$int_D (x^2+xy),dxwedge dy$ with $D$ unit disk.

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Show that $w=(x^2+xy),dxwedge dy$ is closed.
Compute $int_Dw$ if $D=$unit disk in $mathbbR^2.$



I have this:



$dw=((2x+y),dx+x,dy)wedge dxwedge dy=0$ therefore is closed.



Now, $x=rcos(t)$, $y=rsin(t)$ then $dx=-rsin(t),dt+cos(t),dr$ and
$dy=rcos(t),dt+sin(t),dr$



then $dxwedge dy=-r,dtwedge dr$



therefore
beginalign
& int_D w=int_D(r^2cos^2(t)+r^2cos(t)sin(t))(-r),dtwedge dr \[10pt]
= &int_0^1int_0^2pi -r^3 (cos^2(t)+cos(t)sin(t)),dt,dr
=pi/4
endalign



It is correct?



pd: They told me that with Stokes the result is 0. Is it true?
I have already tested the result. :) is $pi/4$




differential-forms




share|cite|improve this question





















  • Did you try and apply Stokes' theorem?
    – Pedro Tamaroff♦
    Jul 31 at 5:04















up vote
0
down vote

favorite












Show that $w=(x^2+xy),dxwedge dy$ is closed.
Compute $int_Dw$ if $D=$unit disk in $mathbbR^2.$



I have this:



$dw=((2x+y),dx+x,dy)wedge dxwedge dy=0$ therefore is closed.



Now, $x=rcos(t)$, $y=rsin(t)$ then $dx=-rsin(t),dt+cos(t),dr$ and
$dy=rcos(t),dt+sin(t),dr$



then $dxwedge dy=-r,dtwedge dr$



therefore
beginalign
& int_D w=int_D(r^2cos^2(t)+r^2cos(t)sin(t))(-r),dtwedge dr \[10pt]
= &int_0^1int_0^2pi -r^3 (cos^2(t)+cos(t)sin(t)),dt,dr
=pi/4
endalign



It is correct?



pd: They told me that with Stokes the result is 0. Is it true?
I have already tested the result. :) is $pi/4$




differential-forms




share|cite|improve this question





















  • Did you try and apply Stokes' theorem?
    – Pedro Tamaroff♦
    Jul 31 at 5:04













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Show that $w=(x^2+xy),dxwedge dy$ is closed.
Compute $int_Dw$ if $D=$unit disk in $mathbbR^2.$



I have this:



$dw=((2x+y),dx+x,dy)wedge dxwedge dy=0$ therefore is closed.



Now, $x=rcos(t)$, $y=rsin(t)$ then $dx=-rsin(t),dt+cos(t),dr$ and
$dy=rcos(t),dt+sin(t),dr$



then $dxwedge dy=-r,dtwedge dr$



therefore
beginalign
& int_D w=int_D(r^2cos^2(t)+r^2cos(t)sin(t))(-r),dtwedge dr \[10pt]
= &int_0^1int_0^2pi -r^3 (cos^2(t)+cos(t)sin(t)),dt,dr
=pi/4
endalign



It is correct?



pd: They told me that with Stokes the result is 0. Is it true?
I have already tested the result. :) is $pi/4$




differential-forms




share|cite|improve this question













Show that $w=(x^2+xy),dxwedge dy$ is closed.
Compute $int_Dw$ if $D=$unit disk in $mathbbR^2.$



I have this:



$dw=((2x+y),dx+x,dy)wedge dxwedge dy=0$ therefore is closed.



Now, $x=rcos(t)$, $y=rsin(t)$ then $dx=-rsin(t),dt+cos(t),dr$ and
$dy=rcos(t),dt+sin(t),dr$



then $dxwedge dy=-r,dtwedge dr$



therefore
beginalign
& int_D w=int_D(r^2cos^2(t)+r^2cos(t)sin(t))(-r),dtwedge dr \[10pt]
= &int_0^1int_0^2pi -r^3 (cos^2(t)+cos(t)sin(t)),dt,dr
=pi/4
endalign



It is correct?



pd: They told me that with Stokes the result is 0. Is it true?
I have already tested the result. :) is $pi/4$





differential-forms





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 31 at 3:12
























asked Jul 31 at 2:23









eraldcoil

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  • Did you try and apply Stokes' theorem?
    – Pedro Tamaroff♦
    Jul 31 at 5:04

















  • Did you try and apply Stokes' theorem?
    – Pedro Tamaroff♦
    Jul 31 at 5:04
















Did you try and apply Stokes' theorem?
– Pedro Tamaroff♦
Jul 31 at 5:04





Did you try and apply Stokes' theorem?
– Pedro Tamaroff♦
Jul 31 at 5:04
















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