Flux through the surface of an ellipsoidCalculate the flux through a closed surfaceFlux Integral in spherical coordinatesFlux through a SurfaceEvaluating surface integral (1) directly and (2) by applying Divergence Theorem give different resolutsCompute the flux of the vector field $vecF$ through the surface SThe flux of the vector field $u = x hatimath + y hatjmath + z hatk$ through the surface of the ellipsoidFlux through surface (correct or not?)Calculation of flux through sphere when the vector field is not defined at the origincalculate flux through surfaceFlux of a hemisphere
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Flux through the surface of an ellipsoid
Calculate the flux through a closed surfaceFlux Integral in spherical coordinatesFlux through a SurfaceEvaluating surface integral (1) directly and (2) by applying Divergence Theorem give different resolutsCompute the flux of the vector field $vecF$ through the surface SThe flux of the vector field $u = x hatimath + y hatjmath + z hatk$ through the surface of the ellipsoidFlux through surface (correct or not?)Calculation of flux through sphere when the vector field is not defined at the origincalculate flux through surfaceFlux of a hemisphere
$begingroup$
I was asked to calculate the flux of the field $$mathbf A = (1/R^2)hat r$$
where $R$ is the radius, through the surface of the ellipsoid $$left(fracx^2a^2right) + left(fracy^2b^2right) + z^2 = 1$$
Without doing the formal calculations (I have never encountered a problem like this), I neverless suspect that it's equivalent to the flux of the same field through the surface of a sphere whose radius is the mean radius of the ellipsoid (i.e. $4 pi$)...am I wrong and/or how to solve?
integration multivariable-calculus vector-analysis surface-integrals
$endgroup$
add a comment |
$begingroup$
I was asked to calculate the flux of the field $$mathbf A = (1/R^2)hat r$$
where $R$ is the radius, through the surface of the ellipsoid $$left(fracx^2a^2right) + left(fracy^2b^2right) + z^2 = 1$$
Without doing the formal calculations (I have never encountered a problem like this), I neverless suspect that it's equivalent to the flux of the same field through the surface of a sphere whose radius is the mean radius of the ellipsoid (i.e. $4 pi$)...am I wrong and/or how to solve?
integration multivariable-calculus vector-analysis surface-integrals
$endgroup$
add a comment |
$begingroup$
I was asked to calculate the flux of the field $$mathbf A = (1/R^2)hat r$$
where $R$ is the radius, through the surface of the ellipsoid $$left(fracx^2a^2right) + left(fracy^2b^2right) + z^2 = 1$$
Without doing the formal calculations (I have never encountered a problem like this), I neverless suspect that it's equivalent to the flux of the same field through the surface of a sphere whose radius is the mean radius of the ellipsoid (i.e. $4 pi$)...am I wrong and/or how to solve?
integration multivariable-calculus vector-analysis surface-integrals
$endgroup$
I was asked to calculate the flux of the field $$mathbf A = (1/R^2)hat r$$
where $R$ is the radius, through the surface of the ellipsoid $$left(fracx^2a^2right) + left(fracy^2b^2right) + z^2 = 1$$
Without doing the formal calculations (I have never encountered a problem like this), I neverless suspect that it's equivalent to the flux of the same field through the surface of a sphere whose radius is the mean radius of the ellipsoid (i.e. $4 pi$)...am I wrong and/or how to solve?
integration multivariable-calculus vector-analysis surface-integrals
integration multivariable-calculus vector-analysis surface-integrals
asked Mar 13 at 10:43
Lo ScrondoLo Scrondo
19410
19410
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$begingroup$
The field you are asked about is the prominent gravitational / Coulomb field which main characteristics is the fact that its divergence is $0$ everywhere except for the origin point $(0,0,0)$:
$$
nablacdotmathbf A=4pidelta(mathbf r),
$$
which is easy to check.
Therefore by divergence theorem the flux over the surface is $4pi$ since the origin point is inside the surface.
$endgroup$
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1 Answer
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1 Answer
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$begingroup$
The field you are asked about is the prominent gravitational / Coulomb field which main characteristics is the fact that its divergence is $0$ everywhere except for the origin point $(0,0,0)$:
$$
nablacdotmathbf A=4pidelta(mathbf r),
$$
which is easy to check.
Therefore by divergence theorem the flux over the surface is $4pi$ since the origin point is inside the surface.
$endgroup$
add a comment |
$begingroup$
The field you are asked about is the prominent gravitational / Coulomb field which main characteristics is the fact that its divergence is $0$ everywhere except for the origin point $(0,0,0)$:
$$
nablacdotmathbf A=4pidelta(mathbf r),
$$
which is easy to check.
Therefore by divergence theorem the flux over the surface is $4pi$ since the origin point is inside the surface.
$endgroup$
add a comment |
$begingroup$
The field you are asked about is the prominent gravitational / Coulomb field which main characteristics is the fact that its divergence is $0$ everywhere except for the origin point $(0,0,0)$:
$$
nablacdotmathbf A=4pidelta(mathbf r),
$$
which is easy to check.
Therefore by divergence theorem the flux over the surface is $4pi$ since the origin point is inside the surface.
$endgroup$
The field you are asked about is the prominent gravitational / Coulomb field which main characteristics is the fact that its divergence is $0$ everywhere except for the origin point $(0,0,0)$:
$$
nablacdotmathbf A=4pidelta(mathbf r),
$$
which is easy to check.
Therefore by divergence theorem the flux over the surface is $4pi$ since the origin point is inside the surface.
edited Mar 13 at 14:54
answered Mar 13 at 14:49
useruser
5,45411030
5,45411030
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