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Integral of 1-form $omega=dfrac-y ,dx + x ,dyx^2 +y^2$ over a triangle.
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I'm trying to evaluate the integral of the $1$-form
$$omega=dfrac-y ,dx +x,dyx^2 +y^2$$
through the corners of a triangle with the vertices $A= (-5,-2)$, $B=(5,-2)$, $C=(0,3)$.
I've tried to use Green's theorem but it didn't work because the 1-form isn't differentiable on (0,0) so I had to parametrize the paths from each point and got three line integrals but it got so complicated at the end, so it doesn't seem to be the right solution or at least the one I'm supposed to calculate. Is there anyway else to evaluate the integral?
integration multivariable-calculus differential-forms line-integrals
edited yesterday
Michael Hardy
204k23185460
asked yesterday
Mohamed Mossad
102
add a comment |Â
up vote
0
down vote
favorite
I'm trying to evaluate the integral of the $1$-form
$$omega=dfrac-y ,dx +x,dyx^2 +y^2$$
through the corners of a triangle with the vertices $A= (-5,-2)$, $B=(5,-2)$, $C=(0,3)$.
I've tried to use Green's theorem but it didn't work because the 1-form isn't differentiable on (0,0) so I had to parametrize the paths from each point and got three line integrals but it got so complicated at the end, so it doesn't seem to be the right solution or at least the one I'm supposed to calculate. Is there anyway else to evaluate the integral?
integration multivariable-calculus differential-forms line-integrals
edited yesterday
Michael Hardy
204k23185460
asked yesterday
Mohamed Mossad
102
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to evaluate the integral of the $1$-form
$$omega=dfrac-y ,dx +x,dyx^2 +y^2$$
through the corners of a triangle with the vertices $A= (-5,-2)$, $B=(5,-2)$, $C=(0,3)$.
I've tried to use Green's theorem but it didn't work because the 1-form isn't differentiable on (0,0) so I had to parametrize the paths from each point and got three line integrals but it got so complicated at the end, so it doesn't seem to be the right solution or at least the one I'm supposed to calculate. Is there anyway else to evaluate the integral?
integration multivariable-calculus differential-forms line-integrals
edited yesterday
Michael Hardy
204k23185460
asked yesterday
Mohamed Mossad
102
I'm trying to evaluate the integral of the $1$-form
$$omega=dfrac-y ,dx +x,dyx^2 +y^2$$
through the corners of a triangle with the vertices $A= (-5,-2)$, $B=(5,-2)$, $C=(0,3)$.
I've tried to use Green's theorem but it didn't work because the 1-form isn't differentiable on (0,0) so I had to parametrize the paths from each point and got three line integrals but it got so complicated at the end, so it doesn't seem to be the right solution or at least the one I'm supposed to calculate. Is there anyway else to evaluate the integral?
integration multivariable-calculus differential-forms line-integrals
edited yesterday
Michael Hardy
204k23185460
asked yesterday
Mohamed Mossad
102
edited yesterday
Michael Hardy
204k23185460
edited yesterday
Michael Hardy
204k23185460
edited yesterday
Michael Hardy
204k23185460
204k23185460
asked yesterday
Mohamed Mossad
102
asked yesterday
Mohamed Mossad
102
asked yesterday
Mohamed Mossad
102
102
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
We know that
$$omega=dfrac-y ,dx +x,dyx^2 +y^2=darctandfracyx$$
is an exact differential and the area $Delta$ includes the origin, since the $arctandfracyx$ winds around the origin one time then
$$int_Delta omega=2pi$$
edited yesterday
Michael Hardy
204k23185460
answered yesterday
user 108128
18.6k41544
May I ask how did you come to that answer? like what's intuitive about it? I know that 1-form is quite well-known but I am supposed to calculate the integral not use a trigonometric function which I don't know.
– Mohamed Mossad
yesterday
When you have an exact differential over a closed area, and while it winds around the origin $n$ times then it's integral is $2pi n$. Also as you said you can't apply green thm, except these you are forced to parametrize the triangle and substitude them in the exact form.
– user 108128
yesterday
My script says $omega$ is not exact because there's no function f such that $df= omega$ even though $domega = 0$, for $omega$ on $R^2 0$
– Mohamed Mossad
yesterday
add a comment |Â
up vote
1
down vote
This is essentially a long comment to the other answer, but worth making an answer of its own, since it highlights a useful trick which also works for Stoke's Theorem, as well as explains the $2 pi n$ answer. It's true that you can't apply Green's theorem to this problem directly, since the function contains a pole at the origin, but this doesn't stop you from applying Green's theorem on other regions that may also be useful.
Observe that your triangle $T$ does not pass through the origin. This means that you can find a small disk inside the triangle containing the origin $D$. Now consider the region $T - D$. This has two boundary components with opposite orientations, and we can consider the integral over $T - D$.
This gives us:
$$int_textinterior of T - D d omega = int_T omega - int_D omega$$ by Green's Theorem (or the Stoke's Theorem for differential forms, if you prefer to call it this). You can check that on this region, the left integral is $0$, as $d omega$ is $0$ everywhere here. This means that you can replace the integral of $T$ with the integral over any nice circle you like encompassing the origin, since the left side is $0$.
Now recognize that this integral is super friendly to polar coordinates, and convert to these. You will get the correct answer.
answered yesterday
Alfred Yerger
9,2761743
I know it's friendly to calculate in polar coordinates because I had an example of it on $c:[0,2pi] rightarrow R^2$ with $t rightarrow (cos t, sin t)$ which is equal $2pi$ when $omega$ isn't defined on 0, but is it the same with the triangle which doesn't pass through the origin?
– Mohamed Mossad
yesterday
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
We know that
$$omega=dfrac-y ,dx +x,dyx^2 +y^2=darctandfracyx$$
is an exact differential and the area $Delta$ includes the origin, since the $arctandfracyx$ winds around the origin one time then
$$int_Delta omega=2pi$$
edited yesterday
Michael Hardy
204k23185460
answered yesterday
user 108128
18.6k41544
May I ask how did you come to that answer? like what's intuitive about it? I know that 1-form is quite well-known but I am supposed to calculate the integral not use a trigonometric function which I don't know.
– Mohamed Mossad
yesterday
When you have an exact differential over a closed area, and while it winds around the origin $n$ times then it's integral is $2pi n$. Also as you said you can't apply green thm, except these you are forced to parametrize the triangle and substitude them in the exact form.
– user 108128
yesterday
My script says $omega$ is not exact because there's no function f such that $df= omega$ even though $domega = 0$, for $omega$ on $R^2 0$
– Mohamed Mossad
yesterday
add a comment |Â
up vote
2
down vote
accepted
We know that
$$omega=dfrac-y ,dx +x,dyx^2 +y^2=darctandfracyx$$
is an exact differential and the area $Delta$ includes the origin, since the $arctandfracyx$ winds around the origin one time then
$$int_Delta omega=2pi$$
edited yesterday
Michael Hardy
204k23185460
answered yesterday
user 108128
18.6k41544
May I ask how did you come to that answer? like what's intuitive about it? I know that 1-form is quite well-known but I am supposed to calculate the integral not use a trigonometric function which I don't know.
– Mohamed Mossad
yesterday
When you have an exact differential over a closed area, and while it winds around the origin $n$ times then it's integral is $2pi n$. Also as you said you can't apply green thm, except these you are forced to parametrize the triangle and substitude them in the exact form.
– user 108128
yesterday
My script says $omega$ is not exact because there's no function f such that $df= omega$ even though $domega = 0$, for $omega$ on $R^2 0$
– Mohamed Mossad
yesterday
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
We know that
$$omega=dfrac-y ,dx +x,dyx^2 +y^2=darctandfracyx$$
is an exact differential and the area $Delta$ includes the origin, since the $arctandfracyx$ winds around the origin one time then
$$int_Delta omega=2pi$$
edited yesterday
Michael Hardy
204k23185460
answered yesterday
user 108128
18.6k41544
We know that
$$omega=dfrac-y ,dx +x,dyx^2 +y^2=darctandfracyx$$
is an exact differential and the area $Delta$ includes the origin, since the $arctandfracyx$ winds around the origin one time then
$$int_Delta omega=2pi$$
edited yesterday
Michael Hardy
204k23185460
answered yesterday
user 108128
18.6k41544
edited yesterday
Michael Hardy
204k23185460
edited yesterday
Michael Hardy
204k23185460
edited yesterday
Michael Hardy
204k23185460
204k23185460
answered yesterday
user 108128
18.6k41544
answered yesterday
user 108128
18.6k41544
answered yesterday
user 108128
18.6k41544
18.6k41544
May I ask how did you come to that answer? like what's intuitive about it? I know that 1-form is quite well-known but I am supposed to calculate the integral not use a trigonometric function which I don't know.
– Mohamed Mossad
yesterday
When you have an exact differential over a closed area, and while it winds around the origin $n$ times then it's integral is $2pi n$. Also as you said you can't apply green thm, except these you are forced to parametrize the triangle and substitude them in the exact form.
– user 108128
yesterday
My script says $omega$ is not exact because there's no function f such that $df= omega$ even though $domega = 0$, for $omega$ on $R^2 0$
– Mohamed Mossad
yesterday
add a comment |Â
May I ask how did you come to that answer? like what's intuitive about it? I know that 1-form is quite well-known but I am supposed to calculate the integral not use a trigonometric function which I don't know.
– Mohamed Mossad
yesterday
When you have an exact differential over a closed area, and while it winds around the origin $n$ times then it's integral is $2pi n$. Also as you said you can't apply green thm, except these you are forced to parametrize the triangle and substitude them in the exact form.
– user 108128
yesterday
My script says $omega$ is not exact because there's no function f such that $df= omega$ even though $domega = 0$, for $omega$ on $R^2 0$
– Mohamed Mossad
yesterday
May I ask how did you come to that answer? like what's intuitive about it? I know that 1-form is quite well-known but I am supposed to calculate the integral not use a trigonometric function which I don't know.
– Mohamed Mossad
yesterday
May I ask how did you come to that answer? like what's intuitive about it? I know that 1-form is quite well-known but I am supposed to calculate the integral not use a trigonometric function which I don't know.
– Mohamed Mossad
yesterday
When you have an exact differential over a closed area, and while it winds around the origin $n$ times then it's integral is $2pi n$. Also as you said you can't apply green thm, except these you are forced to parametrize the triangle and substitude them in the exact form.
– user 108128
yesterday
When you have an exact differential over a closed area, and while it winds around the origin $n$ times then it's integral is $2pi n$. Also as you said you can't apply green thm, except these you are forced to parametrize the triangle and substitude them in the exact form.
– user 108128
yesterday
My script says $omega$ is not exact because there's no function f such that $df= omega$ even though $domega = 0$, for $omega$ on $R^2 0$
– Mohamed Mossad
yesterday
My script says $omega$ is not exact because there's no function f such that $df= omega$ even though $domega = 0$, for $omega$ on $R^2 0$
– Mohamed Mossad
yesterday
add a comment |Â
up vote
1
down vote
This is essentially a long comment to the other answer, but worth making an answer of its own, since it highlights a useful trick which also works for Stoke's Theorem, as well as explains the $2 pi n$ answer. It's true that you can't apply Green's theorem to this problem directly, since the function contains a pole at the origin, but this doesn't stop you from applying Green's theorem on other regions that may also be useful.
Observe that your triangle $T$ does not pass through the origin. This means that you can find a small disk inside the triangle containing the origin $D$. Now consider the region $T - D$. This has two boundary components with opposite orientations, and we can consider the integral over $T - D$.
This gives us:
$$int_textinterior of T - D d omega = int_T omega - int_D omega$$ by Green's Theorem (or the Stoke's Theorem for differential forms, if you prefer to call it this). You can check that on this region, the left integral is $0$, as $d omega$ is $0$ everywhere here. This means that you can replace the integral of $T$ with the integral over any nice circle you like encompassing the origin, since the left side is $0$.
Now recognize that this integral is super friendly to polar coordinates, and convert to these. You will get the correct answer.
answered yesterday
Alfred Yerger
9,2761743
I know it's friendly to calculate in polar coordinates because I had an example of it on $c:[0,2pi] rightarrow R^2$ with $t rightarrow (cos t, sin t)$ which is equal $2pi$ when $omega$ isn't defined on 0, but is it the same with the triangle which doesn't pass through the origin?
– Mohamed Mossad
yesterday
add a comment |Â
up vote
1
down vote
This is essentially a long comment to the other answer, but worth making an answer of its own, since it highlights a useful trick which also works for Stoke's Theorem, as well as explains the $2 pi n$ answer. It's true that you can't apply Green's theorem to this problem directly, since the function contains a pole at the origin, but this doesn't stop you from applying Green's theorem on other regions that may also be useful.
Observe that your triangle $T$ does not pass through the origin. This means that you can find a small disk inside the triangle containing the origin $D$. Now consider the region $T - D$. This has two boundary components with opposite orientations, and we can consider the integral over $T - D$.
This gives us:
$$int_textinterior of T - D d omega = int_T omega - int_D omega$$ by Green's Theorem (or the Stoke's Theorem for differential forms, if you prefer to call it this). You can check that on this region, the left integral is $0$, as $d omega$ is $0$ everywhere here. This means that you can replace the integral of $T$ with the integral over any nice circle you like encompassing the origin, since the left side is $0$.
Now recognize that this integral is super friendly to polar coordinates, and convert to these. You will get the correct answer.
answered yesterday
Alfred Yerger
9,2761743
I know it's friendly to calculate in polar coordinates because I had an example of it on $c:[0,2pi] rightarrow R^2$ with $t rightarrow (cos t, sin t)$ which is equal $2pi$ when $omega$ isn't defined on 0, but is it the same with the triangle which doesn't pass through the origin?
– Mohamed Mossad
yesterday
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This is essentially a long comment to the other answer, but worth making an answer of its own, since it highlights a useful trick which also works for Stoke's Theorem, as well as explains the $2 pi n$ answer. It's true that you can't apply Green's theorem to this problem directly, since the function contains a pole at the origin, but this doesn't stop you from applying Green's theorem on other regions that may also be useful.
Observe that your triangle $T$ does not pass through the origin. This means that you can find a small disk inside the triangle containing the origin $D$. Now consider the region $T - D$. This has two boundary components with opposite orientations, and we can consider the integral over $T - D$.
This gives us:
$$int_textinterior of T - D d omega = int_T omega - int_D omega$$ by Green's Theorem (or the Stoke's Theorem for differential forms, if you prefer to call it this). You can check that on this region, the left integral is $0$, as $d omega$ is $0$ everywhere here. This means that you can replace the integral of $T$ with the integral over any nice circle you like encompassing the origin, since the left side is $0$.
Now recognize that this integral is super friendly to polar coordinates, and convert to these. You will get the correct answer.
answered yesterday
Alfred Yerger
9,2761743
This is essentially a long comment to the other answer, but worth making an answer of its own, since it highlights a useful trick which also works for Stoke's Theorem, as well as explains the $2 pi n$ answer. It's true that you can't apply Green's theorem to this problem directly, since the function contains a pole at the origin, but this doesn't stop you from applying Green's theorem on other regions that may also be useful.
Observe that your triangle $T$ does not pass through the origin. This means that you can find a small disk inside the triangle containing the origin $D$. Now consider the region $T - D$. This has two boundary components with opposite orientations, and we can consider the integral over $T - D$.
This gives us:
$$int_textinterior of T - D d omega = int_T omega - int_D omega$$ by Green's Theorem (or the Stoke's Theorem for differential forms, if you prefer to call it this). You can check that on this region, the left integral is $0$, as $d omega$ is $0$ everywhere here. This means that you can replace the integral of $T$ with the integral over any nice circle you like encompassing the origin, since the left side is $0$.
Now recognize that this integral is super friendly to polar coordinates, and convert to these. You will get the correct answer.
answered yesterday
Alfred Yerger
9,2761743
answered yesterday
Alfred Yerger
9,2761743
answered yesterday
Alfred Yerger
9,2761743
answered yesterday
Alfred Yerger
9,2761743
9,2761743
I know it's friendly to calculate in polar coordinates because I had an example of it on $c:[0,2pi] rightarrow R^2$ with $t rightarrow (cos t, sin t)$ which is equal $2pi$ when $omega$ isn't defined on 0, but is it the same with the triangle which doesn't pass through the origin?
– Mohamed Mossad
yesterday
add a comment |Â
I know it's friendly to calculate in polar coordinates because I had an example of it on $c:[0,2pi] rightarrow R^2$ with $t rightarrow (cos t, sin t)$ which is equal $2pi$ when $omega$ isn't defined on 0, but is it the same with the triangle which doesn't pass through the origin?
– Mohamed Mossad
yesterday
I know it's friendly to calculate in polar coordinates because I had an example of it on $c:[0,2pi] rightarrow R^2$ with $t rightarrow (cos t, sin t)$ which is equal $2pi$ when $omega$ isn't defined on 0, but is it the same with the triangle which doesn't pass through the origin?
– Mohamed Mossad
yesterday
I know it's friendly to calculate in polar coordinates because I had an example of it on $c:[0,2pi] rightarrow R^2$ with $t rightarrow (cos t, sin t)$ which is equal $2pi$ when $omega$ isn't defined on 0, but is it the same with the triangle which doesn't pass through the origin?
– Mohamed Mossad
yesterday
add a comment |Â
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