how to calculate $int_C frac{2xy^2dx-2yx^2dy}{x^2+y^2}$ using green theorm or directly












0












$begingroup$



Calculate $$int_C frac{2xy^2dx-2yx^2dy}{x^2+y^2},$$ where $C$ is the ellipse $3x^2 +5y^2 = 1$ taken in the positive direction.




I tried to calculate the integral using green theorm.



now i need to build enclosier that doesn't enclose $(0,0)$
i am having hard time guessing what to build.



a circle and ellipse might be perfect but then the domain is not easy to write. can i have hint please ?






multivariable-calculus greens-theorem







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Are you sure it is a minus between the two terms in the numerator? Green's theorem doesn't seem to work if it is a minus, since $$P_y=frac{4xy}{x^2+y^2}\Q_x=-frac{4xy}{x^2+y^2}$$So $P_yne Q_x$
    $endgroup$
    – John Doe
    Jan 7 at 22:19












  • $begingroup$
    oh right , anyway can i apply green therom here ? even if its not $0$ or the integral is too complex to calculate ?
    $endgroup$
    – Mather
    Jan 7 at 22:22












  • $begingroup$
    i have also tried to calculate it with basic parametrization of the ellipse but i got harsh integrand
    $endgroup$
    – Mather
    Jan 7 at 22:23












  • $begingroup$
    @JohnDoe are you sure? Shouldn't there be a $(x^2+y^2)^2$ in the denominator from quotient rule. I don't think it cancels.
    $endgroup$
    – AHusain
    Jan 7 at 22:24










  • $begingroup$
    @AHusain Oh whoops, I didn't differentiate that properly, you''re right! However, even if you do differentiate it properly, it still doesn't give $P_y=Q_x$, the minus sign will still mess this up
    $endgroup$
    – John Doe
    Jan 7 at 22:34


















0












$begingroup$



Calculate $$int_C frac{2xy^2dx-2yx^2dy}{x^2+y^2},$$ where $C$ is the ellipse $3x^2 +5y^2 = 1$ taken in the positive direction.




I tried to calculate the integral using green theorm.



now i need to build enclosier that doesn't enclose $(0,0)$
i am having hard time guessing what to build.



a circle and ellipse might be perfect but then the domain is not easy to write. can i have hint please ?






multivariable-calculus greens-theorem







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Are you sure it is a minus between the two terms in the numerator? Green's theorem doesn't seem to work if it is a minus, since $$P_y=frac{4xy}{x^2+y^2}\Q_x=-frac{4xy}{x^2+y^2}$$So $P_yne Q_x$
    $endgroup$
    – John Doe
    Jan 7 at 22:19












  • $begingroup$
    oh right , anyway can i apply green therom here ? even if its not $0$ or the integral is too complex to calculate ?
    $endgroup$
    – Mather
    Jan 7 at 22:22












  • $begingroup$
    i have also tried to calculate it with basic parametrization of the ellipse but i got harsh integrand
    $endgroup$
    – Mather
    Jan 7 at 22:23












  • $begingroup$
    @JohnDoe are you sure? Shouldn't there be a $(x^2+y^2)^2$ in the denominator from quotient rule. I don't think it cancels.
    $endgroup$
    – AHusain
    Jan 7 at 22:24










  • $begingroup$
    @AHusain Oh whoops, I didn't differentiate that properly, you''re right! However, even if you do differentiate it properly, it still doesn't give $P_y=Q_x$, the minus sign will still mess this up
    $endgroup$
    – John Doe
    Jan 7 at 22:34
















0












0








0





$begingroup$



Calculate $$int_C frac{2xy^2dx-2yx^2dy}{x^2+y^2},$$ where $C$ is the ellipse $3x^2 +5y^2 = 1$ taken in the positive direction.




I tried to calculate the integral using green theorm.



now i need to build enclosier that doesn't enclose $(0,0)$
i am having hard time guessing what to build.



a circle and ellipse might be perfect but then the domain is not easy to write. can i have hint please ?






multivariable-calculus greens-theorem







share|cite|improve this question











$endgroup$





Calculate $$int_C frac{2xy^2dx-2yx^2dy}{x^2+y^2},$$ where $C$ is the ellipse $3x^2 +5y^2 = 1$ taken in the positive direction.




I tried to calculate the integral using green theorm.



now i need to build enclosier that doesn't enclose $(0,0)$
i am having hard time guessing what to build.



a circle and ellipse might be perfect but then the domain is not easy to write. can i have hint please ?






multivariable-calculus greens-theorem




multivariable-calculus greens-theorem






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 22:27







Mather

















asked Jan 7 at 22:11









Mather Mather

4028




4028








  • 3




    $begingroup$
    Are you sure it is a minus between the two terms in the numerator? Green's theorem doesn't seem to work if it is a minus, since $$P_y=frac{4xy}{x^2+y^2}\Q_x=-frac{4xy}{x^2+y^2}$$So $P_yne Q_x$
    $endgroup$
    – John Doe
    Jan 7 at 22:19












  • $begingroup$
    oh right , anyway can i apply green therom here ? even if its not $0$ or the integral is too complex to calculate ?
    $endgroup$
    – Mather
    Jan 7 at 22:22












  • $begingroup$
    i have also tried to calculate it with basic parametrization of the ellipse but i got harsh integrand
    $endgroup$
    – Mather
    Jan 7 at 22:23












  • $begingroup$
    @JohnDoe are you sure? Shouldn't there be a $(x^2+y^2)^2$ in the denominator from quotient rule. I don't think it cancels.
    $endgroup$
    – AHusain
    Jan 7 at 22:24










  • $begingroup$
    @AHusain Oh whoops, I didn't differentiate that properly, you''re right! However, even if you do differentiate it properly, it still doesn't give $P_y=Q_x$, the minus sign will still mess this up
    $endgroup$
    – John Doe
    Jan 7 at 22:34
















  • 3




    $begingroup$
    Are you sure it is a minus between the two terms in the numerator? Green's theorem doesn't seem to work if it is a minus, since $$P_y=frac{4xy}{x^2+y^2}\Q_x=-frac{4xy}{x^2+y^2}$$So $P_yne Q_x$
    $endgroup$
    – John Doe
    Jan 7 at 22:19












  • $begingroup$
    oh right , anyway can i apply green therom here ? even if its not $0$ or the integral is too complex to calculate ?
    $endgroup$
    – Mather
    Jan 7 at 22:22












  • $begingroup$
    i have also tried to calculate it with basic parametrization of the ellipse but i got harsh integrand
    $endgroup$
    – Mather
    Jan 7 at 22:23












  • $begingroup$
    @JohnDoe are you sure? Shouldn't there be a $(x^2+y^2)^2$ in the denominator from quotient rule. I don't think it cancels.
    $endgroup$
    – AHusain
    Jan 7 at 22:24










  • $begingroup$
    @AHusain Oh whoops, I didn't differentiate that properly, you''re right! However, even if you do differentiate it properly, it still doesn't give $P_y=Q_x$, the minus sign will still mess this up
    $endgroup$
    – John Doe
    Jan 7 at 22:34










3




3




$begingroup$
Are you sure it is a minus between the two terms in the numerator? Green's theorem doesn't seem to work if it is a minus, since $$P_y=frac{4xy}{x^2+y^2}\Q_x=-frac{4xy}{x^2+y^2}$$So $P_yne Q_x$
$endgroup$
– John Doe
Jan 7 at 22:19






$begingroup$
Are you sure it is a minus between the two terms in the numerator? Green's theorem doesn't seem to work if it is a minus, since $$P_y=frac{4xy}{x^2+y^2}\Q_x=-frac{4xy}{x^2+y^2}$$So $P_yne Q_x$
$endgroup$
– John Doe
Jan 7 at 22:19














$begingroup$
oh right , anyway can i apply green therom here ? even if its not $0$ or the integral is too complex to calculate ?
$endgroup$
– Mather
Jan 7 at 22:22






$begingroup$
oh right , anyway can i apply green therom here ? even if its not $0$ or the integral is too complex to calculate ?
$endgroup$
– Mather
Jan 7 at 22:22














$begingroup$
i have also tried to calculate it with basic parametrization of the ellipse but i got harsh integrand
$endgroup$
– Mather
Jan 7 at 22:23






$begingroup$
i have also tried to calculate it with basic parametrization of the ellipse but i got harsh integrand
$endgroup$
– Mather
Jan 7 at 22:23














$begingroup$
@JohnDoe are you sure? Shouldn't there be a $(x^2+y^2)^2$ in the denominator from quotient rule. I don't think it cancels.
$endgroup$
– AHusain
Jan 7 at 22:24




$begingroup$
@JohnDoe are you sure? Shouldn't there be a $(x^2+y^2)^2$ in the denominator from quotient rule. I don't think it cancels.
$endgroup$
– AHusain
Jan 7 at 22:24












$begingroup$
@AHusain Oh whoops, I didn't differentiate that properly, you''re right! However, even if you do differentiate it properly, it still doesn't give $P_y=Q_x$, the minus sign will still mess this up
$endgroup$
– John Doe
Jan 7 at 22:34






$begingroup$
@AHusain Oh whoops, I didn't differentiate that properly, you''re right! However, even if you do differentiate it properly, it still doesn't give $P_y=Q_x$, the minus sign will still mess this up
$endgroup$
– John Doe
Jan 7 at 22:34












2 Answers
2






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oldest

votes


















0












$begingroup$

We have



$$int_mathcal C frac{2xy^2mathrm dx-2x^2ymathrm dy}{x^2+y^2}=int_{3x^2+5y^2=1}frac{mathrm d(x^2)y^2-x^2mathrm d(y^2)}{x^2+y^2}tag1$$



Let $u=x^2, v=y^2$. Then we have



$$int_{3u+5v=1,quad u,vge0}left(frac{v}{u+v}mathrm du-frac{u}{u+v}mathrm dvright)tag2$$



We compute the first integral. The path is $3u+5v=1,quad u,vge0$, so $v=frac15-frac35 u$ and $uin[0,frac13]$.



$$int_mathcal Cfrac v{u+v}mathrm du=int_0^frac13frac{1-3u}{1+2u}=cdots=frac14left(5logfrac53-2right)$$



We compute the second integral. The path is $u=frac13-frac53 v$ with $vin[0,frac15]$.



$$int_mathcal Cfrac u{u+v}mathrm dv=int_0^frac15frac{1-5v}{1-2v}mathrm dv=cdots=frac14left(2-3logfrac53right)$$



Hence the amount that $(2)$ contributes to the total integral $(1)$ is$$frac14left(5logfrac53-2-2+3logfrac53right)=2logfrac53-1$$




Edit:



As pointed out in the comments, this change of co-ordinates was not a bijective map, so we have not included all the points on the ellipse by doing this calculation, we have only done it for $x,y>0$. For $x,y<0$, we would have the same path, but traversed backwards (i.e. the integration limits will be swapped), which would directly cancel out what we have computed here.



For $x>0,y<0$, we have the same path as in $(2)$, while for $x<0,y>0$, we have this path traversed backwards (you can see this by considering the ellipse traversed clockwise, seeing whether $x$, and correspondingly $u$, is increasing in magnitude or not). So again, these two contributions cancel. This is what gives $0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    shouldn't it be $0$ ?
    $endgroup$
    – Mather
    Jan 8 at 9:40








  • 1




    $begingroup$
    @JohnDoe This isn't entirely correct. Your new coordinate system only maps a quarter of the ellipse. To go all the way around the ellipse, you must go forwards and backwards on the line in $(u,v)$ twice, making the integral $0$.
    $endgroup$
    – Dylan
    Jan 8 at 12:38












  • $begingroup$
    @mather see the above comment ^ I'll edit my answer in a little while
    $endgroup$
    – John Doe
    Jan 8 at 13:20












  • $begingroup$
    @Dylan I have edited the question, thanks for pointing that out
    $endgroup$
    – John Doe
    Jan 8 at 15:40



















0












$begingroup$

You can evaluate the line integral directly by taking $mathbf r(t) = (cos(t)/sqrt 3, sin(t)/sqrt 5)$:
$$I = int_C mathbf F cdot dmathbf r =
-int_0^{2 pi} frac {sin 2 t} {4 + cos 2 t} dt =
-int_0^pi frac {sin 2 t} {4 + cos 2 t} +
int_0^pi frac {sin 2 t} {4 + cos 2 t} dt = 0.$$
Green's theorem still holds for $mathbf F$ even though $mathbf F$ doesn't have continuous partial derivatives at $(0, 0)$:
$$I = -iint_{3 x^2 + 5 y^2 leq 1} frac {4 x y} {x^2 + y^2} dx dy.$$
This form makes it clearer that the result is zero because of the symmetries wrt the coordinate axes.






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    2 Answers
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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes









    0












    $begingroup$

    We have



    $$int_mathcal C frac{2xy^2mathrm dx-2x^2ymathrm dy}{x^2+y^2}=int_{3x^2+5y^2=1}frac{mathrm d(x^2)y^2-x^2mathrm d(y^2)}{x^2+y^2}tag1$$



    Let $u=x^2, v=y^2$. Then we have



    $$int_{3u+5v=1,quad u,vge0}left(frac{v}{u+v}mathrm du-frac{u}{u+v}mathrm dvright)tag2$$



    We compute the first integral. The path is $3u+5v=1,quad u,vge0$, so $v=frac15-frac35 u$ and $uin[0,frac13]$.



    $$int_mathcal Cfrac v{u+v}mathrm du=int_0^frac13frac{1-3u}{1+2u}=cdots=frac14left(5logfrac53-2right)$$



    We compute the second integral. The path is $u=frac13-frac53 v$ with $vin[0,frac15]$.



    $$int_mathcal Cfrac u{u+v}mathrm dv=int_0^frac15frac{1-5v}{1-2v}mathrm dv=cdots=frac14left(2-3logfrac53right)$$



    Hence the amount that $(2)$ contributes to the total integral $(1)$ is$$frac14left(5logfrac53-2-2+3logfrac53right)=2logfrac53-1$$




    Edit:



    As pointed out in the comments, this change of co-ordinates was not a bijective map, so we have not included all the points on the ellipse by doing this calculation, we have only done it for $x,y>0$. For $x,y<0$, we would have the same path, but traversed backwards (i.e. the integration limits will be swapped), which would directly cancel out what we have computed here.



    For $x>0,y<0$, we have the same path as in $(2)$, while for $x<0,y>0$, we have this path traversed backwards (you can see this by considering the ellipse traversed clockwise, seeing whether $x$, and correspondingly $u$, is increasing in magnitude or not). So again, these two contributions cancel. This is what gives $0$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      shouldn't it be $0$ ?
      $endgroup$
      – Mather
      Jan 8 at 9:40








    • 1




      $begingroup$
      @JohnDoe This isn't entirely correct. Your new coordinate system only maps a quarter of the ellipse. To go all the way around the ellipse, you must go forwards and backwards on the line in $(u,v)$ twice, making the integral $0$.
      $endgroup$
      – Dylan
      Jan 8 at 12:38












    • $begingroup$
      @mather see the above comment ^ I'll edit my answer in a little while
      $endgroup$
      – John Doe
      Jan 8 at 13:20












    • $begingroup$
      @Dylan I have edited the question, thanks for pointing that out
      $endgroup$
      – John Doe
      Jan 8 at 15:40
















    0












    $begingroup$

    We have



    $$int_mathcal C frac{2xy^2mathrm dx-2x^2ymathrm dy}{x^2+y^2}=int_{3x^2+5y^2=1}frac{mathrm d(x^2)y^2-x^2mathrm d(y^2)}{x^2+y^2}tag1$$



    Let $u=x^2, v=y^2$. Then we have



    $$int_{3u+5v=1,quad u,vge0}left(frac{v}{u+v}mathrm du-frac{u}{u+v}mathrm dvright)tag2$$



    We compute the first integral. The path is $3u+5v=1,quad u,vge0$, so $v=frac15-frac35 u$ and $uin[0,frac13]$.



    $$int_mathcal Cfrac v{u+v}mathrm du=int_0^frac13frac{1-3u}{1+2u}=cdots=frac14left(5logfrac53-2right)$$



    We compute the second integral. The path is $u=frac13-frac53 v$ with $vin[0,frac15]$.



    $$int_mathcal Cfrac u{u+v}mathrm dv=int_0^frac15frac{1-5v}{1-2v}mathrm dv=cdots=frac14left(2-3logfrac53right)$$



    Hence the amount that $(2)$ contributes to the total integral $(1)$ is$$frac14left(5logfrac53-2-2+3logfrac53right)=2logfrac53-1$$




    Edit:



    As pointed out in the comments, this change of co-ordinates was not a bijective map, so we have not included all the points on the ellipse by doing this calculation, we have only done it for $x,y>0$. For $x,y<0$, we would have the same path, but traversed backwards (i.e. the integration limits will be swapped), which would directly cancel out what we have computed here.



    For $x>0,y<0$, we have the same path as in $(2)$, while for $x<0,y>0$, we have this path traversed backwards (you can see this by considering the ellipse traversed clockwise, seeing whether $x$, and correspondingly $u$, is increasing in magnitude or not). So again, these two contributions cancel. This is what gives $0$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      shouldn't it be $0$ ?
      $endgroup$
      – Mather
      Jan 8 at 9:40








    • 1




      $begingroup$
      @JohnDoe This isn't entirely correct. Your new coordinate system only maps a quarter of the ellipse. To go all the way around the ellipse, you must go forwards and backwards on the line in $(u,v)$ twice, making the integral $0$.
      $endgroup$
      – Dylan
      Jan 8 at 12:38












    • $begingroup$
      @mather see the above comment ^ I'll edit my answer in a little while
      $endgroup$
      – John Doe
      Jan 8 at 13:20












    • $begingroup$
      @Dylan I have edited the question, thanks for pointing that out
      $endgroup$
      – John Doe
      Jan 8 at 15:40














    0












    0








    0





    $begingroup$

    We have



    $$int_mathcal C frac{2xy^2mathrm dx-2x^2ymathrm dy}{x^2+y^2}=int_{3x^2+5y^2=1}frac{mathrm d(x^2)y^2-x^2mathrm d(y^2)}{x^2+y^2}tag1$$



    Let $u=x^2, v=y^2$. Then we have



    $$int_{3u+5v=1,quad u,vge0}left(frac{v}{u+v}mathrm du-frac{u}{u+v}mathrm dvright)tag2$$



    We compute the first integral. The path is $3u+5v=1,quad u,vge0$, so $v=frac15-frac35 u$ and $uin[0,frac13]$.



    $$int_mathcal Cfrac v{u+v}mathrm du=int_0^frac13frac{1-3u}{1+2u}=cdots=frac14left(5logfrac53-2right)$$



    We compute the second integral. The path is $u=frac13-frac53 v$ with $vin[0,frac15]$.



    $$int_mathcal Cfrac u{u+v}mathrm dv=int_0^frac15frac{1-5v}{1-2v}mathrm dv=cdots=frac14left(2-3logfrac53right)$$



    Hence the amount that $(2)$ contributes to the total integral $(1)$ is$$frac14left(5logfrac53-2-2+3logfrac53right)=2logfrac53-1$$




    Edit:



    As pointed out in the comments, this change of co-ordinates was not a bijective map, so we have not included all the points on the ellipse by doing this calculation, we have only done it for $x,y>0$. For $x,y<0$, we would have the same path, but traversed backwards (i.e. the integration limits will be swapped), which would directly cancel out what we have computed here.



    For $x>0,y<0$, we have the same path as in $(2)$, while for $x<0,y>0$, we have this path traversed backwards (you can see this by considering the ellipse traversed clockwise, seeing whether $x$, and correspondingly $u$, is increasing in magnitude or not). So again, these two contributions cancel. This is what gives $0$.






    share|cite|improve this answer











    $endgroup$



    We have



    $$int_mathcal C frac{2xy^2mathrm dx-2x^2ymathrm dy}{x^2+y^2}=int_{3x^2+5y^2=1}frac{mathrm d(x^2)y^2-x^2mathrm d(y^2)}{x^2+y^2}tag1$$



    Let $u=x^2, v=y^2$. Then we have



    $$int_{3u+5v=1,quad u,vge0}left(frac{v}{u+v}mathrm du-frac{u}{u+v}mathrm dvright)tag2$$



    We compute the first integral. The path is $3u+5v=1,quad u,vge0$, so $v=frac15-frac35 u$ and $uin[0,frac13]$.



    $$int_mathcal Cfrac v{u+v}mathrm du=int_0^frac13frac{1-3u}{1+2u}=cdots=frac14left(5logfrac53-2right)$$



    We compute the second integral. The path is $u=frac13-frac53 v$ with $vin[0,frac15]$.



    $$int_mathcal Cfrac u{u+v}mathrm dv=int_0^frac15frac{1-5v}{1-2v}mathrm dv=cdots=frac14left(2-3logfrac53right)$$



    Hence the amount that $(2)$ contributes to the total integral $(1)$ is$$frac14left(5logfrac53-2-2+3logfrac53right)=2logfrac53-1$$




    Edit:



    As pointed out in the comments, this change of co-ordinates was not a bijective map, so we have not included all the points on the ellipse by doing this calculation, we have only done it for $x,y>0$. For $x,y<0$, we would have the same path, but traversed backwards (i.e. the integration limits will be swapped), which would directly cancel out what we have computed here.



    For $x>0,y<0$, we have the same path as in $(2)$, while for $x<0,y>0$, we have this path traversed backwards (you can see this by considering the ellipse traversed clockwise, seeing whether $x$, and correspondingly $u$, is increasing in magnitude or not). So again, these two contributions cancel. This is what gives $0$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 8 at 15:39

























    answered Jan 7 at 22:55









    John DoeJohn Doe

    11.3k11239




    11.3k11239












    • $begingroup$
      shouldn't it be $0$ ?
      $endgroup$
      – Mather
      Jan 8 at 9:40








    • 1




      $begingroup$
      @JohnDoe This isn't entirely correct. Your new coordinate system only maps a quarter of the ellipse. To go all the way around the ellipse, you must go forwards and backwards on the line in $(u,v)$ twice, making the integral $0$.
      $endgroup$
      – Dylan
      Jan 8 at 12:38












    • $begingroup$
      @mather see the above comment ^ I'll edit my answer in a little while
      $endgroup$
      – John Doe
      Jan 8 at 13:20












    • $begingroup$
      @Dylan I have edited the question, thanks for pointing that out
      $endgroup$
      – John Doe
      Jan 8 at 15:40


















    • $begingroup$
      shouldn't it be $0$ ?
      $endgroup$
      – Mather
      Jan 8 at 9:40








    • 1




      $begingroup$
      @JohnDoe This isn't entirely correct. Your new coordinate system only maps a quarter of the ellipse. To go all the way around the ellipse, you must go forwards and backwards on the line in $(u,v)$ twice, making the integral $0$.
      $endgroup$
      – Dylan
      Jan 8 at 12:38












    • $begingroup$
      @mather see the above comment ^ I'll edit my answer in a little while
      $endgroup$
      – John Doe
      Jan 8 at 13:20












    • $begingroup$
      @Dylan I have edited the question, thanks for pointing that out
      $endgroup$
      – John Doe
      Jan 8 at 15:40
















    $begingroup$
    shouldn't it be $0$ ?
    $endgroup$
    – Mather
    Jan 8 at 9:40






    $begingroup$
    shouldn't it be $0$ ?
    $endgroup$
    – Mather
    Jan 8 at 9:40






    1




    1




    $begingroup$
    @JohnDoe This isn't entirely correct. Your new coordinate system only maps a quarter of the ellipse. To go all the way around the ellipse, you must go forwards and backwards on the line in $(u,v)$ twice, making the integral $0$.
    $endgroup$
    – Dylan
    Jan 8 at 12:38






    $begingroup$
    @JohnDoe This isn't entirely correct. Your new coordinate system only maps a quarter of the ellipse. To go all the way around the ellipse, you must go forwards and backwards on the line in $(u,v)$ twice, making the integral $0$.
    $endgroup$
    – Dylan
    Jan 8 at 12:38














    $begingroup$
    @mather see the above comment ^ I'll edit my answer in a little while
    $endgroup$
    – John Doe
    Jan 8 at 13:20






    $begingroup$
    @mather see the above comment ^ I'll edit my answer in a little while
    $endgroup$
    – John Doe
    Jan 8 at 13:20














    $begingroup$
    @Dylan I have edited the question, thanks for pointing that out
    $endgroup$
    – John Doe
    Jan 8 at 15:40




    $begingroup$
    @Dylan I have edited the question, thanks for pointing that out
    $endgroup$
    – John Doe
    Jan 8 at 15:40











    0












    $begingroup$

    You can evaluate the line integral directly by taking $mathbf r(t) = (cos(t)/sqrt 3, sin(t)/sqrt 5)$:
    $$I = int_C mathbf F cdot dmathbf r =
    -int_0^{2 pi} frac {sin 2 t} {4 + cos 2 t} dt =
    -int_0^pi frac {sin 2 t} {4 + cos 2 t} +
    int_0^pi frac {sin 2 t} {4 + cos 2 t} dt = 0.$$
    Green's theorem still holds for $mathbf F$ even though $mathbf F$ doesn't have continuous partial derivatives at $(0, 0)$:
    $$I = -iint_{3 x^2 + 5 y^2 leq 1} frac {4 x y} {x^2 + y^2} dx dy.$$
    This form makes it clearer that the result is zero because of the symmetries wrt the coordinate axes.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      You can evaluate the line integral directly by taking $mathbf r(t) = (cos(t)/sqrt 3, sin(t)/sqrt 5)$:
      $$I = int_C mathbf F cdot dmathbf r =
      -int_0^{2 pi} frac {sin 2 t} {4 + cos 2 t} dt =
      -int_0^pi frac {sin 2 t} {4 + cos 2 t} +
      int_0^pi frac {sin 2 t} {4 + cos 2 t} dt = 0.$$
      Green's theorem still holds for $mathbf F$ even though $mathbf F$ doesn't have continuous partial derivatives at $(0, 0)$:
      $$I = -iint_{3 x^2 + 5 y^2 leq 1} frac {4 x y} {x^2 + y^2} dx dy.$$
      This form makes it clearer that the result is zero because of the symmetries wrt the coordinate axes.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        You can evaluate the line integral directly by taking $mathbf r(t) = (cos(t)/sqrt 3, sin(t)/sqrt 5)$:
        $$I = int_C mathbf F cdot dmathbf r =
        -int_0^{2 pi} frac {sin 2 t} {4 + cos 2 t} dt =
        -int_0^pi frac {sin 2 t} {4 + cos 2 t} +
        int_0^pi frac {sin 2 t} {4 + cos 2 t} dt = 0.$$
        Green's theorem still holds for $mathbf F$ even though $mathbf F$ doesn't have continuous partial derivatives at $(0, 0)$:
        $$I = -iint_{3 x^2 + 5 y^2 leq 1} frac {4 x y} {x^2 + y^2} dx dy.$$
        This form makes it clearer that the result is zero because of the symmetries wrt the coordinate axes.






        share|cite|improve this answer











        $endgroup$



        You can evaluate the line integral directly by taking $mathbf r(t) = (cos(t)/sqrt 3, sin(t)/sqrt 5)$:
        $$I = int_C mathbf F cdot dmathbf r =
        -int_0^{2 pi} frac {sin 2 t} {4 + cos 2 t} dt =
        -int_0^pi frac {sin 2 t} {4 + cos 2 t} +
        int_0^pi frac {sin 2 t} {4 + cos 2 t} dt = 0.$$
        Green's theorem still holds for $mathbf F$ even though $mathbf F$ doesn't have continuous partial derivatives at $(0, 0)$:
        $$I = -iint_{3 x^2 + 5 y^2 leq 1} frac {4 x y} {x^2 + y^2} dx dy.$$
        This form makes it clearer that the result is zero because of the symmetries wrt the coordinate axes.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 8 at 18:52

























        answered Jan 8 at 18:43









        MaximMaxim

        6,0631221




        6,0631221






























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