Mixing
how to calculate $int_C frac{2xy^2dx-2yx^2dy}{x^2+y^2}$ using green theorm or directly
$begingroup$
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
asked Jan 7 at 22:11
Mather Mather
3047
$endgroup$
add a comment |
$begingroup$
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
asked Jan 7 at 22:11
Mather Mather
3047
$endgroup$
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
add a comment |
$begingroup$
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
asked Jan 7 at 22:11
Mather Mather
3047
$endgroup$
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
asked Jan 7 at 22:11
Mather Mather
3047
asked Jan 7 at 22:11
Mather Mather
3047
edited Jan 7 at 22:27
Mather
asked Jan 7 at 22:11
Mather Mather
3047
asked Jan 7 at 22:11
Mather Mather
3047
asked Jan 7 at 22:11
Mather Mather
3047
3047
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $0$.
answered Jan 7 at 22:55
John DoeJohn Doe
11.1k11238
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 8 at 18:43
MaximMaxim
5,1281219
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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 $0$.
answered Jan 7 at 22:55
John DoeJohn Doe
11.1k11238
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $0$.
answered Jan 7 at 22:55
John DoeJohn Doe
11.1k11238
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $0$.
answered Jan 7 at 22:55
John DoeJohn Doe
11.1k11238
$endgroup$
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 $0$.
answered Jan 7 at 22:55
John DoeJohn Doe
11.1k11238
edited Jan 8 at 15:39
answered Jan 7 at 22:55
John DoeJohn Doe
11.1k11238
answered Jan 7 at 22:55
John DoeJohn Doe
11.1k11238
answered Jan 7 at 22:55
John DoeJohn Doe
11.1k11238
11.1k11238
$begingroup$
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
add a comment |
$begingroup$
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
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– John Doe
Jan 8 at 15:40
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@Dylan I have edited the question, thanks for pointing that out
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– John Doe
Jan 8 at 15:40
add a comment |
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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.
answered Jan 8 at 18:43
MaximMaxim
5,1281219
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 8 at 18:43
MaximMaxim
5,1281219
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 8 at 18:43
MaximMaxim
5,1281219
$endgroup$
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.
answered Jan 8 at 18:43
MaximMaxim
5,1281219
edited Jan 8 at 18:52
answered Jan 8 at 18:43
MaximMaxim
5,1281219
answered Jan 8 at 18:43
MaximMaxim
5,1281219
answered Jan 8 at 18:43
MaximMaxim
5,1281219
5,1281219
add a comment |
add a comment |
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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 $P_yne Q_x$
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– John Doe
Jan 7 at 22:19
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oh right , anyway can i apply green therom here ? even if its not $0$ or the integral is too complex to calculate ?
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– Mather
Jan 7 at 22:22
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i have also tried to calculate it with basic parametrization of the ellipse but i got harsh integrand
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– Mather
Jan 7 at 22:23
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@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.
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– AHusain
Jan 7 at 22:24
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@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
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– John Doe
Jan 7 at 22:34