Mixing
Stokes theorem, intersection between cylinder and plane
Clash Royale CLAN TAG#URR8PPP
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$mathcal C$ is the intersection curve between the cylinder $x^2 + y^2 = 2y$ and the plane $y = z$. I tried parameterizing the curve by expanding the cylinder equation $x^2 + (y-1)^2 = 1$. I think I can write the parameterization as follows: $(r cos(theta), r sin(theta) + 1, 0)$? How do I proceed from here, I want to use Stokes theorem. $mathcal C$ is oriented counterclockwise.
vector-analysis stokes-theorem
edited Aug 13 at 17:07
Maxim
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asked Aug 6 at 11:26
StudentMaths
246
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$mathcal C$ is the intersection curve between the cylinder $x^2 + y^2 = 2y$ and the plane $y = z$. I tried parameterizing the curve by expanding the cylinder equation $x^2 + (y-1)^2 = 1$. I think I can write the parameterization as follows: $(r cos(theta), r sin(theta) + 1, 0)$? How do I proceed from here, I want to use Stokes theorem. $mathcal C$ is oriented counterclockwise.
vector-analysis stokes-theorem
edited Aug 13 at 17:07
Maxim
2,150113
asked Aug 6 at 11:26
StudentMaths
246
You’ve parameterized the projection of the cylinder onto the $x$-$y$ plane. Now use $y=z$ to get the correct curve.
– amd
Aug 6 at 22:53
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
$mathcal C$ is the intersection curve between the cylinder $x^2 + y^2 = 2y$ and the plane $y = z$. I tried parameterizing the curve by expanding the cylinder equation $x^2 + (y-1)^2 = 1$. I think I can write the parameterization as follows: $(r cos(theta), r sin(theta) + 1, 0)$? How do I proceed from here, I want to use Stokes theorem. $mathcal C$ is oriented counterclockwise.
vector-analysis stokes-theorem
edited Aug 13 at 17:07
Maxim
2,150113
asked Aug 6 at 11:26
StudentMaths
246
$mathcal C$ is the intersection curve between the cylinder $x^2 + y^2 = 2y$ and the plane $y = z$. I tried parameterizing the curve by expanding the cylinder equation $x^2 + (y-1)^2 = 1$. I think I can write the parameterization as follows: $(r cos(theta), r sin(theta) + 1, 0)$? How do I proceed from here, I want to use Stokes theorem. $mathcal C$ is oriented counterclockwise.
vector-analysis stokes-theorem
edited Aug 13 at 17:07
Maxim
2,150113
asked Aug 6 at 11:26
StudentMaths
246
edited Aug 13 at 17:07
Maxim
2,150113
edited Aug 13 at 17:07
Maxim
2,150113
edited Aug 13 at 17:07
Maxim
2,150113
2,150113
asked Aug 6 at 11:26
StudentMaths
246
asked Aug 6 at 11:26
StudentMaths
246
asked Aug 6 at 11:26
StudentMaths
246
246
You’ve parameterized the projection of the cylinder onto the $x$-$y$ plane. Now use $y=z$ to get the correct curve.
– amd
Aug 6 at 22:53
add a comment |Â
You’ve parameterized the projection of the cylinder onto the $x$-$y$ plane. Now use $y=z$ to get the correct curve.
– amd
Aug 6 at 22:53
You’ve parameterized the projection of the cylinder onto the $x$-$y$ plane. Now use $y=z$ to get the correct curve.
– amd
Aug 6 at 22:53
You’ve parameterized the projection of the cylinder onto the $x$-$y$ plane. Now use $y=z$ to get the correct curve.
– amd
Aug 6 at 22:53
add a comment |Â
2 Answers
2
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0
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HINT
We have
- $F=(y^2,xy,xz)implies operatornamerot(F)=(0,-z,-y)$
then
$$int_C y^2,dx+xy,dy+xz,dz =int_S operatornamerot(F)cdot n ,dS$$
and
$$operatornamerot(F)cdot n=(0,z,-y)cdot (0,-1/sqrt 2,1/sqrt 2)=frac z sqrt 2-frac y sqrt 2$$
answered Aug 6 at 12:39
gimusi
65.4k73684
How do you find the normal vector? And should'nt it be rot(F) = (0, -z, -y) ?
– StudentMaths
Aug 6 at 12:46
@StudentMaths Ops...you are right it is rot(F) = (0, -z, -y) of course. For the normal recall that for a plane ax+by+cz+d=0 the normal n=(a,b,c).
– gimusi
Aug 6 at 12:49
Thank you! I get this 1/sqrt(2) * integral 4y dxdy but what are the boundaries? I feel this would be easier to solve with polar coordinates.
– StudentMaths
Aug 6 at 12:54
Since y=z I suppose rot(F)=0.
– gimusi
Aug 6 at 13:00
add a comment |Â
up vote
0
down vote
Take $S$ to be the region in the plane $y = z$ enclosed by $mathcal C$. Since $nabla times mathbf F = (0, -z, -y)$ is parallel to the plane $y = z$, $(nabla times mathbf F) cdot dmathbf S = 0$.
answered Aug 13 at 17:03
Maxim
2,150113
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
HINT
We have
- $F=(y^2,xy,xz)implies operatornamerot(F)=(0,-z,-y)$
then
$$int_C y^2,dx+xy,dy+xz,dz =int_S operatornamerot(F)cdot n ,dS$$
and
$$operatornamerot(F)cdot n=(0,z,-y)cdot (0,-1/sqrt 2,1/sqrt 2)=frac z sqrt 2-frac y sqrt 2$$
answered Aug 6 at 12:39
gimusi
65.4k73684
How do you find the normal vector? And should'nt it be rot(F) = (0, -z, -y) ?
– StudentMaths
Aug 6 at 12:46
@StudentMaths Ops...you are right it is rot(F) = (0, -z, -y) of course. For the normal recall that for a plane ax+by+cz+d=0 the normal n=(a,b,c).
– gimusi
Aug 6 at 12:49
Thank you! I get this 1/sqrt(2) * integral 4y dxdy but what are the boundaries? I feel this would be easier to solve with polar coordinates.
– StudentMaths
Aug 6 at 12:54
Since y=z I suppose rot(F)=0.
– gimusi
Aug 6 at 13:00
add a comment |Â
up vote
0
down vote
HINT
We have
- $F=(y^2,xy,xz)implies operatornamerot(F)=(0,-z,-y)$
then
$$int_C y^2,dx+xy,dy+xz,dz =int_S operatornamerot(F)cdot n ,dS$$
and
$$operatornamerot(F)cdot n=(0,z,-y)cdot (0,-1/sqrt 2,1/sqrt 2)=frac z sqrt 2-frac y sqrt 2$$
answered Aug 6 at 12:39
gimusi
65.4k73684
How do you find the normal vector? And should'nt it be rot(F) = (0, -z, -y) ?
– StudentMaths
Aug 6 at 12:46
@StudentMaths Ops...you are right it is rot(F) = (0, -z, -y) of course. For the normal recall that for a plane ax+by+cz+d=0 the normal n=(a,b,c).
– gimusi
Aug 6 at 12:49
Thank you! I get this 1/sqrt(2) * integral 4y dxdy but what are the boundaries? I feel this would be easier to solve with polar coordinates.
– StudentMaths
Aug 6 at 12:54
Since y=z I suppose rot(F)=0.
– gimusi
Aug 6 at 13:00
add a comment |Â
up vote
0
down vote
up vote
0
down vote
HINT
We have
- $F=(y^2,xy,xz)implies operatornamerot(F)=(0,-z,-y)$
then
$$int_C y^2,dx+xy,dy+xz,dz =int_S operatornamerot(F)cdot n ,dS$$
and
$$operatornamerot(F)cdot n=(0,z,-y)cdot (0,-1/sqrt 2,1/sqrt 2)=frac z sqrt 2-frac y sqrt 2$$
answered Aug 6 at 12:39
gimusi
65.4k73684
HINT
We have
- $F=(y^2,xy,xz)implies operatornamerot(F)=(0,-z,-y)$
then
$$int_C y^2,dx+xy,dy+xz,dz =int_S operatornamerot(F)cdot n ,dS$$
and
$$operatornamerot(F)cdot n=(0,z,-y)cdot (0,-1/sqrt 2,1/sqrt 2)=frac z sqrt 2-frac y sqrt 2$$
answered Aug 6 at 12:39
gimusi
65.4k73684
edited Aug 6 at 12:50
answered Aug 6 at 12:39
gimusi
65.4k73684
answered Aug 6 at 12:39
gimusi
65.4k73684
answered Aug 6 at 12:39
gimusi
65.4k73684
65.4k73684
How do you find the normal vector? And should'nt it be rot(F) = (0, -z, -y) ?
– StudentMaths
Aug 6 at 12:46
@StudentMaths Ops...you are right it is rot(F) = (0, -z, -y) of course. For the normal recall that for a plane ax+by+cz+d=0 the normal n=(a,b,c).
– gimusi
Aug 6 at 12:49
Thank you! I get this 1/sqrt(2) * integral 4y dxdy but what are the boundaries? I feel this would be easier to solve with polar coordinates.
– StudentMaths
Aug 6 at 12:54
Since y=z I suppose rot(F)=0.
– gimusi
Aug 6 at 13:00
add a comment |Â
How do you find the normal vector? And should'nt it be rot(F) = (0, -z, -y) ?
– StudentMaths
Aug 6 at 12:46
@StudentMaths Ops...you are right it is rot(F) = (0, -z, -y) of course. For the normal recall that for a plane ax+by+cz+d=0 the normal n=(a,b,c).
– gimusi
Aug 6 at 12:49
Thank you! I get this 1/sqrt(2) * integral 4y dxdy but what are the boundaries? I feel this would be easier to solve with polar coordinates.
– StudentMaths
Aug 6 at 12:54
Since y=z I suppose rot(F)=0.
– gimusi
Aug 6 at 13:00
How do you find the normal vector? And should'nt it be rot(F) = (0, -z, -y) ?
– StudentMaths
Aug 6 at 12:46
How do you find the normal vector? And should'nt it be rot(F) = (0, -z, -y) ?
– StudentMaths
Aug 6 at 12:46
@StudentMaths Ops...you are right it is rot(F) = (0, -z, -y) of course. For the normal recall that for a plane ax+by+cz+d=0 the normal n=(a,b,c).
– gimusi
Aug 6 at 12:49
@StudentMaths Ops...you are right it is rot(F) = (0, -z, -y) of course. For the normal recall that for a plane ax+by+cz+d=0 the normal n=(a,b,c).
– gimusi
Aug 6 at 12:49
Thank you! I get this 1/sqrt(2) * integral 4y dxdy but what are the boundaries? I feel this would be easier to solve with polar coordinates.
– StudentMaths
Aug 6 at 12:54
Thank you! I get this 1/sqrt(2) * integral 4y dxdy but what are the boundaries? I feel this would be easier to solve with polar coordinates.
– StudentMaths
Aug 6 at 12:54
Since y=z I suppose rot(F)=0.
– gimusi
Aug 6 at 13:00
Since y=z I suppose rot(F)=0.
– gimusi
Aug 6 at 13:00
add a comment |Â
up vote
0
down vote
Take $S$ to be the region in the plane $y = z$ enclosed by $mathcal C$. Since $nabla times mathbf F = (0, -z, -y)$ is parallel to the plane $y = z$, $(nabla times mathbf F) cdot dmathbf S = 0$.
answered Aug 13 at 17:03
Maxim
2,150113
add a comment |Â
up vote
0
down vote
Take $S$ to be the region in the plane $y = z$ enclosed by $mathcal C$. Since $nabla times mathbf F = (0, -z, -y)$ is parallel to the plane $y = z$, $(nabla times mathbf F) cdot dmathbf S = 0$.
answered Aug 13 at 17:03
Maxim
2,150113
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Take $S$ to be the region in the plane $y = z$ enclosed by $mathcal C$. Since $nabla times mathbf F = (0, -z, -y)$ is parallel to the plane $y = z$, $(nabla times mathbf F) cdot dmathbf S = 0$.
answered Aug 13 at 17:03
Maxim
2,150113
Take $S$ to be the region in the plane $y = z$ enclosed by $mathcal C$. Since $nabla times mathbf F = (0, -z, -y)$ is parallel to the plane $y = z$, $(nabla times mathbf F) cdot dmathbf S = 0$.
answered Aug 13 at 17:03
Maxim
2,150113
edited Aug 13 at 17:33
answered Aug 13 at 17:03
Maxim
2,150113
answered Aug 13 at 17:03
Maxim
2,150113
answered Aug 13 at 17:03
Maxim
2,150113
2,150113
add a comment |Â
add a comment |Â
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You’ve parameterized the projection of the cylinder onto the $x$-$y$ plane. Now use $y=z$ to get the correct curve.
– amd
Aug 6 at 22:53