integrate a differential form over the indicated smooth cube
$begingroup$
I'm supposed to integrate the differential form over the smooth cube and I'm not sure if I'm on the right track.
So suppose,
$ int_c x dy+ y dz$ where $c: [-1,1] rightarrow mathbb{R^{3}}$ is $c(t) = (sin(t), cos(t), t) = (x,y,z)$
I took the derivative of $c(t)$ so I have
$c'(t) = (cos(t), -sin(t), 1) = (dx,dy,dz)$
and then I substituted back into the equation which is
$ int_{[-1,1]} sin(t) (-sin(t))+ cos(t)) (1)$
$ int_{[-1,1]} -(sin(t))^{2}+ cos(t) $
So do I now take the antiderivative of this and evaluate from $-1$ to $1$?
differential-geometry vector-analysis differential-forms
edited Dec 1 '18 at 0:30
amWhy
1
asked Dec 1 '18 at 0:25
usukidollusukidoll
1,1521033
$endgroup$
add a comment |
$begingroup$
I'm supposed to integrate the differential form over the smooth cube and I'm not sure if I'm on the right track.
So suppose,
$ int_c x dy+ y dz$ where $c: [-1,1] rightarrow mathbb{R^{3}}$ is $c(t) = (sin(t), cos(t), t) = (x,y,z)$
I took the derivative of $c(t)$ so I have
$c'(t) = (cos(t), -sin(t), 1) = (dx,dy,dz)$
and then I substituted back into the equation which is
$ int_{[-1,1]} sin(t) (-sin(t))+ cos(t)) (1)$
$ int_{[-1,1]} -(sin(t))^{2}+ cos(t) $
So do I now take the antiderivative of this and evaluate from $-1$ to $1$?
differential-geometry vector-analysis differential-forms
edited Dec 1 '18 at 0:30
amWhy
1
asked Dec 1 '18 at 0:25
usukidollusukidoll
1,1521033
$endgroup$
add a comment |
$begingroup$
I'm supposed to integrate the differential form over the smooth cube and I'm not sure if I'm on the right track.
So suppose,
$ int_c x dy+ y dz$ where $c: [-1,1] rightarrow mathbb{R^{3}}$ is $c(t) = (sin(t), cos(t), t) = (x,y,z)$
I took the derivative of $c(t)$ so I have
$c'(t) = (cos(t), -sin(t), 1) = (dx,dy,dz)$
and then I substituted back into the equation which is
$ int_{[-1,1]} sin(t) (-sin(t))+ cos(t)) (1)$
$ int_{[-1,1]} -(sin(t))^{2}+ cos(t) $
So do I now take the antiderivative of this and evaluate from $-1$ to $1$?
differential-geometry vector-analysis differential-forms
edited Dec 1 '18 at 0:30
amWhy
1
asked Dec 1 '18 at 0:25
usukidollusukidoll
1,1521033
$endgroup$
I'm supposed to integrate the differential form over the smooth cube and I'm not sure if I'm on the right track.
So suppose,
$ int_c x dy+ y dz$ where $c: [-1,1] rightarrow mathbb{R^{3}}$ is $c(t) = (sin(t), cos(t), t) = (x,y,z)$
I took the derivative of $c(t)$ so I have
$c'(t) = (cos(t), -sin(t), 1) = (dx,dy,dz)$
and then I substituted back into the equation which is
$ int_{[-1,1]} sin(t) (-sin(t))+ cos(t)) (1)$
$ int_{[-1,1]} -(sin(t))^{2}+ cos(t) $
So do I now take the antiderivative of this and evaluate from $-1$ to $1$?
differential-geometry vector-analysis differential-forms
differential-geometry vector-analysis differential-forms
edited Dec 1 '18 at 0:30
amWhy
1
asked Dec 1 '18 at 0:25
usukidollusukidoll
1,1521033
edited Dec 1 '18 at 0:30
amWhy
1
asked Dec 1 '18 at 0:25
usukidollusukidoll
1,1521033
edited Dec 1 '18 at 0:30
amWhy
1
edited Dec 1 '18 at 0:30
amWhy
1
edited Dec 1 '18 at 0:30
amWhy
1
1
asked Dec 1 '18 at 0:25
usukidollusukidoll
1,1521033
asked Dec 1 '18 at 0:25
usukidollusukidoll
1,1521033
asked Dec 1 '18 at 0:25
usukidollusukidoll
1,1521033
1,1521033
add a comment |
add a comment |
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