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Evaluate Integral Using Stokes Theorem
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Using Stokes theorem evaluate
$$int_Gamma(z-y),dx-(x+z),dy-(x+y),dz$$
where $Gamma$ is the intersection of $x^2+y^2+z^2=4$ with the plane $z=y$ with anticlockwise orientation when looking on $z$ from the positive side
If we take $nablatimes F=(0,2,0)$ but how can we find $hatn$
calculus multivariable-calculus stokes-theorem
asked Aug 6 at 17:58
newhere
786310
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up vote
2
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Using Stokes theorem evaluate
$$int_Gamma(z-y),dx-(x+z),dy-(x+y),dz$$
where $Gamma$ is the intersection of $x^2+y^2+z^2=4$ with the plane $z=y$ with anticlockwise orientation when looking on $z$ from the positive side
If we take $nablatimes F=(0,2,0)$ but how can we find $hatn$
calculus multivariable-calculus stokes-theorem
asked Aug 6 at 17:58
newhere
786310
Are you sure the curl is 0?
– Davey
Aug 6 at 18:06
@Davey correct, edited
– newhere
Aug 6 at 18:18
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Using Stokes theorem evaluate
$$int_Gamma(z-y),dx-(x+z),dy-(x+y),dz$$
where $Gamma$ is the intersection of $x^2+y^2+z^2=4$ with the plane $z=y$ with anticlockwise orientation when looking on $z$ from the positive side
If we take $nablatimes F=(0,2,0)$ but how can we find $hatn$
calculus multivariable-calculus stokes-theorem
asked Aug 6 at 17:58
newhere
786310
Using Stokes theorem evaluate
$$int_Gamma(z-y),dx-(x+z),dy-(x+y),dz$$
where $Gamma$ is the intersection of $x^2+y^2+z^2=4$ with the plane $z=y$ with anticlockwise orientation when looking on $z$ from the positive side
If we take $nablatimes F=(0,2,0)$ but how can we find $hatn$
calculus multivariable-calculus stokes-theorem
asked Aug 6 at 17:58
newhere
786310
edited Aug 6 at 18:18
asked Aug 6 at 17:58
newhere
786310
asked Aug 6 at 17:58
newhere
786310
asked Aug 6 at 17:58
newhere
786310
786310
Are you sure the curl is 0?
– Davey
Aug 6 at 18:06
@Davey correct, edited
– newhere
Aug 6 at 18:18
add a comment |Â
Are you sure the curl is 0?
– Davey
Aug 6 at 18:06
@Davey correct, edited
– newhere
Aug 6 at 18:18
Are you sure the curl is 0?
– Davey
Aug 6 at 18:06
Are you sure the curl is 0?
– Davey
Aug 6 at 18:06
@Davey correct, edited
– newhere
Aug 6 at 18:18
@Davey correct, edited
– newhere
Aug 6 at 18:18
add a comment |Â
2 Answers
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Since $Gamma$ is a circle in a plane, we can apply Stokes' Theorem to the disk $D$ that it encloses to get that $int_Gamma F cdot ds = int_D$ curl $F cdot dA$. At each point of $D$, the normal to the disk is the same as the normal to the plane. In general, a normal to a plane is a vector $v$ such that the equation of the plane can be written in the form $(x - x_0, y - y_0, z - z_0) cdot v = 0$, where $(x_0, y_0, z_o)$ is any fixed point on the plane. In this case, we can take $(x_0, y_0, z_0)$ to be the origin. So you just need to find a unit vector $v$ with this property, pointing in whichever direction agrees with the orientation of $Gamma$.
answered Aug 6 at 18:36
Davey
43827
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1
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$vec n$ is normal to plane. So, $ vec n = <0,1,-1>/sqrt 2 \ iint Curl (vec F). vec n dS = iint <0,2,0> cdot (<0,1,-1>/sqrt 2) dS \ = iint
sqrt 2 dS = sqrt 2 (Area space of space circle) = sqrt 2 pi(2^2) \ = 4 sqrt 2pi$
answered Aug 6 at 19:30
Magneto
787213
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Since $Gamma$ is a circle in a plane, we can apply Stokes' Theorem to the disk $D$ that it encloses to get that $int_Gamma F cdot ds = int_D$ curl $F cdot dA$. At each point of $D$, the normal to the disk is the same as the normal to the plane. In general, a normal to a plane is a vector $v$ such that the equation of the plane can be written in the form $(x - x_0, y - y_0, z - z_0) cdot v = 0$, where $(x_0, y_0, z_o)$ is any fixed point on the plane. In this case, we can take $(x_0, y_0, z_0)$ to be the origin. So you just need to find a unit vector $v$ with this property, pointing in whichever direction agrees with the orientation of $Gamma$.
answered Aug 6 at 18:36
Davey
43827
add a comment |Â
up vote
2
down vote
accepted
Since $Gamma$ is a circle in a plane, we can apply Stokes' Theorem to the disk $D$ that it encloses to get that $int_Gamma F cdot ds = int_D$ curl $F cdot dA$. At each point of $D$, the normal to the disk is the same as the normal to the plane. In general, a normal to a plane is a vector $v$ such that the equation of the plane can be written in the form $(x - x_0, y - y_0, z - z_0) cdot v = 0$, where $(x_0, y_0, z_o)$ is any fixed point on the plane. In this case, we can take $(x_0, y_0, z_0)$ to be the origin. So you just need to find a unit vector $v$ with this property, pointing in whichever direction agrees with the orientation of $Gamma$.
answered Aug 6 at 18:36
Davey
43827
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Since $Gamma$ is a circle in a plane, we can apply Stokes' Theorem to the disk $D$ that it encloses to get that $int_Gamma F cdot ds = int_D$ curl $F cdot dA$. At each point of $D$, the normal to the disk is the same as the normal to the plane. In general, a normal to a plane is a vector $v$ such that the equation of the plane can be written in the form $(x - x_0, y - y_0, z - z_0) cdot v = 0$, where $(x_0, y_0, z_o)$ is any fixed point on the plane. In this case, we can take $(x_0, y_0, z_0)$ to be the origin. So you just need to find a unit vector $v$ with this property, pointing in whichever direction agrees with the orientation of $Gamma$.
answered Aug 6 at 18:36
Davey
43827
Since $Gamma$ is a circle in a plane, we can apply Stokes' Theorem to the disk $D$ that it encloses to get that $int_Gamma F cdot ds = int_D$ curl $F cdot dA$. At each point of $D$, the normal to the disk is the same as the normal to the plane. In general, a normal to a plane is a vector $v$ such that the equation of the plane can be written in the form $(x - x_0, y - y_0, z - z_0) cdot v = 0$, where $(x_0, y_0, z_o)$ is any fixed point on the plane. In this case, we can take $(x_0, y_0, z_0)$ to be the origin. So you just need to find a unit vector $v$ with this property, pointing in whichever direction agrees with the orientation of $Gamma$.
answered Aug 6 at 18:36
Davey
43827
answered Aug 6 at 18:36
Davey
43827
answered Aug 6 at 18:36
Davey
43827
answered Aug 6 at 18:36
Davey
43827
43827
add a comment |Â
add a comment |Â
up vote
1
down vote
$vec n$ is normal to plane. So, $ vec n = <0,1,-1>/sqrt 2 \ iint Curl (vec F). vec n dS = iint <0,2,0> cdot (<0,1,-1>/sqrt 2) dS \ = iint
sqrt 2 dS = sqrt 2 (Area space of space circle) = sqrt 2 pi(2^2) \ = 4 sqrt 2pi$
answered Aug 6 at 19:30
Magneto
787213
add a comment |Â
up vote
1
down vote
$vec n$ is normal to plane. So, $ vec n = <0,1,-1>/sqrt 2 \ iint Curl (vec F). vec n dS = iint <0,2,0> cdot (<0,1,-1>/sqrt 2) dS \ = iint
sqrt 2 dS = sqrt 2 (Area space of space circle) = sqrt 2 pi(2^2) \ = 4 sqrt 2pi$
answered Aug 6 at 19:30
Magneto
787213
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$vec n$ is normal to plane. So, $ vec n = <0,1,-1>/sqrt 2 \ iint Curl (vec F). vec n dS = iint <0,2,0> cdot (<0,1,-1>/sqrt 2) dS \ = iint
sqrt 2 dS = sqrt 2 (Area space of space circle) = sqrt 2 pi(2^2) \ = 4 sqrt 2pi$
answered Aug 6 at 19:30
Magneto
787213
$vec n$ is normal to plane. So, $ vec n = <0,1,-1>/sqrt 2 \ iint Curl (vec F). vec n dS = iint <0,2,0> cdot (<0,1,-1>/sqrt 2) dS \ = iint
sqrt 2 dS = sqrt 2 (Area space of space circle) = sqrt 2 pi(2^2) \ = 4 sqrt 2pi$
answered Aug 6 at 19:30
Magneto
787213
answered Aug 6 at 19:30
Magneto
787213
answered Aug 6 at 19:30
Magneto
787213
answered Aug 6 at 19:30
Magneto
787213
787213
add a comment |Â
add a comment |Â
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Are you sure the curl is 0?
– Davey
Aug 6 at 18:06
@Davey correct, edited
– newhere
Aug 6 at 18:18