I'm trying to prove that $iint_U dxdy=iint_V J_u,v(x(u,v),y(u,v))mathrmdumathrmdv$
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First on $mathbb R^2$, the space of the 2-form has dimension 1 and thus $$mathrmdxmathrmdy=alpha mathrmdumathrmdv.$$
Let find $a$. I denote $partial _u$ for $fracpartial partial u$ and $x_u$ for $fracpartial xpartial u$.
We have that $mathrmdx=amathrmdu+bmathrmdy$ and thus $x_u=mathrmdx(partial u)=a$ and $x_v=mathrmdx(partial _v)=b$. Therefore $$mathrmdx=x_umathrmdu+x_vmathrmdv$$
and $$mathrmdy=y_umathrmdu+y_vmathrmdv.$$
Now $$mathrmdxwedge mathrmdy=(x_umathrmdu+x_vmathrmdv)wedge (y_vmathrmdu+y_vmathrmdv)=\=(x_uy_v-x_vy_y)mathrmduwedge mathrmdv=det J_u,v(x(u,v),y(u,v))mathrmduwedge mathrmdv.$$
Q1) Could someone explain way my proof is not rigorous ? Because I used exactly definition and theorem I have in my course of differentiable manifold, but my teacher said that what I did is not formal.
Q2) Supposed I proved that $$mathrmdxwedge mathrmdy=det J_u,v(x,y)mathrmduwedge mathrmdv$$
can I conclude directly that $$iint_U mathrmdxmathrmdy=iint_V |det J_u,v(x,y)|mathrmdumathrmdv ?$$
(if of course I have the right domain $U$ and $V$).
Q3) In fact, before I had to write $mathrmdxwedge mathrmdy$ and $mathrmduwedge mathrmdv$ that disappeared in the integral. I don't really why we don't write it in the integral. May be $mathrmdxmathrmdy=|det J_u,v(x,y)|mathrmdumathrmdv$ rather than $mathrmdxwedge mathrmdy=|det J_u,v(x,y)|mathrmduwedge mathrmdv$ ? But if so, I know that $$mathrmdxmathrmdy=-mathrmdymathrmdv$$
but $$iint mathrmdxmathrmdy=iint mathrmdumathrmdv,$$
so since in the integral it commute, it should have an other reason.
integration manifolds
edited Aug 10 at 17:45
Davide Morgante
2,245322
asked Aug 10 at 17:20
user380364
964214
add a comment |Â
up vote
2
down vote
favorite
First on $mathbb R^2$, the space of the 2-form has dimension 1 and thus $$mathrmdxmathrmdy=alpha mathrmdumathrmdv.$$
Let find $a$. I denote $partial _u$ for $fracpartial partial u$ and $x_u$ for $fracpartial xpartial u$.
We have that $mathrmdx=amathrmdu+bmathrmdy$ and thus $x_u=mathrmdx(partial u)=a$ and $x_v=mathrmdx(partial _v)=b$. Therefore $$mathrmdx=x_umathrmdu+x_vmathrmdv$$
and $$mathrmdy=y_umathrmdu+y_vmathrmdv.$$
Now $$mathrmdxwedge mathrmdy=(x_umathrmdu+x_vmathrmdv)wedge (y_vmathrmdu+y_vmathrmdv)=\=(x_uy_v-x_vy_y)mathrmduwedge mathrmdv=det J_u,v(x(u,v),y(u,v))mathrmduwedge mathrmdv.$$
Q1) Could someone explain way my proof is not rigorous ? Because I used exactly definition and theorem I have in my course of differentiable manifold, but my teacher said that what I did is not formal.
Q2) Supposed I proved that $$mathrmdxwedge mathrmdy=det J_u,v(x,y)mathrmduwedge mathrmdv$$
can I conclude directly that $$iint_U mathrmdxmathrmdy=iint_V |det J_u,v(x,y)|mathrmdumathrmdv ?$$
(if of course I have the right domain $U$ and $V$).
Q3) In fact, before I had to write $mathrmdxwedge mathrmdy$ and $mathrmduwedge mathrmdv$ that disappeared in the integral. I don't really why we don't write it in the integral. May be $mathrmdxmathrmdy=|det J_u,v(x,y)|mathrmdumathrmdv$ rather than $mathrmdxwedge mathrmdy=|det J_u,v(x,y)|mathrmduwedge mathrmdv$ ? But if so, I know that $$mathrmdxmathrmdy=-mathrmdymathrmdv$$
but $$iint mathrmdxmathrmdy=iint mathrmdumathrmdv,$$
so since in the integral it commute, it should have an other reason.
integration manifolds
edited Aug 10 at 17:45
Davide Morgante
2,245322
asked Aug 10 at 17:20
user380364
964214
In things like $f(z),dz$ it is standard to put a small space before $dz$ and in things like $a,dx,dy$ one has small spaces before both $dx$ and $dy.$ I edited the question accordingly.
– Michael Hardy
Aug 10 at 17:26
Also, in $displaystyle iint_V | det J_u,v(x,y)|,du,dv,$ notice the odd space before $det.$ I changed it to $displaystyle iint_V left| det J_u,v(x,y)right|,du,dv,$ coded as iint_V left| det J_u,v(x,y)right|,du,dv, using left and right. $qquad$
– Michael Hardy
Aug 10 at 17:28
@MichaelHardy: thank you.
– user380364
Aug 10 at 17:29
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
First on $mathbb R^2$, the space of the 2-form has dimension 1 and thus $$mathrmdxmathrmdy=alpha mathrmdumathrmdv.$$
Let find $a$. I denote $partial _u$ for $fracpartial partial u$ and $x_u$ for $fracpartial xpartial u$.
We have that $mathrmdx=amathrmdu+bmathrmdy$ and thus $x_u=mathrmdx(partial u)=a$ and $x_v=mathrmdx(partial _v)=b$. Therefore $$mathrmdx=x_umathrmdu+x_vmathrmdv$$
and $$mathrmdy=y_umathrmdu+y_vmathrmdv.$$
Now $$mathrmdxwedge mathrmdy=(x_umathrmdu+x_vmathrmdv)wedge (y_vmathrmdu+y_vmathrmdv)=\=(x_uy_v-x_vy_y)mathrmduwedge mathrmdv=det J_u,v(x(u,v),y(u,v))mathrmduwedge mathrmdv.$$
Q1) Could someone explain way my proof is not rigorous ? Because I used exactly definition and theorem I have in my course of differentiable manifold, but my teacher said that what I did is not formal.
Q2) Supposed I proved that $$mathrmdxwedge mathrmdy=det J_u,v(x,y)mathrmduwedge mathrmdv$$
can I conclude directly that $$iint_U mathrmdxmathrmdy=iint_V |det J_u,v(x,y)|mathrmdumathrmdv ?$$
(if of course I have the right domain $U$ and $V$).
Q3) In fact, before I had to write $mathrmdxwedge mathrmdy$ and $mathrmduwedge mathrmdv$ that disappeared in the integral. I don't really why we don't write it in the integral. May be $mathrmdxmathrmdy=|det J_u,v(x,y)|mathrmdumathrmdv$ rather than $mathrmdxwedge mathrmdy=|det J_u,v(x,y)|mathrmduwedge mathrmdv$ ? But if so, I know that $$mathrmdxmathrmdy=-mathrmdymathrmdv$$
but $$iint mathrmdxmathrmdy=iint mathrmdumathrmdv,$$
so since in the integral it commute, it should have an other reason.
integration manifolds
edited Aug 10 at 17:45
Davide Morgante
2,245322
asked Aug 10 at 17:20
user380364
964214
First on $mathbb R^2$, the space of the 2-form has dimension 1 and thus $$mathrmdxmathrmdy=alpha mathrmdumathrmdv.$$
Let find $a$. I denote $partial _u$ for $fracpartial partial u$ and $x_u$ for $fracpartial xpartial u$.
We have that $mathrmdx=amathrmdu+bmathrmdy$ and thus $x_u=mathrmdx(partial u)=a$ and $x_v=mathrmdx(partial _v)=b$. Therefore $$mathrmdx=x_umathrmdu+x_vmathrmdv$$
and $$mathrmdy=y_umathrmdu+y_vmathrmdv.$$
Now $$mathrmdxwedge mathrmdy=(x_umathrmdu+x_vmathrmdv)wedge (y_vmathrmdu+y_vmathrmdv)=\=(x_uy_v-x_vy_y)mathrmduwedge mathrmdv=det J_u,v(x(u,v),y(u,v))mathrmduwedge mathrmdv.$$
Q1) Could someone explain way my proof is not rigorous ? Because I used exactly definition and theorem I have in my course of differentiable manifold, but my teacher said that what I did is not formal.
Q2) Supposed I proved that $$mathrmdxwedge mathrmdy=det J_u,v(x,y)mathrmduwedge mathrmdv$$
can I conclude directly that $$iint_U mathrmdxmathrmdy=iint_V |det J_u,v(x,y)|mathrmdumathrmdv ?$$
(if of course I have the right domain $U$ and $V$).
Q3) In fact, before I had to write $mathrmdxwedge mathrmdy$ and $mathrmduwedge mathrmdv$ that disappeared in the integral. I don't really why we don't write it in the integral. May be $mathrmdxmathrmdy=|det J_u,v(x,y)|mathrmdumathrmdv$ rather than $mathrmdxwedge mathrmdy=|det J_u,v(x,y)|mathrmduwedge mathrmdv$ ? But if so, I know that $$mathrmdxmathrmdy=-mathrmdymathrmdv$$
but $$iint mathrmdxmathrmdy=iint mathrmdumathrmdv,$$
so since in the integral it commute, it should have an other reason.
integration manifolds
edited Aug 10 at 17:45
Davide Morgante
2,245322
asked Aug 10 at 17:20
user380364
964214
edited Aug 10 at 17:45
Davide Morgante
2,245322
edited Aug 10 at 17:45
Davide Morgante
2,245322
edited Aug 10 at 17:45
Davide Morgante
2,245322
2,245322
asked Aug 10 at 17:20
user380364
964214
asked Aug 10 at 17:20
user380364
964214
asked Aug 10 at 17:20
user380364
964214
964214
In things like $f(z),dz$ it is standard to put a small space before $dz$ and in things like $a,dx,dy$ one has small spaces before both $dx$ and $dy.$ I edited the question accordingly.
– Michael Hardy
Aug 10 at 17:26
Also, in $displaystyle iint_V | det J_u,v(x,y)|,du,dv,$ notice the odd space before $det.$ I changed it to $displaystyle iint_V left| det J_u,v(x,y)right|,du,dv,$ coded as iint_V left| det J_u,v(x,y)right|,du,dv, using left and right. $qquad$
– Michael Hardy
Aug 10 at 17:28
@MichaelHardy: thank you.
– user380364
Aug 10 at 17:29
add a comment |Â
In things like $f(z),dz$ it is standard to put a small space before $dz$ and in things like $a,dx,dy$ one has small spaces before both $dx$ and $dy.$ I edited the question accordingly.
– Michael Hardy
Aug 10 at 17:26
Also, in $displaystyle iint_V | det J_u,v(x,y)|,du,dv,$ notice the odd space before $det.$ I changed it to $displaystyle iint_V left| det J_u,v(x,y)right|,du,dv,$ coded as iint_V left| det J_u,v(x,y)right|,du,dv, using left and right. $qquad$
– Michael Hardy
Aug 10 at 17:28
@MichaelHardy: thank you.
– user380364
Aug 10 at 17:29
In things like $f(z),dz$ it is standard to put a small space before $dz$ and in things like $a,dx,dy$ one has small spaces before both $dx$ and $dy.$ I edited the question accordingly.
– Michael Hardy
Aug 10 at 17:26
In things like $f(z),dz$ it is standard to put a small space before $dz$ and in things like $a,dx,dy$ one has small spaces before both $dx$ and $dy.$ I edited the question accordingly.
– Michael Hardy
Aug 10 at 17:26
Also, in $displaystyle iint_V | det J_u,v(x,y)|,du,dv,$ notice the odd space before $det.$ I changed it to $displaystyle iint_V left| det J_u,v(x,y)right|,du,dv,$ coded as iint_V left| det J_u,v(x,y)right|,du,dv, using left and right. $qquad$
– Michael Hardy
Aug 10 at 17:28
Also, in $displaystyle iint_V | det J_u,v(x,y)|,du,dv,$ notice the odd space before $det.$ I changed it to $displaystyle iint_V left| det J_u,v(x,y)right|,du,dv,$ coded as iint_V left| det J_u,v(x,y)right|,du,dv, using left and right. $qquad$
– Michael Hardy
Aug 10 at 17:28
@MichaelHardy: thank you.
– user380364
Aug 10 at 17:29
@MichaelHardy: thank you.
– user380364
Aug 10 at 17:29
add a comment |Â
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In things like $f(z),dz$ it is standard to put a small space before $dz$ and in things like $a,dx,dy$ one has small spaces before both $dx$ and $dy.$ I edited the question accordingly.
– Michael Hardy
Aug 10 at 17:26
Also, in $displaystyle iint_V | det J_u,v(x,y)|,du,dv,$ notice the odd space before $det.$ I changed it to $displaystyle iint_V left| det J_u,v(x,y)right|,du,dv,$ coded as iint_V left| det J_u,v(x,y)right|,du,dv, using left and right. $qquad$
– Michael Hardy
Aug 10 at 17:28
@MichaelHardy: thank you.
– user380364
Aug 10 at 17:29