Calculate Flux of Vector Field in Closed Surface | Divergence Problem Solution

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SUMMARY

The discussion focuses on calculating the flux of the vector field F = x^3 + y^3 + z^3 out of a closed surface bounding the solid region defined by x^2 + y^2 ≤ 16 and 0 ≤ z ≤ 9. The divergence of the vector field is determined to be 3x^2 + 3y^2 + 3z^2. To find the flux, the divergence theorem is applied, which simplifies the calculation to integrating the divergence over the volume, utilizing cylindrical coordinates and the appropriate Jacobian for the transformation.

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Homework Statement



Find the flux of the vector field out of the closed surface bounding the solid region x^2 + y^2 ≤ 16, 0 ≤ z ≤ 9, oriented outward.

F = x^3 + y^3 + z^3


Homework Equations





The Attempt at a Solution


I found the divergence which is 3x^2+3y^2+3z^2.

And then I'm stuck. I know Flux is divergence * Volume, in a simplifed way. So i factored 3 out, and i got 3(x^2+y^2) + 3z^2. I do not know where to go from here. Thanks in advance
 
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Integrate over the surface using the transformation to cylindrical co-ordinates, using x = square_root(16) cos(theta) and similarly for y, and the divergence theorem gives support to your integrating over the vector field, once you have added in the jacobian, from zero to two pi for theta, and from zero to 9 for z.
 

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