Finding the flux (Divergence Theorem)

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Homework Help Overview

The discussion revolves around applying the divergence theorem to find the flux of a vector field out of a surface defined by a paraboloid. The vector field in question is given as F = (y^3)i + (x^3)j + (3z^2)k, and the surface is bounded by z = x^2 + y^2, with z ≤ 9.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the integration limits for the z-integral and how to express the maximum radius in terms of z. There are questions about integrating with variable limits and concerns regarding the surface integral and potential errors in calculations.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to the problem, including the implications of changing integration limits and the complexities of applying the divergence theorem. Some guidance has been offered regarding the integration setup, but no consensus has been reached on the surface integral or the identification of errors.

Contextual Notes

Participants note the need to clarify the maximum radius for integration based on the variable z and express uncertainty about the surface integral calculations. There is an acknowledgment of potential errors in previous attempts, particularly regarding the volume integral.

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Homework Statement


By using divergence theorem find the flux of vector F out of the surface of the paraboloid z = x^2 + y^2, z<=9, when F = (y^3)i + (x^3)j + (3z^2)k

Homework Equations


Divergence theorem equation stated in the attempt part

The Attempt at a Solution


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Suppose you are doing the z-integral first, so we are looking at a slice of fixed z. Then the slice looks like a disk of radius rmax. Instead of 0 to 3, you want to integrate each slice over r from 0 to rmax. You can express rmax in terms of z.
So you will get
\int_0^9 \int_0^{r_\mathrm{max}(z)} \int_0^{2\pi} 6 z r \, \mathrm{d}\varphi \, \mathrm{d}r \, \mathrm d{z}
where rmax(z) depends on z instead of being identically equal to 3 as you have now.
 
Last edited:
CompuChip said:
Suppose you are doing the z-integral first, so we are looking at a slice of fixed z. Then the slice looks like a disk of radius rmax. Instead of 0 to 3, you want to integrate each slice over r from 0 to rmax. You can express rmax in terms of z.
So you will get
\int_0^9 \int_0^{r_\mathrm{max}(z)} \int_0^{2\pi} 6 z r \, \mathrm{d}\varphi \, \mathrm{d}r \, \mathrm d{z}
where rmax(z) depends on z instead of being identically equal to 3 as you have now.
I understand what you mean about the radius changing with z, but how would i integrate dr with limits rmax(z)?
Also what about the Fn ds double integral?
 
For some given z, what is the maximum radius (i.e. the upper boundary of your r-integration)?

I haven't looked into the other approach (where you first apply the divergence theorem) but it does look a bit more complicated to me (finding the correct normal unit vector and all).
 
Just thought id bump this question. I understand my error when calculating the volume integral, the problem I'm left with is the surface integral. I still can't spot the mistake there. Any suggestions? Thanks.
 

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