Integral of a differential form along a non-defined path












1












$begingroup$


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







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$endgroup$












  • $begingroup$
    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
















1












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














1












1








1





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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1 Answer
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oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 4 at 22:33

























answered Jan 4 at 22:25









Kenny WongKenny Wong

19.1k21441




19.1k21441












  • $begingroup$
    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


















  • $begingroup$
    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
















$begingroup$
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




$begingroup$
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






$begingroup$
@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






$begingroup$
… 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






$begingroup$
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




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




$begingroup$
@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


















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