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Given the union of a cylinder and a sphere calculate the line integral using Stoke's theorem
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Given the surface $S=S_1cup S_2$ where $S_1= (x,y,z) in mathbbR^3 / x^2+y^2=1 ; space 0 leq zleq1 $ and $S_2= (x,y,z) in mathbbR^3 / x^2+y^2+(z-1)^2=1 ; space zgeq1 $, orientated with the normal pointing outside the cylinder (with cap) and the sphere, respectively.
Let $F(x,y,z)=(zx + z^2y+x, z^3yx+y,z^4x^2)$. Calculate $int_S (nabla times F)space dS$
Then, because Stoke's theorem we have that:
$int_S (nabla times F)space dS= int_C_1 F space dS + int_C_2 F space dS$
Where, $C_1$ is the simple-closed curve of the cylinder, and $C_2$ of the half of the sphere.
I'm having troubles about defining the appropriate orientation of the curve to make the normal vector pointing outside both surfaces, and also I wonder if I should or not parametrize them with polar coordinates. Can someone help me with this?
calculus vector-analysis parametrization line-integrals stokes-theorem
edited 2 days ago
Maxim
2,075113
asked Jul 20 at 13:42
Neisy SofÃa Vadori
302213
 |Â
show 2 more comments
up vote
0
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Given the surface $S=S_1cup S_2$ where $S_1= (x,y,z) in mathbbR^3 / x^2+y^2=1 ; space 0 leq zleq1 $ and $S_2= (x,y,z) in mathbbR^3 / x^2+y^2+(z-1)^2=1 ; space zgeq1 $, orientated with the normal pointing outside the cylinder (with cap) and the sphere, respectively.
Let $F(x,y,z)=(zx + z^2y+x, z^3yx+y,z^4x^2)$. Calculate $int_S (nabla times F)space dS$
Then, because Stoke's theorem we have that:
$int_S (nabla times F)space dS= int_C_1 F space dS + int_C_2 F space dS$
Where, $C_1$ is the simple-closed curve of the cylinder, and $C_2$ of the half of the sphere.
I'm having troubles about defining the appropriate orientation of the curve to make the normal vector pointing outside both surfaces, and also I wonder if I should or not parametrize them with polar coordinates. Can someone help me with this?
calculus vector-analysis parametrization line-integrals stokes-theorem
edited 2 days ago
Maxim
2,075113
asked Jul 20 at 13:42
Neisy SofÃa Vadori
302213
I think that, according to Stokes, you should have $int_S (nabla times F) dS = int_C F ds,$ where $C$ is the circle $x^2+y^2 = 1, , z=0.$ Try drawing your surface.
– Sobi
Jul 20 at 14:00
I already drew my surface, and is a cylinder from $z=0$ up to $z=1$ and then you have as the rest of $S$ the half part of a sphere, where both figures have the same radius, equal to 1.
– Neisy SofÃa Vadori
Jul 20 at 14:10
Great! Can you see that the boundary of the surface is the circle $C$ from my first comment?
– Sobi
Jul 20 at 14:11
Yes, but what about the plane $z=0$, that is like the cylinder base? That also give as another curve for the cylinder
– Neisy SofÃa Vadori
Jul 20 at 14:12
1
Polar coordinates should work great here, yes!
– Sobi
Jul 20 at 14:16
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given the surface $S=S_1cup S_2$ where $S_1= (x,y,z) in mathbbR^3 / x^2+y^2=1 ; space 0 leq zleq1 $ and $S_2= (x,y,z) in mathbbR^3 / x^2+y^2+(z-1)^2=1 ; space zgeq1 $, orientated with the normal pointing outside the cylinder (with cap) and the sphere, respectively.
Let $F(x,y,z)=(zx + z^2y+x, z^3yx+y,z^4x^2)$. Calculate $int_S (nabla times F)space dS$
Then, because Stoke's theorem we have that:
$int_S (nabla times F)space dS= int_C_1 F space dS + int_C_2 F space dS$
Where, $C_1$ is the simple-closed curve of the cylinder, and $C_2$ of the half of the sphere.
I'm having troubles about defining the appropriate orientation of the curve to make the normal vector pointing outside both surfaces, and also I wonder if I should or not parametrize them with polar coordinates. Can someone help me with this?
calculus vector-analysis parametrization line-integrals stokes-theorem
edited 2 days ago
Maxim
2,075113
asked Jul 20 at 13:42
Neisy SofÃa Vadori
302213
Given the surface $S=S_1cup S_2$ where $S_1= (x,y,z) in mathbbR^3 / x^2+y^2=1 ; space 0 leq zleq1 $ and $S_2= (x,y,z) in mathbbR^3 / x^2+y^2+(z-1)^2=1 ; space zgeq1 $, orientated with the normal pointing outside the cylinder (with cap) and the sphere, respectively.
Let $F(x,y,z)=(zx + z^2y+x, z^3yx+y,z^4x^2)$. Calculate $int_S (nabla times F)space dS$
Then, because Stoke's theorem we have that:
$int_S (nabla times F)space dS= int_C_1 F space dS + int_C_2 F space dS$
Where, $C_1$ is the simple-closed curve of the cylinder, and $C_2$ of the half of the sphere.
I'm having troubles about defining the appropriate orientation of the curve to make the normal vector pointing outside both surfaces, and also I wonder if I should or not parametrize them with polar coordinates. Can someone help me with this?
calculus vector-analysis parametrization line-integrals stokes-theorem
edited 2 days ago
Maxim
2,075113
asked Jul 20 at 13:42
Neisy SofÃa Vadori
302213
edited 2 days ago
Maxim
2,075113
edited 2 days ago
Maxim
2,075113
edited 2 days ago
Maxim
2,075113
2,075113
asked Jul 20 at 13:42
Neisy SofÃa Vadori
302213
asked Jul 20 at 13:42
Neisy SofÃa Vadori
302213
asked Jul 20 at 13:42
Neisy SofÃa Vadori
302213
302213
I think that, according to Stokes, you should have $int_S (nabla times F) dS = int_C F ds,$ where $C$ is the circle $x^2+y^2 = 1, , z=0.$ Try drawing your surface.
– Sobi
Jul 20 at 14:00
I already drew my surface, and is a cylinder from $z=0$ up to $z=1$ and then you have as the rest of $S$ the half part of a sphere, where both figures have the same radius, equal to 1.
– Neisy SofÃa Vadori
Jul 20 at 14:10
Great! Can you see that the boundary of the surface is the circle $C$ from my first comment?
– Sobi
Jul 20 at 14:11
Yes, but what about the plane $z=0$, that is like the cylinder base? That also give as another curve for the cylinder
– Neisy SofÃa Vadori
Jul 20 at 14:12
1
Polar coordinates should work great here, yes!
– Sobi
Jul 20 at 14:16
 |Â
show 2 more comments
I think that, according to Stokes, you should have $int_S (nabla times F) dS = int_C F ds,$ where $C$ is the circle $x^2+y^2 = 1, , z=0.$ Try drawing your surface.
– Sobi
Jul 20 at 14:00
I already drew my surface, and is a cylinder from $z=0$ up to $z=1$ and then you have as the rest of $S$ the half part of a sphere, where both figures have the same radius, equal to 1.
– Neisy SofÃa Vadori
Jul 20 at 14:10
Great! Can you see that the boundary of the surface is the circle $C$ from my first comment?
– Sobi
Jul 20 at 14:11
Yes, but what about the plane $z=0$, that is like the cylinder base? That also give as another curve for the cylinder
– Neisy SofÃa Vadori
Jul 20 at 14:12
1
Polar coordinates should work great here, yes!
– Sobi
Jul 20 at 14:16
I think that, according to Stokes, you should have $int_S (nabla times F) dS = int_C F ds,$ where $C$ is the circle $x^2+y^2 = 1, , z=0.$ Try drawing your surface.
– Sobi
Jul 20 at 14:00
I think that, according to Stokes, you should have $int_S (nabla times F) dS = int_C F ds,$ where $C$ is the circle $x^2+y^2 = 1, , z=0.$ Try drawing your surface.
– Sobi
Jul 20 at 14:00
I already drew my surface, and is a cylinder from $z=0$ up to $z=1$ and then you have as the rest of $S$ the half part of a sphere, where both figures have the same radius, equal to 1.
– Neisy SofÃa Vadori
Jul 20 at 14:10
I already drew my surface, and is a cylinder from $z=0$ up to $z=1$ and then you have as the rest of $S$ the half part of a sphere, where both figures have the same radius, equal to 1.
– Neisy SofÃa Vadori
Jul 20 at 14:10
Great! Can you see that the boundary of the surface is the circle $C$ from my first comment?
– Sobi
Jul 20 at 14:11
Great! Can you see that the boundary of the surface is the circle $C$ from my first comment?
– Sobi
Jul 20 at 14:11
Yes, but what about the plane $z=0$, that is like the cylinder base? That also give as another curve for the cylinder
– Neisy SofÃa Vadori
Jul 20 at 14:12
Yes, but what about the plane $z=0$, that is like the cylinder base? That also give as another curve for the cylinder
– Neisy SofÃa Vadori
Jul 20 at 14:12
1
1
Polar coordinates should work great here, yes!
– Sobi
Jul 20 at 14:16
Polar coordinates should work great here, yes!
– Sobi
Jul 20 at 14:16
 |Â
show 2 more comments
1 Answer
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Notice that $mathbf F(x, y, 0) = (x, y, 0)$ and the boundary of the surface is $(cos t, sin t, 0)$, therefore $mathbf F cdot dmathbf s = 0$.
answered 2 days ago
Maxim
2,075113
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Notice that $mathbf F(x, y, 0) = (x, y, 0)$ and the boundary of the surface is $(cos t, sin t, 0)$, therefore $mathbf F cdot dmathbf s = 0$.
answered 2 days ago
Maxim
2,075113
add a comment |Â
up vote
0
down vote
Notice that $mathbf F(x, y, 0) = (x, y, 0)$ and the boundary of the surface is $(cos t, sin t, 0)$, therefore $mathbf F cdot dmathbf s = 0$.
answered 2 days ago
Maxim
2,075113
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Notice that $mathbf F(x, y, 0) = (x, y, 0)$ and the boundary of the surface is $(cos t, sin t, 0)$, therefore $mathbf F cdot dmathbf s = 0$.
answered 2 days ago
Maxim
2,075113
Notice that $mathbf F(x, y, 0) = (x, y, 0)$ and the boundary of the surface is $(cos t, sin t, 0)$, therefore $mathbf F cdot dmathbf s = 0$.
answered 2 days ago
Maxim
2,075113
answered 2 days ago
Maxim
2,075113
answered 2 days ago
Maxim
2,075113
answered 2 days ago
Maxim
2,075113
2,075113
add a comment |Â
add a comment |Â
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I think that, according to Stokes, you should have $int_S (nabla times F) dS = int_C F ds,$ where $C$ is the circle $x^2+y^2 = 1, , z=0.$ Try drawing your surface.
– Sobi
Jul 20 at 14:00
I already drew my surface, and is a cylinder from $z=0$ up to $z=1$ and then you have as the rest of $S$ the half part of a sphere, where both figures have the same radius, equal to 1.
– Neisy SofÃa Vadori
Jul 20 at 14:10
Great! Can you see that the boundary of the surface is the circle $C$ from my first comment?
– Sobi
Jul 20 at 14:11
Yes, but what about the plane $z=0$, that is like the cylinder base? That also give as another curve for the cylinder
– Neisy SofÃa Vadori
Jul 20 at 14:12
1
Polar coordinates should work great here, yes!
– Sobi
Jul 20 at 14:16