Flux through a parabolic cylinder Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Green's theorem and fluxFormulas for calculating fluxFlux integral through elliptic cylinderImplications of zero divergence ($nabla cdot F$) when finding the fluxFlux integral using Stokes' TheoremEvaluate Flux of Field through Open CylinderFind flux through a sphereEvaluating surface integral (1) directly and (2) by applying Divergence Theorem give different resolutsFundamental calculation of flux using Green and Gauss's TheoremDivergence theorem to calculate flux through an open cylinder
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Flux through a parabolic cylinder
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Green's theorem and fluxFormulas for calculating fluxFlux integral through elliptic cylinderImplications of zero divergence ($nabla cdot F$) when finding the fluxFlux integral using Stokes' TheoremEvaluate Flux of Field through Open CylinderFind flux through a sphereEvaluating surface integral (1) directly and (2) by applying Divergence Theorem give different resolutsFundamental calculation of flux using Green and Gauss's TheoremDivergence theorem to calculate flux through an open cylinder
$begingroup$
I just want to check if what I've done is correct.
We know by the Divergence Theorem that the integral of the divergence over a volume yields the value of the function at the boundary, which is a surface:
$$int_V (nabla cdot vec F )dV= int_S vec F cdot d vec S$$
Then we can use it so as to compute the flux through a surface:
$$nabla cdot vec F = -2z$$
$$-2iiint zdV = -2int_0^1int_1^x^2int_0^1zdV = frac323$$
multivariable-calculus
asked Apr 2 at 10:09
JD_PMJD_PM
292111
$endgroup$
add a comment |
$begingroup$
I just want to check if what I've done is correct.
We know by the Divergence Theorem that the integral of the divergence over a volume yields the value of the function at the boundary, which is a surface:
$$int_V (nabla cdot vec F )dV= int_S vec F cdot d vec S$$
Then we can use it so as to compute the flux through a surface:
$$nabla cdot vec F = -2z$$
$$-2iiint zdV = -2int_0^1int_1^x^2int_0^1zdV = frac323$$
multivariable-calculus
asked Apr 2 at 10:09
JD_PMJD_PM
292111
$endgroup$
$begingroup$
Not quite, $z$ varies between $0$ and $4$
$endgroup$
– Rafa Budría
Apr 2 at 10:18
$begingroup$
@RafaBudría Typo, my bad. The integral is solved using the boundaries: $$-2iiint zdV = -2int_0^1int_1^x^2int_0^4zdV = frac323$$
$endgroup$
– JD_PM
Apr 2 at 11:07
add a comment |
$begingroup$
I just want to check if what I've done is correct.
We know by the Divergence Theorem that the integral of the divergence over a volume yields the value of the function at the boundary, which is a surface:
$$int_V (nabla cdot vec F )dV= int_S vec F cdot d vec S$$
Then we can use it so as to compute the flux through a surface:
$$nabla cdot vec F = -2z$$
$$-2iiint zdV = -2int_0^1int_1^x^2int_0^1zdV = frac323$$
multivariable-calculus
asked Apr 2 at 10:09
JD_PMJD_PM
292111
$endgroup$
I just want to check if what I've done is correct.
We know by the Divergence Theorem that the integral of the divergence over a volume yields the value of the function at the boundary, which is a surface:
$$int_V (nabla cdot vec F )dV= int_S vec F cdot d vec S$$
Then we can use it so as to compute the flux through a surface:
$$nabla cdot vec F = -2z$$
$$-2iiint zdV = -2int_0^1int_1^x^2int_0^1zdV = frac323$$
multivariable-calculus
multivariable-calculus
asked Apr 2 at 10:09
JD_PMJD_PM
292111
asked Apr 2 at 10:09
JD_PMJD_PM
292111
asked Apr 2 at 10:09
JD_PMJD_PM
292111
asked Apr 2 at 10:09
JD_PMJD_PM
292111
asked Apr 2 at 10:09
JD_PMJD_PM
292111
292111
$begingroup$
Not quite, $z$ varies between $0$ and $4$
$endgroup$
– Rafa Budría
Apr 2 at 10:18
$begingroup$
@RafaBudría Typo, my bad. The integral is solved using the boundaries: $$-2iiint zdV = -2int_0^1int_1^x^2int_0^4zdV = frac323$$
$endgroup$
– JD_PM
Apr 2 at 11:07
add a comment |
$begingroup$
Not quite, $z$ varies between $0$ and $4$
$endgroup$
– Rafa Budría
Apr 2 at 10:18
$begingroup$
@RafaBudría Typo, my bad. The integral is solved using the boundaries: $$-2iiint zdV = -2int_0^1int_1^x^2int_0^4zdV = frac323$$
$endgroup$
– JD_PM
Apr 2 at 11:07
$begingroup$
Not quite, $z$ varies between $0$ and $4$
$endgroup$
– Rafa Budría
Apr 2 at 10:18
$begingroup$
Not quite, $z$ varies between $0$ and $4$
$endgroup$
– Rafa Budría
Apr 2 at 10:18
$begingroup$
@RafaBudría Typo, my bad. The integral is solved using the boundaries: $$-2iiint zdV = -2int_0^1int_1^x^2int_0^4zdV = frac323$$
$endgroup$
– JD_PM
Apr 2 at 11:07
$begingroup$
@RafaBudría Typo, my bad. The integral is solved using the boundaries: $$-2iiint zdV = -2int_0^1int_1^x^2int_0^4zdV = frac323$$
$endgroup$
– JD_PM
Apr 2 at 11:07
add a comment |
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$begingroup$
Not quite, $z$ varies between $0$ and $4$
$endgroup$
– Rafa Budría
Apr 2 at 10:18
$begingroup$
@RafaBudría Typo, my bad. The integral is solved using the boundaries: $$-2iiint zdV = -2int_0^1int_1^x^2int_0^4zdV = frac323$$
$endgroup$
– JD_PM
Apr 2 at 11:07