Calculate Divergence Theorem for F with S and Q

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SUMMARY

The discussion focuses on calculating the Divergence Theorem for the vector field F defined as F(x,y,z) = (xy^2 + cosz)i + (x^2*y + sinz)j + e^(z)k, with the divergence calculated as divF = y^2 + x^2 + e^z. The closed surface is described as an ice-cream cone with its tip at the origin and capped by the plane z=8. To evaluate the triple integral Q = ∫∫∫ (y^2 + x^2 + e^z)dV, cylindrical coordinates are recommended for simplification, with limits for r set from 0 to 2z, θ from 0 to 2π, and z from 0 to 8.

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S\int\int
F*Nds
F(x,y,z) = (xy^2 + cosz)i + (x^2*y + sinz)j + e^(z)*k

s: z = 1/2\sqrt{x^2 + y^2} , z = 8

divF = y^2 + x^2 +e^z

Q\int\int\int (y^2 + x^2 + e^k)dV

This is as far as I got, I have no idea how to do the limits for this triple integral

thanks in advance guys.
 
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You can see that the closed surface concerned is an ice-cream cone with its tip centered at the origin and capped by the plane z=8. With this in mind, note that it would be easiest to evaluate the triple integral in either cylindrical or spherical coordinates. Cylindrical coordinates is simpler, in my opinion. Set up the integral and integrate in the order drdθdz.

Try to express r in terms of z. You already have the equation for a cone. And since x^2 + y^2 = r^2 = 4z^2, you can do so easily. Bearing that in mind, imagine a ray from the origin passing through the cone, it has minimum value 0 and maximum value where it leaves the cone at z=8, so your limits for r should be 0 and 2z. Limits for θ should be 0 to 2pi. And finally you are given the limits for z in the question, right? It should be 0 and 8.
 

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