Mixing
Integration of a $2$-form
$begingroup$
What is
$$int_C{omega}$$
where $omega=frac{dx wedge dy}{x^2+y^2}$ and $C(t_1,t_2)=(t_1+1)(cos(2pi t_2),sin(2pi t_2)) : I_2 rightarrow mathbb{R}^2 - text{{(0,0)}}$?
The integrals of $1$-forms I understood as we take $dx$ and $dy$ as the derivatives of the parametrisation $C(t)$ w.r.t $t$. However for $2$-forms we now have the wedge product $dx wedge dy$ which is where my problem lies in this question. Is this a matter of differentiation in two variables when computing this wedge? For example, before even calculating the wedge product, what is $dx(t_1,t_2)$?
The denominator isn't a problem: $x^2+y^2 = (t_1+1)^2$
differential-geometry manifolds differential-forms
edited Jan 10 at 21:56
mechanodroid
27.3k62446
asked Jan 10 '14 at 13:41
PhibertPhibert
348212
$endgroup$
add a comment |
$begingroup$
What is
$$int_C{omega}$$
where $omega=frac{dx wedge dy}{x^2+y^2}$ and $C(t_1,t_2)=(t_1+1)(cos(2pi t_2),sin(2pi t_2)) : I_2 rightarrow mathbb{R}^2 - text{{(0,0)}}$?
The integrals of $1$-forms I understood as we take $dx$ and $dy$ as the derivatives of the parametrisation $C(t)$ w.r.t $t$. However for $2$-forms we now have the wedge product $dx wedge dy$ which is where my problem lies in this question. Is this a matter of differentiation in two variables when computing this wedge? For example, before even calculating the wedge product, what is $dx(t_1,t_2)$?
The denominator isn't a problem: $x^2+y^2 = (t_1+1)^2$
differential-geometry manifolds differential-forms
edited Jan 10 at 21:56
mechanodroid
27.3k62446
asked Jan 10 '14 at 13:41
PhibertPhibert
348212
$endgroup$
$begingroup$
Have you tried something? If so include it in your post.
$endgroup$
– Test123
Jan 10 '14 at 13:44
$begingroup$
I have updated to show where I'm up to; not far.
$endgroup$
– Phibert
Jan 10 '14 at 16:25
add a comment |
$begingroup$
What is
$$int_C{omega}$$
where $omega=frac{dx wedge dy}{x^2+y^2}$ and $C(t_1,t_2)=(t_1+1)(cos(2pi t_2),sin(2pi t_2)) : I_2 rightarrow mathbb{R}^2 - text{{(0,0)}}$?
The integrals of $1$-forms I understood as we take $dx$ and $dy$ as the derivatives of the parametrisation $C(t)$ w.r.t $t$. However for $2$-forms we now have the wedge product $dx wedge dy$ which is where my problem lies in this question. Is this a matter of differentiation in two variables when computing this wedge? For example, before even calculating the wedge product, what is $dx(t_1,t_2)$?
The denominator isn't a problem: $x^2+y^2 = (t_1+1)^2$
differential-geometry manifolds differential-forms
edited Jan 10 at 21:56
mechanodroid
27.3k62446
asked Jan 10 '14 at 13:41
PhibertPhibert
348212
$endgroup$
What is
$$int_C{omega}$$
where $omega=frac{dx wedge dy}{x^2+y^2}$ and $C(t_1,t_2)=(t_1+1)(cos(2pi t_2),sin(2pi t_2)) : I_2 rightarrow mathbb{R}^2 - text{{(0,0)}}$?
The integrals of $1$-forms I understood as we take $dx$ and $dy$ as the derivatives of the parametrisation $C(t)$ w.r.t $t$. However for $2$-forms we now have the wedge product $dx wedge dy$ which is where my problem lies in this question. Is this a matter of differentiation in two variables when computing this wedge? For example, before even calculating the wedge product, what is $dx(t_1,t_2)$?
The denominator isn't a problem: $x^2+y^2 = (t_1+1)^2$
differential-geometry manifolds differential-forms
differential-geometry manifolds differential-forms
edited Jan 10 at 21:56
mechanodroid
27.3k62446
asked Jan 10 '14 at 13:41
PhibertPhibert
348212
edited Jan 10 at 21:56
mechanodroid
27.3k62446
asked Jan 10 '14 at 13:41
PhibertPhibert
348212
edited Jan 10 at 21:56
mechanodroid
27.3k62446
edited Jan 10 at 21:56
mechanodroid
27.3k62446
edited Jan 10 at 21:56
mechanodroid
27.3k62446
27.3k62446
asked Jan 10 '14 at 13:41
PhibertPhibert
348212
asked Jan 10 '14 at 13:41
PhibertPhibert
348212
asked Jan 10 '14 at 13:41
PhibertPhibert
348212
348212
$begingroup$
Have you tried something? If so include it in your post.
$endgroup$
– Test123
Jan 10 '14 at 13:44
$begingroup$
I have updated to show where I'm up to; not far.
$endgroup$
– Phibert
Jan 10 '14 at 16:25
add a comment |
$begingroup$
Have you tried something? If so include it in your post.
$endgroup$
– Test123
Jan 10 '14 at 13:44
$begingroup$
I have updated to show where I'm up to; not far.
$endgroup$
– Phibert
Jan 10 '14 at 16:25
$begingroup$
Have you tried something? If so include it in your post.
$endgroup$
– Test123
Jan 10 '14 at 13:44
$begingroup$
Have you tried something? If so include it in your post.
$endgroup$
– Test123
Jan 10 '14 at 13:44
$begingroup$
I have updated to show where I'm up to; not far.
$endgroup$
– Phibert
Jan 10 '14 at 16:25
$begingroup$
I have updated to show where I'm up to; not far.
$endgroup$
– Phibert
Jan 10 '14 at 16:25
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I've never seen a definition of the integral of a $2$-form along a "2D path" $C: I^2 to mathbb{R}^2$ (has anyone?), but it seems clear to me that the sensible definition should be
$$ int_C omega := int_{I^2} C^*omega$$
where $C^*omega$ is the pull-back of $omega$ by $C$. Recall that it is defined by
$$ (C^*omega)_{|p} (u, v) := omega_{|C(p)}(dC_{|p} (u), dC_{|p} (v))~.$$
NB: More generally, we could define the same way integrals $int_C omega$ where $omega$ is a $k$-form on a $k$-dimensional manifold $M$ and $c$ is a smooth map $U subset mathbb{R}^k to M$.
Let's come back to your problem. Practically, you can compute $C^* omega$ by letting $(x,y) = C(t_1,t_2)$ in the expression of $omega$, you get:
$$
begin{align*}
x &= (t_1 + 1) cos (2pi t_2)\
y &= (t_1 + 1) sin (2pi t_2) \
end{align*}
$$
hence
$$
begin{align*}
x^2 + y^2 &= (t_1+1)^2\
dx &= cos (2pi t_2),dt_1 - 2pi (t_1 + 1)sin(2pi t_2) dt_2\
dy &= sin (2pi t_2),dt_1 + 2pi (t_1 + 1)cos(2pi t_2) dt_2\
dx wedge dy &= 2pi (t_1+1), dt_1wedge dt_2\
end{align*}
$$
thus
$$
C^* omega = {2pi, dt_1wedge dt_2over t_1 +1}
$$
You can now easily compute your integral: assuming $I = [0,1]$ (you don't say what I is)
$$
int_C omega = int_{I^2}{2pi, dt_1wedge dt_2over t_1 +1} = int_{I^2}{2pi, dt_1 dt_2over t_1 +1}
$$
which gives us by Tonelli's theorem
$$
int_C omega = 2pileft(int_{0}^1{dt_1over t_1 +1}right)left(int_{0}^1 dt_2right) = 2pi log (2)
$$
NB: Another approach ("just for fun") would be to work in polar coordinates, I'll let you try to figure out how do that.
answered Jan 10 '14 at 20:11
SeubSeub
3,8671128
$endgroup$
$begingroup$
Having thought about your last point about polar coordinates, I wouldn't even know where to start. Any hint?
$endgroup$
– Phibert
Jan 12 '14 at 21:55
$begingroup$
Well, given that $x = rcos theta$ and $y = rintheta$, can you express $C$, $dx$, $dy$ and $omega$ in terms of $r$ and $theta$ ?
$endgroup$
– Seub
Jan 13 '14 at 1:47
$begingroup$
The only way I can interpret what you mean is $r=(t_1+1)$ and $theta = 2pi t_2$. Am I being naive here or are you just suggesting substitutions? From what I can see, we're already working in polar coordinates looking at our $C$.
$endgroup$
– Phibert
Jan 13 '14 at 22:54
$begingroup$
That's right, now you should try to express everything in those coordinates as I suggested above
$endgroup$
– Seub
Jan 13 '14 at 23:18
add a comment |
$begingroup$
Changing to polar coordinates, let $t_1+1=rin[1,2]$, and $2pi t_2=phiin[0,2pi]$ (full circle). The set $C$ is an annulus in the $x,y$ plane bounded by the two concentric circles $r=1,2$, with $x^2+y^2=(rcosphi)^2+(rsinphi)^2=r^2$ The surface element $dxwedge dy = r dr d phi$. The function depends only on $r$, so the integration over the angle gives us simply $2pi$, and finally the result is
$$2piint_1^2 frac{rdr}{r^2}=2pilog2$$
answered Jan 21 '14 at 4:55
rychrych
2,4611717
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
oldest
votes
$begingroup$
I've never seen a definition of the integral of a $2$-form along a "2D path" $C: I^2 to mathbb{R}^2$ (has anyone?), but it seems clear to me that the sensible definition should be
$$ int_C omega := int_{I^2} C^*omega$$
where $C^*omega$ is the pull-back of $omega$ by $C$. Recall that it is defined by
$$ (C^*omega)_{|p} (u, v) := omega_{|C(p)}(dC_{|p} (u), dC_{|p} (v))~.$$
NB: More generally, we could define the same way integrals $int_C omega$ where $omega$ is a $k$-form on a $k$-dimensional manifold $M$ and $c$ is a smooth map $U subset mathbb{R}^k to M$.
Let's come back to your problem. Practically, you can compute $C^* omega$ by letting $(x,y) = C(t_1,t_2)$ in the expression of $omega$, you get:
$$
begin{align*}
x &= (t_1 + 1) cos (2pi t_2)\
y &= (t_1 + 1) sin (2pi t_2) \
end{align*}
$$
hence
$$
begin{align*}
x^2 + y^2 &= (t_1+1)^2\
dx &= cos (2pi t_2),dt_1 - 2pi (t_1 + 1)sin(2pi t_2) dt_2\
dy &= sin (2pi t_2),dt_1 + 2pi (t_1 + 1)cos(2pi t_2) dt_2\
dx wedge dy &= 2pi (t_1+1), dt_1wedge dt_2\
end{align*}
$$
thus
$$
C^* omega = {2pi, dt_1wedge dt_2over t_1 +1}
$$
You can now easily compute your integral: assuming $I = [0,1]$ (you don't say what I is)
$$
int_C omega = int_{I^2}{2pi, dt_1wedge dt_2over t_1 +1} = int_{I^2}{2pi, dt_1 dt_2over t_1 +1}
$$
which gives us by Tonelli's theorem
$$
int_C omega = 2pileft(int_{0}^1{dt_1over t_1 +1}right)left(int_{0}^1 dt_2right) = 2pi log (2)
$$
NB: Another approach ("just for fun") would be to work in polar coordinates, I'll let you try to figure out how do that.
answered Jan 10 '14 at 20:11
SeubSeub
3,8671128
$endgroup$
$begingroup$
Having thought about your last point about polar coordinates, I wouldn't even know where to start. Any hint?
$endgroup$
– Phibert
Jan 12 '14 at 21:55
$begingroup$
Well, given that $x = rcos theta$ and $y = rintheta$, can you express $C$, $dx$, $dy$ and $omega$ in terms of $r$ and $theta$ ?
$endgroup$
– Seub
Jan 13 '14 at 1:47
$begingroup$
The only way I can interpret what you mean is $r=(t_1+1)$ and $theta = 2pi t_2$. Am I being naive here or are you just suggesting substitutions? From what I can see, we're already working in polar coordinates looking at our $C$.
$endgroup$
– Phibert
Jan 13 '14 at 22:54
$begingroup$
That's right, now you should try to express everything in those coordinates as I suggested above
$endgroup$
– Seub
Jan 13 '14 at 23:18
add a comment |
$begingroup$
I've never seen a definition of the integral of a $2$-form along a "2D path" $C: I^2 to mathbb{R}^2$ (has anyone?), but it seems clear to me that the sensible definition should be
$$ int_C omega := int_{I^2} C^*omega$$
where $C^*omega$ is the pull-back of $omega$ by $C$. Recall that it is defined by
$$ (C^*omega)_{|p} (u, v) := omega_{|C(p)}(dC_{|p} (u), dC_{|p} (v))~.$$
NB: More generally, we could define the same way integrals $int_C omega$ where $omega$ is a $k$-form on a $k$-dimensional manifold $M$ and $c$ is a smooth map $U subset mathbb{R}^k to M$.
Let's come back to your problem. Practically, you can compute $C^* omega$ by letting $(x,y) = C(t_1,t_2)$ in the expression of $omega$, you get:
$$
begin{align*}
x &= (t_1 + 1) cos (2pi t_2)\
y &= (t_1 + 1) sin (2pi t_2) \
end{align*}
$$
hence
$$
begin{align*}
x^2 + y^2 &= (t_1+1)^2\
dx &= cos (2pi t_2),dt_1 - 2pi (t_1 + 1)sin(2pi t_2) dt_2\
dy &= sin (2pi t_2),dt_1 + 2pi (t_1 + 1)cos(2pi t_2) dt_2\
dx wedge dy &= 2pi (t_1+1), dt_1wedge dt_2\
end{align*}
$$
thus
$$
C^* omega = {2pi, dt_1wedge dt_2over t_1 +1}
$$
You can now easily compute your integral: assuming $I = [0,1]$ (you don't say what I is)
$$
int_C omega = int_{I^2}{2pi, dt_1wedge dt_2over t_1 +1} = int_{I^2}{2pi, dt_1 dt_2over t_1 +1}
$$
which gives us by Tonelli's theorem
$$
int_C omega = 2pileft(int_{0}^1{dt_1over t_1 +1}right)left(int_{0}^1 dt_2right) = 2pi log (2)
$$
NB: Another approach ("just for fun") would be to work in polar coordinates, I'll let you try to figure out how do that.
answered Jan 10 '14 at 20:11
SeubSeub
3,8671128
$endgroup$
$begingroup$
Having thought about your last point about polar coordinates, I wouldn't even know where to start. Any hint?
$endgroup$
– Phibert
Jan 12 '14 at 21:55
$begingroup$
Well, given that $x = rcos theta$ and $y = rintheta$, can you express $C$, $dx$, $dy$ and $omega$ in terms of $r$ and $theta$ ?
$endgroup$
– Seub
Jan 13 '14 at 1:47
$begingroup$
The only way I can interpret what you mean is $r=(t_1+1)$ and $theta = 2pi t_2$. Am I being naive here or are you just suggesting substitutions? From what I can see, we're already working in polar coordinates looking at our $C$.
$endgroup$
– Phibert
Jan 13 '14 at 22:54
$begingroup$
That's right, now you should try to express everything in those coordinates as I suggested above
$endgroup$
– Seub
Jan 13 '14 at 23:18
add a comment |
$begingroup$
I've never seen a definition of the integral of a $2$-form along a "2D path" $C: I^2 to mathbb{R}^2$ (has anyone?), but it seems clear to me that the sensible definition should be
$$ int_C omega := int_{I^2} C^*omega$$
where $C^*omega$ is the pull-back of $omega$ by $C$. Recall that it is defined by
$$ (C^*omega)_{|p} (u, v) := omega_{|C(p)}(dC_{|p} (u), dC_{|p} (v))~.$$
NB: More generally, we could define the same way integrals $int_C omega$ where $omega$ is a $k$-form on a $k$-dimensional manifold $M$ and $c$ is a smooth map $U subset mathbb{R}^k to M$.
Let's come back to your problem. Practically, you can compute $C^* omega$ by letting $(x,y) = C(t_1,t_2)$ in the expression of $omega$, you get:
$$
begin{align*}
x &= (t_1 + 1) cos (2pi t_2)\
y &= (t_1 + 1) sin (2pi t_2) \
end{align*}
$$
hence
$$
begin{align*}
x^2 + y^2 &= (t_1+1)^2\
dx &= cos (2pi t_2),dt_1 - 2pi (t_1 + 1)sin(2pi t_2) dt_2\
dy &= sin (2pi t_2),dt_1 + 2pi (t_1 + 1)cos(2pi t_2) dt_2\
dx wedge dy &= 2pi (t_1+1), dt_1wedge dt_2\
end{align*}
$$
thus
$$
C^* omega = {2pi, dt_1wedge dt_2over t_1 +1}
$$
You can now easily compute your integral: assuming $I = [0,1]$ (you don't say what I is)
$$
int_C omega = int_{I^2}{2pi, dt_1wedge dt_2over t_1 +1} = int_{I^2}{2pi, dt_1 dt_2over t_1 +1}
$$
which gives us by Tonelli's theorem
$$
int_C omega = 2pileft(int_{0}^1{dt_1over t_1 +1}right)left(int_{0}^1 dt_2right) = 2pi log (2)
$$
NB: Another approach ("just for fun") would be to work in polar coordinates, I'll let you try to figure out how do that.
answered Jan 10 '14 at 20:11
SeubSeub
3,8671128
$endgroup$
I've never seen a definition of the integral of a $2$-form along a "2D path" $C: I^2 to mathbb{R}^2$ (has anyone?), but it seems clear to me that the sensible definition should be
$$ int_C omega := int_{I^2} C^*omega$$
where $C^*omega$ is the pull-back of $omega$ by $C$. Recall that it is defined by
$$ (C^*omega)_{|p} (u, v) := omega_{|C(p)}(dC_{|p} (u), dC_{|p} (v))~.$$
NB: More generally, we could define the same way integrals $int_C omega$ where $omega$ is a $k$-form on a $k$-dimensional manifold $M$ and $c$ is a smooth map $U subset mathbb{R}^k to M$.
Let's come back to your problem. Practically, you can compute $C^* omega$ by letting $(x,y) = C(t_1,t_2)$ in the expression of $omega$, you get:
$$
begin{align*}
x &= (t_1 + 1) cos (2pi t_2)\
y &= (t_1 + 1) sin (2pi t_2) \
end{align*}
$$
hence
$$
begin{align*}
x^2 + y^2 &= (t_1+1)^2\
dx &= cos (2pi t_2),dt_1 - 2pi (t_1 + 1)sin(2pi t_2) dt_2\
dy &= sin (2pi t_2),dt_1 + 2pi (t_1 + 1)cos(2pi t_2) dt_2\
dx wedge dy &= 2pi (t_1+1), dt_1wedge dt_2\
end{align*}
$$
thus
$$
C^* omega = {2pi, dt_1wedge dt_2over t_1 +1}
$$
You can now easily compute your integral: assuming $I = [0,1]$ (you don't say what I is)
$$
int_C omega = int_{I^2}{2pi, dt_1wedge dt_2over t_1 +1} = int_{I^2}{2pi, dt_1 dt_2over t_1 +1}
$$
which gives us by Tonelli's theorem
$$
int_C omega = 2pileft(int_{0}^1{dt_1over t_1 +1}right)left(int_{0}^1 dt_2right) = 2pi log (2)
$$
NB: Another approach ("just for fun") would be to work in polar coordinates, I'll let you try to figure out how do that.
answered Jan 10 '14 at 20:11
SeubSeub
3,8671128
answered Jan 10 '14 at 20:11
SeubSeub
3,8671128
answered Jan 10 '14 at 20:11
SeubSeub
3,8671128
answered Jan 10 '14 at 20:11
SeubSeub
3,8671128
3,8671128
$begingroup$
Having thought about your last point about polar coordinates, I wouldn't even know where to start. Any hint?
$endgroup$
– Phibert
Jan 12 '14 at 21:55
$begingroup$
Well, given that $x = rcos theta$ and $y = rintheta$, can you express $C$, $dx$, $dy$ and $omega$ in terms of $r$ and $theta$ ?
$endgroup$
– Seub
Jan 13 '14 at 1:47
$begingroup$
The only way I can interpret what you mean is $r=(t_1+1)$ and $theta = 2pi t_2$. Am I being naive here or are you just suggesting substitutions? From what I can see, we're already working in polar coordinates looking at our $C$.
$endgroup$
– Phibert
Jan 13 '14 at 22:54
$begingroup$
That's right, now you should try to express everything in those coordinates as I suggested above
$endgroup$
– Seub
Jan 13 '14 at 23:18
add a comment |
$begingroup$
Having thought about your last point about polar coordinates, I wouldn't even know where to start. Any hint?
$endgroup$
– Phibert
Jan 12 '14 at 21:55
$begingroup$
Well, given that $x = rcos theta$ and $y = rintheta$, can you express $C$, $dx$, $dy$ and $omega$ in terms of $r$ and $theta$ ?
$endgroup$
– Seub
Jan 13 '14 at 1:47
$begingroup$
The only way I can interpret what you mean is $r=(t_1+1)$ and $theta = 2pi t_2$. Am I being naive here or are you just suggesting substitutions? From what I can see, we're already working in polar coordinates looking at our $C$.
$endgroup$
– Phibert
Jan 13 '14 at 22:54
$begingroup$
That's right, now you should try to express everything in those coordinates as I suggested above
$endgroup$
– Seub
Jan 13 '14 at 23:18
$begingroup$
Having thought about your last point about polar coordinates, I wouldn't even know where to start. Any hint?
$endgroup$
– Phibert
Jan 12 '14 at 21:55
$begingroup$
Having thought about your last point about polar coordinates, I wouldn't even know where to start. Any hint?
$endgroup$
– Phibert
Jan 12 '14 at 21:55
$begingroup$
Well, given that $x = rcos theta$ and $y = rintheta$, can you express $C$, $dx$, $dy$ and $omega$ in terms of $r$ and $theta$ ?
$endgroup$
– Seub
Jan 13 '14 at 1:47
$begingroup$
Well, given that $x = rcos theta$ and $y = rintheta$, can you express $C$, $dx$, $dy$ and $omega$ in terms of $r$ and $theta$ ?
$endgroup$
– Seub
Jan 13 '14 at 1:47
$begingroup$
The only way I can interpret what you mean is $r=(t_1+1)$ and $theta = 2pi t_2$. Am I being naive here or are you just suggesting substitutions? From what I can see, we're already working in polar coordinates looking at our $C$.
$endgroup$
– Phibert
Jan 13 '14 at 22:54
$begingroup$
The only way I can interpret what you mean is $r=(t_1+1)$ and $theta = 2pi t_2$. Am I being naive here or are you just suggesting substitutions? From what I can see, we're already working in polar coordinates looking at our $C$.
$endgroup$
– Phibert
Jan 13 '14 at 22:54
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That's right, now you should try to express everything in those coordinates as I suggested above
$endgroup$
– Seub
Jan 13 '14 at 23:18
$begingroup$
That's right, now you should try to express everything in those coordinates as I suggested above
$endgroup$
– Seub
Jan 13 '14 at 23:18
add a comment |
$begingroup$
Changing to polar coordinates, let $t_1+1=rin[1,2]$, and $2pi t_2=phiin[0,2pi]$ (full circle). The set $C$ is an annulus in the $x,y$ plane bounded by the two concentric circles $r=1,2$, with $x^2+y^2=(rcosphi)^2+(rsinphi)^2=r^2$ The surface element $dxwedge dy = r dr d phi$. The function depends only on $r$, so the integration over the angle gives us simply $2pi$, and finally the result is
$$2piint_1^2 frac{rdr}{r^2}=2pilog2$$
answered Jan 21 '14 at 4:55
rychrych
2,4611717
$endgroup$
add a comment |
$begingroup$
Changing to polar coordinates, let $t_1+1=rin[1,2]$, and $2pi t_2=phiin[0,2pi]$ (full circle). The set $C$ is an annulus in the $x,y$ plane bounded by the two concentric circles $r=1,2$, with $x^2+y^2=(rcosphi)^2+(rsinphi)^2=r^2$ The surface element $dxwedge dy = r dr d phi$. The function depends only on $r$, so the integration over the angle gives us simply $2pi$, and finally the result is
$$2piint_1^2 frac{rdr}{r^2}=2pilog2$$
answered Jan 21 '14 at 4:55
rychrych
2,4611717
$endgroup$
add a comment |
$begingroup$
Changing to polar coordinates, let $t_1+1=rin[1,2]$, and $2pi t_2=phiin[0,2pi]$ (full circle). The set $C$ is an annulus in the $x,y$ plane bounded by the two concentric circles $r=1,2$, with $x^2+y^2=(rcosphi)^2+(rsinphi)^2=r^2$ The surface element $dxwedge dy = r dr d phi$. The function depends only on $r$, so the integration over the angle gives us simply $2pi$, and finally the result is
$$2piint_1^2 frac{rdr}{r^2}=2pilog2$$
answered Jan 21 '14 at 4:55
rychrych
2,4611717
$endgroup$
Changing to polar coordinates, let $t_1+1=rin[1,2]$, and $2pi t_2=phiin[0,2pi]$ (full circle). The set $C$ is an annulus in the $x,y$ plane bounded by the two concentric circles $r=1,2$, with $x^2+y^2=(rcosphi)^2+(rsinphi)^2=r^2$ The surface element $dxwedge dy = r dr d phi$. The function depends only on $r$, so the integration over the angle gives us simply $2pi$, and finally the result is
$$2piint_1^2 frac{rdr}{r^2}=2pilog2$$
answered Jan 21 '14 at 4:55
rychrych
2,4611717
answered Jan 21 '14 at 4:55
rychrych
2,4611717
answered Jan 21 '14 at 4:55
rychrych
2,4611717
answered Jan 21 '14 at 4:55
rychrych
2,4611717
2,4611717
add a comment |
add a comment |
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Have you tried something? If so include it in your post.
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– Test123
Jan 10 '14 at 13:44
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I have updated to show where I'm up to; not far.
$endgroup$
– Phibert
Jan 10 '14 at 16:25