Integral of a differential form along a non-defined path
$begingroup$
Let $R>0$ and $Omega={(x,y)in mathbb{R^2}:x^2+y^2<R^2,y>0}$. Consider also $omega (x,y)=x^2dx+2xydy$. My goal is to prove that $int_{partial_{+}Omega}omega=frac{4}{3}R^3$, where $partial _+ Omega$ denotes the "edge" of $Omega$ positively oriented. I know that using the Green-Riemann I can show that, but could I use the definition? That is:
$$int_{partial_{+}Omega}omega=int_0^{pi}(omega spacecircspacegamma)(theta)gamma'(theta)dtheta$$
Where $gamma(theta)=(Rcos(theta),Rsin(theta))$. On impulse I'd say it's not possible since it's not defined, but then again, the edge has $0$ volume, hence any semi circle suficiently together would be a pretty good approximation, thus it should work, what is your opinion?
differential-geometry greens-theorem
asked Jan 4 at 21:48
BidonBidon
967
$endgroup$
add a comment |
$begingroup$
Let $R>0$ and $Omega={(x,y)in mathbb{R^2}:x^2+y^2<R^2,y>0}$. Consider also $omega (x,y)=x^2dx+2xydy$. My goal is to prove that $int_{partial_{+}Omega}omega=frac{4}{3}R^3$, where $partial _+ Omega$ denotes the "edge" of $Omega$ positively oriented. I know that using the Green-Riemann I can show that, but could I use the definition? That is:
$$int_{partial_{+}Omega}omega=int_0^{pi}(omega spacecircspacegamma)(theta)gamma'(theta)dtheta$$
Where $gamma(theta)=(Rcos(theta),Rsin(theta))$. On impulse I'd say it's not possible since it's not defined, but then again, the edge has $0$ volume, hence any semi circle suficiently together would be a pretty good approximation, thus it should work, what is your opinion?
differential-geometry greens-theorem
asked Jan 4 at 21:48
BidonBidon
967
$endgroup$
$begingroup$
What do you mean when you say the path is non-defined?!
$endgroup$
– Ted Shifrin
Jan 4 at 23:11
$begingroup$
Let $Gamma_1={(x,y) in mathbb{R^2}:x^2+y^2=R^2}$ and $Gamma_2={(x,y)in mathbb{R^2}: y=0, -Rgeq x leq R}$. I mean that $Gamma_1 cup Gamma_2 notin Omega$.
$endgroup$
– Bidon
Jan 4 at 23:15
1
$begingroup$
The boundary of the open region is still a perfectly defined curve. Some textbooks (professors) take the boundary of the closed region (where you include the boundary curve); either way, the curve is the same.
$endgroup$
– Ted Shifrin
Jan 4 at 23:18
add a comment |
$begingroup$
Let $R>0$ and $Omega={(x,y)in mathbb{R^2}:x^2+y^2<R^2,y>0}$. Consider also $omega (x,y)=x^2dx+2xydy$. My goal is to prove that $int_{partial_{+}Omega}omega=frac{4}{3}R^3$, where $partial _+ Omega$ denotes the "edge" of $Omega$ positively oriented. I know that using the Green-Riemann I can show that, but could I use the definition? That is:
$$int_{partial_{+}Omega}omega=int_0^{pi}(omega spacecircspacegamma)(theta)gamma'(theta)dtheta$$
Where $gamma(theta)=(Rcos(theta),Rsin(theta))$. On impulse I'd say it's not possible since it's not defined, but then again, the edge has $0$ volume, hence any semi circle suficiently together would be a pretty good approximation, thus it should work, what is your opinion?
differential-geometry greens-theorem
asked Jan 4 at 21:48
BidonBidon
967
$endgroup$
Let $R>0$ and $Omega={(x,y)in mathbb{R^2}:x^2+y^2<R^2,y>0}$. Consider also $omega (x,y)=x^2dx+2xydy$. My goal is to prove that $int_{partial_{+}Omega}omega=frac{4}{3}R^3$, where $partial _+ Omega$ denotes the "edge" of $Omega$ positively oriented. I know that using the Green-Riemann I can show that, but could I use the definition? That is:
$$int_{partial_{+}Omega}omega=int_0^{pi}(omega spacecircspacegamma)(theta)gamma'(theta)dtheta$$
Where $gamma(theta)=(Rcos(theta),Rsin(theta))$. On impulse I'd say it's not possible since it's not defined, but then again, the edge has $0$ volume, hence any semi circle suficiently together would be a pretty good approximation, thus it should work, what is your opinion?
differential-geometry greens-theorem
differential-geometry greens-theorem
asked Jan 4 at 21:48
BidonBidon
967
asked Jan 4 at 21:48
BidonBidon
967
asked Jan 4 at 21:48
BidonBidon
967
asked Jan 4 at 21:48
BidonBidon
967
asked Jan 4 at 21:48
BidonBidon
967
967
$begingroup$
What do you mean when you say the path is non-defined?!
$endgroup$
– Ted Shifrin
Jan 4 at 23:11
$begingroup$
Let $Gamma_1={(x,y) in mathbb{R^2}:x^2+y^2=R^2}$ and $Gamma_2={(x,y)in mathbb{R^2}: y=0, -Rgeq x leq R}$. I mean that $Gamma_1 cup Gamma_2 notin Omega$.
$endgroup$
– Bidon
Jan 4 at 23:15
1
$begingroup$
The boundary of the open region is still a perfectly defined curve. Some textbooks (professors) take the boundary of the closed region (where you include the boundary curve); either way, the curve is the same.
$endgroup$
– Ted Shifrin
Jan 4 at 23:18
add a comment |
$begingroup$
What do you mean when you say the path is non-defined?!
$endgroup$
– Ted Shifrin
Jan 4 at 23:11
$begingroup$
Let $Gamma_1={(x,y) in mathbb{R^2}:x^2+y^2=R^2}$ and $Gamma_2={(x,y)in mathbb{R^2}: y=0, -Rgeq x leq R}$. I mean that $Gamma_1 cup Gamma_2 notin Omega$.
$endgroup$
– Bidon
Jan 4 at 23:15
1
$begingroup$
The boundary of the open region is still a perfectly defined curve. Some textbooks (professors) take the boundary of the closed region (where you include the boundary curve); either way, the curve is the same.
$endgroup$
– Ted Shifrin
Jan 4 at 23:18
$begingroup$
What do you mean when you say the path is non-defined?!
$endgroup$
– Ted Shifrin
Jan 4 at 23:11
$begingroup$
What do you mean when you say the path is non-defined?!
$endgroup$
– Ted Shifrin
Jan 4 at 23:11
$begingroup$
Let $Gamma_1={(x,y) in mathbb{R^2}:x^2+y^2=R^2}$ and $Gamma_2={(x,y)in mathbb{R^2}: y=0, -Rgeq x leq R}$. I mean that $Gamma_1 cup Gamma_2 notin Omega$.
$endgroup$
– Bidon
Jan 4 at 23:15
$begingroup$
Let $Gamma_1={(x,y) in mathbb{R^2}:x^2+y^2=R^2}$ and $Gamma_2={(x,y)in mathbb{R^2}: y=0, -Rgeq x leq R}$. I mean that $Gamma_1 cup Gamma_2 notin Omega$.
$endgroup$
– Bidon
Jan 4 at 23:15
1
1
$begingroup$
The boundary of the open region is still a perfectly defined curve. Some textbooks (professors) take the boundary of the closed region (where you include the boundary curve); either way, the curve is the same.
$endgroup$
– Ted Shifrin
Jan 4 at 23:18
$begingroup$
The boundary of the open region is still a perfectly defined curve. Some textbooks (professors) take the boundary of the closed region (where you include the boundary curve); either way, the curve is the same.
$endgroup$
– Ted Shifrin
Jan 4 at 23:18
add a comment |
1 Answer
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active
oldest
votes
$begingroup$
Yes, we can do the computation that way. The boundary actually has two components:
The straight component $C_1$, parameterised by $gamma_1 : [-1, 1] to C_1$ sending $t mapsto (Rt, 0)$.
The curved component $C_2$, parameterised by $gamma_2 : [0, pi] to C_2$ sending $t mapsto (Rcos t, Rsin t)$.
We can now evaluate the integral by pulling back the $1$-form $omega$ to $[-1, 1]$ and $[0, pi]$ via $gamma_1$ and $gamma_2$.
begin{align} oint_{partial_+ Omega} omega &= int_{[-1, 1]} gamma_1^star (omega)+int_{[0,pi]}gamma_2^star (omega) \ &= int_{-1}^1 (Rt)^2(Rdt) + int_0^pi left( (R^2cos^2 t).(-Rsin t dt) + (2R^2 sin t cos t).(Rcos t dt) right) \ &= int_{-1}^1 R^3 t^2 dt + int_{0}^{pi} R^3 sin t cos^2 t dt \ &= frac{4}{3} R^3end{align}
[To spell it out:
$$ int_{[-1,1]} gamma_1^star (x^2 dx) = int_{[-1, 1]} ((x circ gamma_1)(t))^2 left( frac{d(x circ gamma_1)}{dt}(t). dt right) = int_{-1}^1 (Rt)^2 left( frac{d(Rt)}{dt} dtright)$$
and so on...]
A final remark: these integrals are not ill-defined at all. $omega$ is a $1$-form, so you should expect to be able to integrate it along one-dimensional things like $partial_+ Omega$. The fact that $partial_+ Omega $ has zero measure as a subset of $mathbb R^2$ should not bother you.
answered Jan 4 at 22:25
Kenny WongKenny Wong
18.7k21439
$endgroup$
$begingroup$
I'm confortable with the fact that it has zero measure, what was confusing me is that the border of $Omega$ was not explicitly defined, if it were $x^2+y^2leq R^2$ then I would have no questions, but since math has such delicate details I just wanted to make sure.
$endgroup$
– Bidon
Jan 4 at 22:43
$begingroup$
@Bidon For the purposes of evaluating $int_{Omega} d omega$, the fact that the boundary of $Omega$ has zero measure is relevant! That's because here, we're integrating a $2$-form on a $2$-dimensional space. So the two-dimensional measure becomes relevant.
$endgroup$
– Kenny Wong
Jan 4 at 22:45
$begingroup$
… and therefore, $int_{Omega} domega = int_{bar Omega} domega$ (where $barOmega$ is the closure of $Omega$ in $mathbb R^2$, and $omega$ has of course been extended smoothly to include the boundary).
$endgroup$
– Kenny Wong
Jan 4 at 22:46
$begingroup$
Yes, but doesn't the boundary always have zero measure? Because in $mathbb{R^2}$ is simply a line, hence no area, in $mathbb{R^3}$ is a surface thus has no volume, and so forth into higher dimensions. What am I missing? PS: if at some point this starts to be annoying just ignore, if not, thank you so much for the patience
$endgroup$
– Bidon
Jan 4 at 22:58
1
$begingroup$
@Bidon So I would do this computation in two steps. (1) First, I would argue that $int_{Omega} domega = int_{bar Omega} domega$, using the fact that $barOmega - Omega$ has zero measure. (2) Next, notice that $barOmega$ is a manifold-with-boundary! So $int_{barOmega} domega = int_{partial barOmega} omega$ by Stoke's theorem.
$endgroup$
– Kenny Wong
Jan 4 at 23:17
|
show 4 more comments
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1 Answer
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$begingroup$
Yes, we can do the computation that way. The boundary actually has two components:
The straight component $C_1$, parameterised by $gamma_1 : [-1, 1] to C_1$ sending $t mapsto (Rt, 0)$.
The curved component $C_2$, parameterised by $gamma_2 : [0, pi] to C_2$ sending $t mapsto (Rcos t, Rsin t)$.
We can now evaluate the integral by pulling back the $1$-form $omega$ to $[-1, 1]$ and $[0, pi]$ via $gamma_1$ and $gamma_2$.
begin{align} oint_{partial_+ Omega} omega &= int_{[-1, 1]} gamma_1^star (omega)+int_{[0,pi]}gamma_2^star (omega) \ &= int_{-1}^1 (Rt)^2(Rdt) + int_0^pi left( (R^2cos^2 t).(-Rsin t dt) + (2R^2 sin t cos t).(Rcos t dt) right) \ &= int_{-1}^1 R^3 t^2 dt + int_{0}^{pi} R^3 sin t cos^2 t dt \ &= frac{4}{3} R^3end{align}
[To spell it out:
$$ int_{[-1,1]} gamma_1^star (x^2 dx) = int_{[-1, 1]} ((x circ gamma_1)(t))^2 left( frac{d(x circ gamma_1)}{dt}(t). dt right) = int_{-1}^1 (Rt)^2 left( frac{d(Rt)}{dt} dtright)$$
and so on...]
A final remark: these integrals are not ill-defined at all. $omega$ is a $1$-form, so you should expect to be able to integrate it along one-dimensional things like $partial_+ Omega$. The fact that $partial_+ Omega $ has zero measure as a subset of $mathbb R^2$ should not bother you.
answered Jan 4 at 22:25
Kenny WongKenny Wong
18.7k21439
$endgroup$
$begingroup$
I'm confortable with the fact that it has zero measure, what was confusing me is that the border of $Omega$ was not explicitly defined, if it were $x^2+y^2leq R^2$ then I would have no questions, but since math has such delicate details I just wanted to make sure.
$endgroup$
– Bidon
Jan 4 at 22:43
$begingroup$
@Bidon For the purposes of evaluating $int_{Omega} d omega$, the fact that the boundary of $Omega$ has zero measure is relevant! That's because here, we're integrating a $2$-form on a $2$-dimensional space. So the two-dimensional measure becomes relevant.
$endgroup$
– Kenny Wong
Jan 4 at 22:45
$begingroup$
… and therefore, $int_{Omega} domega = int_{bar Omega} domega$ (where $barOmega$ is the closure of $Omega$ in $mathbb R^2$, and $omega$ has of course been extended smoothly to include the boundary).
$endgroup$
– Kenny Wong
Jan 4 at 22:46
$begingroup$
Yes, but doesn't the boundary always have zero measure? Because in $mathbb{R^2}$ is simply a line, hence no area, in $mathbb{R^3}$ is a surface thus has no volume, and so forth into higher dimensions. What am I missing? PS: if at some point this starts to be annoying just ignore, if not, thank you so much for the patience
$endgroup$
– Bidon
Jan 4 at 22:58
1
$begingroup$
@Bidon So I would do this computation in two steps. (1) First, I would argue that $int_{Omega} domega = int_{bar Omega} domega$, using the fact that $barOmega - Omega$ has zero measure. (2) Next, notice that $barOmega$ is a manifold-with-boundary! So $int_{barOmega} domega = int_{partial barOmega} omega$ by Stoke's theorem.
$endgroup$
– Kenny Wong
Jan 4 at 23:17
|
show 4 more comments
$begingroup$
Yes, we can do the computation that way. The boundary actually has two components:
The straight component $C_1$, parameterised by $gamma_1 : [-1, 1] to C_1$ sending $t mapsto (Rt, 0)$.
The curved component $C_2$, parameterised by $gamma_2 : [0, pi] to C_2$ sending $t mapsto (Rcos t, Rsin t)$.
We can now evaluate the integral by pulling back the $1$-form $omega$ to $[-1, 1]$ and $[0, pi]$ via $gamma_1$ and $gamma_2$.
begin{align} oint_{partial_+ Omega} omega &= int_{[-1, 1]} gamma_1^star (omega)+int_{[0,pi]}gamma_2^star (omega) \ &= int_{-1}^1 (Rt)^2(Rdt) + int_0^pi left( (R^2cos^2 t).(-Rsin t dt) + (2R^2 sin t cos t).(Rcos t dt) right) \ &= int_{-1}^1 R^3 t^2 dt + int_{0}^{pi} R^3 sin t cos^2 t dt \ &= frac{4}{3} R^3end{align}
[To spell it out:
$$ int_{[-1,1]} gamma_1^star (x^2 dx) = int_{[-1, 1]} ((x circ gamma_1)(t))^2 left( frac{d(x circ gamma_1)}{dt}(t). dt right) = int_{-1}^1 (Rt)^2 left( frac{d(Rt)}{dt} dtright)$$
and so on...]
A final remark: these integrals are not ill-defined at all. $omega$ is a $1$-form, so you should expect to be able to integrate it along one-dimensional things like $partial_+ Omega$. The fact that $partial_+ Omega $ has zero measure as a subset of $mathbb R^2$ should not bother you.
answered Jan 4 at 22:25
Kenny WongKenny Wong
18.7k21439
$endgroup$
$begingroup$
I'm confortable with the fact that it has zero measure, what was confusing me is that the border of $Omega$ was not explicitly defined, if it were $x^2+y^2leq R^2$ then I would have no questions, but since math has such delicate details I just wanted to make sure.
$endgroup$
– Bidon
Jan 4 at 22:43
$begingroup$
@Bidon For the purposes of evaluating $int_{Omega} d omega$, the fact that the boundary of $Omega$ has zero measure is relevant! That's because here, we're integrating a $2$-form on a $2$-dimensional space. So the two-dimensional measure becomes relevant.
$endgroup$
– Kenny Wong
Jan 4 at 22:45
$begingroup$
… and therefore, $int_{Omega} domega = int_{bar Omega} domega$ (where $barOmega$ is the closure of $Omega$ in $mathbb R^2$, and $omega$ has of course been extended smoothly to include the boundary).
$endgroup$
– Kenny Wong
Jan 4 at 22:46
$begingroup$
Yes, but doesn't the boundary always have zero measure? Because in $mathbb{R^2}$ is simply a line, hence no area, in $mathbb{R^3}$ is a surface thus has no volume, and so forth into higher dimensions. What am I missing? PS: if at some point this starts to be annoying just ignore, if not, thank you so much for the patience
$endgroup$
– Bidon
Jan 4 at 22:58
1
$begingroup$
@Bidon So I would do this computation in two steps. (1) First, I would argue that $int_{Omega} domega = int_{bar Omega} domega$, using the fact that $barOmega - Omega$ has zero measure. (2) Next, notice that $barOmega$ is a manifold-with-boundary! So $int_{barOmega} domega = int_{partial barOmega} omega$ by Stoke's theorem.
$endgroup$
– Kenny Wong
Jan 4 at 23:17
|
show 4 more comments
$begingroup$
Yes, we can do the computation that way. The boundary actually has two components:
The straight component $C_1$, parameterised by $gamma_1 : [-1, 1] to C_1$ sending $t mapsto (Rt, 0)$.
The curved component $C_2$, parameterised by $gamma_2 : [0, pi] to C_2$ sending $t mapsto (Rcos t, Rsin t)$.
We can now evaluate the integral by pulling back the $1$-form $omega$ to $[-1, 1]$ and $[0, pi]$ via $gamma_1$ and $gamma_2$.
begin{align} oint_{partial_+ Omega} omega &= int_{[-1, 1]} gamma_1^star (omega)+int_{[0,pi]}gamma_2^star (omega) \ &= int_{-1}^1 (Rt)^2(Rdt) + int_0^pi left( (R^2cos^2 t).(-Rsin t dt) + (2R^2 sin t cos t).(Rcos t dt) right) \ &= int_{-1}^1 R^3 t^2 dt + int_{0}^{pi} R^3 sin t cos^2 t dt \ &= frac{4}{3} R^3end{align}
[To spell it out:
$$ int_{[-1,1]} gamma_1^star (x^2 dx) = int_{[-1, 1]} ((x circ gamma_1)(t))^2 left( frac{d(x circ gamma_1)}{dt}(t). dt right) = int_{-1}^1 (Rt)^2 left( frac{d(Rt)}{dt} dtright)$$
and so on...]
A final remark: these integrals are not ill-defined at all. $omega$ is a $1$-form, so you should expect to be able to integrate it along one-dimensional things like $partial_+ Omega$. The fact that $partial_+ Omega $ has zero measure as a subset of $mathbb R^2$ should not bother you.
answered Jan 4 at 22:25
Kenny WongKenny Wong
18.7k21439
$endgroup$
Yes, we can do the computation that way. The boundary actually has two components:
The straight component $C_1$, parameterised by $gamma_1 : [-1, 1] to C_1$ sending $t mapsto (Rt, 0)$.
The curved component $C_2$, parameterised by $gamma_2 : [0, pi] to C_2$ sending $t mapsto (Rcos t, Rsin t)$.
We can now evaluate the integral by pulling back the $1$-form $omega$ to $[-1, 1]$ and $[0, pi]$ via $gamma_1$ and $gamma_2$.
begin{align} oint_{partial_+ Omega} omega &= int_{[-1, 1]} gamma_1^star (omega)+int_{[0,pi]}gamma_2^star (omega) \ &= int_{-1}^1 (Rt)^2(Rdt) + int_0^pi left( (R^2cos^2 t).(-Rsin t dt) + (2R^2 sin t cos t).(Rcos t dt) right) \ &= int_{-1}^1 R^3 t^2 dt + int_{0}^{pi} R^3 sin t cos^2 t dt \ &= frac{4}{3} R^3end{align}
[To spell it out:
$$ int_{[-1,1]} gamma_1^star (x^2 dx) = int_{[-1, 1]} ((x circ gamma_1)(t))^2 left( frac{d(x circ gamma_1)}{dt}(t). dt right) = int_{-1}^1 (Rt)^2 left( frac{d(Rt)}{dt} dtright)$$
and so on...]
A final remark: these integrals are not ill-defined at all. $omega$ is a $1$-form, so you should expect to be able to integrate it along one-dimensional things like $partial_+ Omega$. The fact that $partial_+ Omega $ has zero measure as a subset of $mathbb R^2$ should not bother you.
answered Jan 4 at 22:25
Kenny WongKenny Wong
18.7k21439
edited Jan 4 at 22:33
answered Jan 4 at 22:25
Kenny WongKenny Wong
18.7k21439
answered Jan 4 at 22:25
Kenny WongKenny Wong
18.7k21439
answered Jan 4 at 22:25
Kenny WongKenny Wong
18.7k21439
18.7k21439
$begingroup$
I'm confortable with the fact that it has zero measure, what was confusing me is that the border of $Omega$ was not explicitly defined, if it were $x^2+y^2leq R^2$ then I would have no questions, but since math has such delicate details I just wanted to make sure.
$endgroup$
– Bidon
Jan 4 at 22:43
$begingroup$
@Bidon For the purposes of evaluating $int_{Omega} d omega$, the fact that the boundary of $Omega$ has zero measure is relevant! That's because here, we're integrating a $2$-form on a $2$-dimensional space. So the two-dimensional measure becomes relevant.
$endgroup$
– Kenny Wong
Jan 4 at 22:45
$begingroup$
… and therefore, $int_{Omega} domega = int_{bar Omega} domega$ (where $barOmega$ is the closure of $Omega$ in $mathbb R^2$, and $omega$ has of course been extended smoothly to include the boundary).
$endgroup$
– Kenny Wong
Jan 4 at 22:46
$begingroup$
Yes, but doesn't the boundary always have zero measure? Because in $mathbb{R^2}$ is simply a line, hence no area, in $mathbb{R^3}$ is a surface thus has no volume, and so forth into higher dimensions. What am I missing? PS: if at some point this starts to be annoying just ignore, if not, thank you so much for the patience
$endgroup$
– Bidon
Jan 4 at 22:58
1
$begingroup$
@Bidon So I would do this computation in two steps. (1) First, I would argue that $int_{Omega} domega = int_{bar Omega} domega$, using the fact that $barOmega - Omega$ has zero measure. (2) Next, notice that $barOmega$ is a manifold-with-boundary! So $int_{barOmega} domega = int_{partial barOmega} omega$ by Stoke's theorem.
$endgroup$
– Kenny Wong
Jan 4 at 23:17
|
show 4 more comments
$begingroup$
I'm confortable with the fact that it has zero measure, what was confusing me is that the border of $Omega$ was not explicitly defined, if it were $x^2+y^2leq R^2$ then I would have no questions, but since math has such delicate details I just wanted to make sure.
$endgroup$
– Bidon
Jan 4 at 22:43
$begingroup$
@Bidon For the purposes of evaluating $int_{Omega} d omega$, the fact that the boundary of $Omega$ has zero measure is relevant! That's because here, we're integrating a $2$-form on a $2$-dimensional space. So the two-dimensional measure becomes relevant.
$endgroup$
– Kenny Wong
Jan 4 at 22:45
$begingroup$
… and therefore, $int_{Omega} domega = int_{bar Omega} domega$ (where $barOmega$ is the closure of $Omega$ in $mathbb R^2$, and $omega$ has of course been extended smoothly to include the boundary).
$endgroup$
– Kenny Wong
Jan 4 at 22:46
$begingroup$
Yes, but doesn't the boundary always have zero measure? Because in $mathbb{R^2}$ is simply a line, hence no area, in $mathbb{R^3}$ is a surface thus has no volume, and so forth into higher dimensions. What am I missing? PS: if at some point this starts to be annoying just ignore, if not, thank you so much for the patience
$endgroup$
– Bidon
Jan 4 at 22:58
1
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@Bidon So I would do this computation in two steps. (1) First, I would argue that $int_{Omega} domega = int_{bar Omega} domega$, using the fact that $barOmega - Omega$ has zero measure. (2) Next, notice that $barOmega$ is a manifold-with-boundary! So $int_{barOmega} domega = int_{partial barOmega} omega$ by Stoke's theorem.
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– Kenny Wong
Jan 4 at 23:17
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I'm confortable with the fact that it has zero measure, what was confusing me is that the border of $Omega$ was not explicitly defined, if it were $x^2+y^2leq R^2$ then I would have no questions, but since math has such delicate details I just wanted to make sure.
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– Bidon
Jan 4 at 22:43
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I'm confortable with the fact that it has zero measure, what was confusing me is that the border of $Omega$ was not explicitly defined, if it were $x^2+y^2leq R^2$ then I would have no questions, but since math has such delicate details I just wanted to make sure.
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– Bidon
Jan 4 at 22:43
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@Bidon For the purposes of evaluating $int_{Omega} d omega$, the fact that the boundary of $Omega$ has zero measure is relevant! That's because here, we're integrating a $2$-form on a $2$-dimensional space. So the two-dimensional measure becomes relevant.
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– Kenny Wong
Jan 4 at 22:45
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@Bidon For the purposes of evaluating $int_{Omega} d omega$, the fact that the boundary of $Omega$ has zero measure is relevant! That's because here, we're integrating a $2$-form on a $2$-dimensional space. So the two-dimensional measure becomes relevant.
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– Kenny Wong
Jan 4 at 22:45
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… and therefore, $int_{Omega} domega = int_{bar Omega} domega$ (where $barOmega$ is the closure of $Omega$ in $mathbb R^2$, and $omega$ has of course been extended smoothly to include the boundary).
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– Kenny Wong
Jan 4 at 22:46
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… and therefore, $int_{Omega} domega = int_{bar Omega} domega$ (where $barOmega$ is the closure of $Omega$ in $mathbb R^2$, and $omega$ has of course been extended smoothly to include the boundary).
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– Kenny Wong
Jan 4 at 22:46
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Yes, but doesn't the boundary always have zero measure? Because in $mathbb{R^2}$ is simply a line, hence no area, in $mathbb{R^3}$ is a surface thus has no volume, and so forth into higher dimensions. What am I missing? PS: if at some point this starts to be annoying just ignore, if not, thank you so much for the patience
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– Bidon
Jan 4 at 22:58
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Yes, but doesn't the boundary always have zero measure? Because in $mathbb{R^2}$ is simply a line, hence no area, in $mathbb{R^3}$ is a surface thus has no volume, and so forth into higher dimensions. What am I missing? PS: if at some point this starts to be annoying just ignore, if not, thank you so much for the patience
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– Bidon
Jan 4 at 22:58
1
1
$begingroup$
@Bidon So I would do this computation in two steps. (1) First, I would argue that $int_{Omega} domega = int_{bar Omega} domega$, using the fact that $barOmega - Omega$ has zero measure. (2) Next, notice that $barOmega$ is a manifold-with-boundary! So $int_{barOmega} domega = int_{partial barOmega} omega$ by Stoke's theorem.
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– Kenny Wong
Jan 4 at 23:17
$begingroup$
@Bidon So I would do this computation in two steps. (1) First, I would argue that $int_{Omega} domega = int_{bar Omega} domega$, using the fact that $barOmega - Omega$ has zero measure. (2) Next, notice that $barOmega$ is a manifold-with-boundary! So $int_{barOmega} domega = int_{partial barOmega} omega$ by Stoke's theorem.
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– Kenny Wong
Jan 4 at 23:17
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What do you mean when you say the path is non-defined?!
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– Ted Shifrin
Jan 4 at 23:11
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Let $Gamma_1={(x,y) in mathbb{R^2}:x^2+y^2=R^2}$ and $Gamma_2={(x,y)in mathbb{R^2}: y=0, -Rgeq x leq R}$. I mean that $Gamma_1 cup Gamma_2 notin Omega$.
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– Bidon
Jan 4 at 23:15
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The boundary of the open region is still a perfectly defined curve. Some textbooks (professors) take the boundary of the closed region (where you include the boundary curve); either way, the curve is the same.
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– Ted Shifrin
Jan 4 at 23:18