis so the line integral on the WHOLE curve $C$, $int_C xsqrt{x^2-y^2}ds$ should be zero.



    With the condition $xgeq 0$, then by using polar coordinates the curve becomes $r^2(theta)=a^2cos(2theta)$. By symmetry with respect to $x$-axis, the integral can be taken along only one quarter of the whole curve:
    $$int_{C^+} xsqrt{x^2-y^2}ds=2int_0^{theta_1} r(theta)cos(theta)frac{r(theta)^2}{|a|}sqrt{r(theta)^2+r'(theta)^2}dtheta.$$
    where $cos(2theta_1)=0$ that is $theta_1=pi/4$.
    Can you take it from here?







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 25 at 9:43

























    answered Jan 25 at 9:19









    Robert ZRobert Z

    100k1069140




    100k1069140












    • $begingroup$
      uhm... How should I know that this is symettric what made you think of that? You plugged $(-x,-y)$ and it's the same expression ?, Wait there shuold be another thing, $xgeq 0$
      $endgroup$
      – C. Cristi
      Jan 25 at 9:22










    • $begingroup$
      Okay, I know the curve becaomes that but how do I continue? How do I determine the interval?
      $endgroup$
      – C. Cristi
      Jan 25 at 9:31










    • $begingroup$
      Yes, indeed, thanks
      $endgroup$
      – C. Cristi
      Jan 25 at 9:37


















    • $begingroup$
      uhm... How should I know that this is symettric what made you think of that? You plugged $(-x,-y)$ and it's the same expression ?, Wait there shuold be another thing, $xgeq 0$
      $endgroup$
      – C. Cristi
      Jan 25 at 9:22










    • $begingroup$
      Okay, I know the curve becaomes that but how do I continue? How do I determine the interval?
      $endgroup$
      – C. Cristi
      Jan 25 at 9:31










    • $begingroup$
      Yes, indeed, thanks
      $endgroup$
      – C. Cristi
      Jan 25 at 9:37
















    $begingroup$
    uhm... How should I know that this is symettric what made you think of that? You plugged $(-x,-y)$ and it's the same expression ?, Wait there shuold be another thing, $xgeq 0$
    $endgroup$
    – C. Cristi
    Jan 25 at 9:22




    $begingroup$
    uhm... How should I know that this is symettric what made you think of that? You plugged $(-x,-y)$ and it's the same expression ?, Wait there shuold be another thing, $xgeq 0$
    $endgroup$
    – C. Cristi
    Jan 25 at 9:22












    $begingroup$
    Okay, I know the curve becaomes that but how do I continue? How do I determine the interval?
    $endgroup$
    – C. Cristi
    Jan 25 at 9:31




    $begingroup$
    Okay, I know the curve becaomes that but how do I continue? How do I determine the interval?
    $endgroup$
    – C. Cristi
    Jan 25 at 9:31












    $begingroup$
    Yes, indeed, thanks
    $endgroup$
    – C. Cristi
    Jan 25 at 9:37




    $begingroup$
    Yes, indeed, thanks
    $endgroup$
    – C. Cristi
    Jan 25 at 9:37











    0












    $begingroup$

    Hint:



    WLOG, $a=1$, for comfort. Using the polar form, the curve has the equation



    $$r^4=r^2cos2t,$$ or $$r=sqrt{cos2t}.$$



    Then,



    $$dot r=-frac{sin2t}{sqrt{cos 2t}}$$



    The integral now reads



    $$int rcos tsqrt{cos2t},ds=int rcos tsqrt{cos2t}sqrt{r^2+dot r^2},dt=int cos tsqrt{cos2t},dt$$
    after simplification. It is easier to integrate in the form



    $$int cos tsqrt{1-2sin^2t},dt$$



    and that leads to the area of an ellipse.



    The answer will be of the form $pisqrt2$ with a small integer or rational coefficient.




    For $ane1$, the result just needs to be multiplied by $a^3$.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Hint:



      WLOG, $a=1$, for comfort. Using the polar form, the curve has the equation



      $$r^4=r^2cos2t,$$ or $$r=sqrt{cos2t}.$$



      Then,



      $$dot r=-frac{sin2t}{sqrt{cos 2t}}$$



      The integral now reads



      $$int rcos tsqrt{cos2t},ds=int rcos tsqrt{cos2t}sqrt{r^2+dot r^2},dt=int cos tsqrt{cos2t},dt$$
      after simplification. It is easier to integrate in the form



      $$int cos tsqrt{1-2sin^2t},dt$$



      and that leads to the area of an ellipse.



      The answer will be of the form $pisqrt2$ with a small integer or rational coefficient.




      For $ane1$, the result just needs to be multiplied by $a^3$.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Hint:



        WLOG, $a=1$, for comfort. Using the polar form, the curve has the equation



        $$r^4=r^2cos2t,$$ or $$r=sqrt{cos2t}.$$



        Then,



        $$dot r=-frac{sin2t}{sqrt{cos 2t}}$$



        The integral now reads



        $$int rcos tsqrt{cos2t},ds=int rcos tsqrt{cos2t}sqrt{r^2+dot r^2},dt=int cos tsqrt{cos2t},dt$$
        after simplification. It is easier to integrate in the form



        $$int cos tsqrt{1-2sin^2t},dt$$



        and that leads to the area of an ellipse.



        The answer will be of the form $pisqrt2$ with a small integer or rational coefficient.




        For $ane1$, the result just needs to be multiplied by $a^3$.






        share|cite|improve this answer











        $endgroup$



        Hint:



        WLOG, $a=1$, for comfort. Using the polar form, the curve has the equation



        $$r^4=r^2cos2t,$$ or $$r=sqrt{cos2t}.$$



        Then,



        $$dot r=-frac{sin2t}{sqrt{cos 2t}}$$



        The integral now reads



        $$int rcos tsqrt{cos2t},ds=int rcos tsqrt{cos2t}sqrt{r^2+dot r^2},dt=int cos tsqrt{cos2t},dt$$
        after simplification. It is easier to integrate in the form



        $$int cos tsqrt{1-2sin^2t},dt$$



        and that leads to the area of an ellipse.



        The answer will be of the form $pisqrt2$ with a small integer or rational coefficient.




        For $ane1$, the result just needs to be multiplied by $a^3$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 25 at 9:30

























        answered Jan 25 at 9:24









        Yves DaoustYves Daoust

        130k676227




        130k676227






























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