Solve the following integral using Stokes Theorem.
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I am asked to evaluate the following integral:
$$intint text{curl} vec{F} cdot dvec{S}$$
where $F = (y,-x,zx^3y^2)$ and $S$ is the surface given by $x^2 + y^2 + 3z^2 = 1$ with $z leq 0$.
I did not have any problem with any other exercises of this kind. But this one is hard.
Any hint would be appreciated.
surface-integrals stokes-theorem
asked Nov 16 at 19:10
TheNicouU
171211
add a comment |
up vote
0
down vote
favorite
I am asked to evaluate the following integral:
$$intint text{curl} vec{F} cdot dvec{S}$$
where $F = (y,-x,zx^3y^2)$ and $S$ is the surface given by $x^2 + y^2 + 3z^2 = 1$ with $z leq 0$.
I did not have any problem with any other exercises of this kind. But this one is hard.
Any hint would be appreciated.
surface-integrals stokes-theorem
asked Nov 16 at 19:10
TheNicouU
171211
What exactly are you having difficulties with? Have you found the boundary of the surface?
– Maxim
Nov 17 at 19:24
I am having difficulties with all the exercie. I dont know how to parametrize that boundary. Do you have any idea?
– TheNicouU
Nov 18 at 0:16
The surface is the lower half of the ellipsoid, lying below the plane $z = 0$. The boundary lies in that plane. Substitute $z = 0$ into the equation of the ellipsoid.
– Maxim
Nov 18 at 1:04
So I just parametrize $x^2 + y^2 = 1$ and then calculate $F$ over that parametrization?
– TheNicouU
Nov 18 at 1:16
1
Correct, then you integrate $mathbf F cdot dmathbf s$.
– Maxim
Nov 18 at 1:53
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am asked to evaluate the following integral:
$$intint text{curl} vec{F} cdot dvec{S}$$
where $F = (y,-x,zx^3y^2)$ and $S$ is the surface given by $x^2 + y^2 + 3z^2 = 1$ with $z leq 0$.
I did not have any problem with any other exercises of this kind. But this one is hard.
Any hint would be appreciated.
surface-integrals stokes-theorem
asked Nov 16 at 19:10
TheNicouU
171211
I am asked to evaluate the following integral:
$$intint text{curl} vec{F} cdot dvec{S}$$
where $F = (y,-x,zx^3y^2)$ and $S$ is the surface given by $x^2 + y^2 + 3z^2 = 1$ with $z leq 0$.
I did not have any problem with any other exercises of this kind. But this one is hard.
Any hint would be appreciated.
surface-integrals stokes-theorem
surface-integrals stokes-theorem
asked Nov 16 at 19:10
TheNicouU
171211
asked Nov 16 at 19:10
TheNicouU
171211
asked Nov 16 at 19:10
TheNicouU
171211
asked Nov 16 at 19:10
TheNicouU
171211
asked Nov 16 at 19:10
TheNicouU
171211
171211
What exactly are you having difficulties with? Have you found the boundary of the surface?
– Maxim
Nov 17 at 19:24
I am having difficulties with all the exercie. I dont know how to parametrize that boundary. Do you have any idea?
– TheNicouU
Nov 18 at 0:16
The surface is the lower half of the ellipsoid, lying below the plane $z = 0$. The boundary lies in that plane. Substitute $z = 0$ into the equation of the ellipsoid.
– Maxim
Nov 18 at 1:04
So I just parametrize $x^2 + y^2 = 1$ and then calculate $F$ over that parametrization?
– TheNicouU
Nov 18 at 1:16
1
Correct, then you integrate $mathbf F cdot dmathbf s$.
– Maxim
Nov 18 at 1:53
add a comment |
What exactly are you having difficulties with? Have you found the boundary of the surface?
– Maxim
Nov 17 at 19:24
I am having difficulties with all the exercie. I dont know how to parametrize that boundary. Do you have any idea?
– TheNicouU
Nov 18 at 0:16
The surface is the lower half of the ellipsoid, lying below the plane $z = 0$. The boundary lies in that plane. Substitute $z = 0$ into the equation of the ellipsoid.
– Maxim
Nov 18 at 1:04
So I just parametrize $x^2 + y^2 = 1$ and then calculate $F$ over that parametrization?
– TheNicouU
Nov 18 at 1:16
1
Correct, then you integrate $mathbf F cdot dmathbf s$.
– Maxim
Nov 18 at 1:53
What exactly are you having difficulties with? Have you found the boundary of the surface?
– Maxim
Nov 17 at 19:24
What exactly are you having difficulties with? Have you found the boundary of the surface?
– Maxim
Nov 17 at 19:24
I am having difficulties with all the exercie. I dont know how to parametrize that boundary. Do you have any idea?
– TheNicouU
Nov 18 at 0:16
I am having difficulties with all the exercie. I dont know how to parametrize that boundary. Do you have any idea?
– TheNicouU
Nov 18 at 0:16
The surface is the lower half of the ellipsoid, lying below the plane $z = 0$. The boundary lies in that plane. Substitute $z = 0$ into the equation of the ellipsoid.
– Maxim
Nov 18 at 1:04
The surface is the lower half of the ellipsoid, lying below the plane $z = 0$. The boundary lies in that plane. Substitute $z = 0$ into the equation of the ellipsoid.
– Maxim
Nov 18 at 1:04
So I just parametrize $x^2 + y^2 = 1$ and then calculate $F$ over that parametrization?
– TheNicouU
Nov 18 at 1:16
So I just parametrize $x^2 + y^2 = 1$ and then calculate $F$ over that parametrization?
– TheNicouU
Nov 18 at 1:16
1
1
Correct, then you integrate $mathbf F cdot dmathbf s$.
– Maxim
Nov 18 at 1:53
Correct, then you integrate $mathbf F cdot dmathbf s$.
– Maxim
Nov 18 at 1:53
add a comment |
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What exactly are you having difficulties with? Have you found the boundary of the surface?
– Maxim
Nov 17 at 19:24
I am having difficulties with all the exercie. I dont know how to parametrize that boundary. Do you have any idea?
– TheNicouU
Nov 18 at 0:16
The surface is the lower half of the ellipsoid, lying below the plane $z = 0$. The boundary lies in that plane. Substitute $z = 0$ into the equation of the ellipsoid.
– Maxim
Nov 18 at 1:04
So I just parametrize $x^2 + y^2 = 1$ and then calculate $F$ over that parametrization?
– TheNicouU
Nov 18 at 1:16
1
Correct, then you integrate $mathbf F cdot dmathbf s$.
– Maxim
Nov 18 at 1:53