Open Cylinder Flux Calculations using Divergence Theorem

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Homework Help Overview

The problem involves calculating the flux of a vector field \(\vec{G}\) through an open cylinder using the divergence theorem. The vector field is defined outside the cylinder \(x^2 + y^2 = 4\) and the flux is to be evaluated over the surface of the cylinder defined by \(x^2 + y^2 = 16\) and \(0 \leq z \leq 7\).

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the applicability of the divergence theorem and the use of cylindrical coordinates. There is uncertainty regarding the calculation of divergence and how to parametrize the open cylinder.

Discussion Status

Some participants are exploring the implications of the vector field being defined outside the cylinder, with differing interpretations of what that entails. Guidance has been offered regarding the divergence calculation and the need for clarification on the field's definition.

Contextual Notes

There is confusion about the meaning of the vector field being defined outside the cylinder, with suggestions that it may imply the field is null inside the cylinder and on its boundaries. Participants are seeking further clarification on this point.

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Homework Statement



A vector field [tex]\vec{G}[/tex] in 3-space is defined outside the cylinder [tex]x^2 + y^2 = 4[/tex][tex]\vec{G} = \frac{6y\vec{i}-6x\vec{j}}{x^2+y^2}[/tex]

Find [tex]\int\limits_S \vec{G} \cdot · d\vec{A}[/tex] where S is the open cylinder [tex]x^2 + y^2 = 16 , 0 \leq z \leq 7[/tex] oriented outward.

Homework Equations


The Attempt at a Solution



I am planning to use divergence theorem here... can I use it?
 
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use also cilndric coordinates ;)
 
it's kind of hard to find the divergence here... is it just:

[tex]\frac{24xy}{(x^2+y^2)^2}[/tex]

at least that's what I got. How do I parametrize the open cylinder?
 
What does it mean defined outside the cylinder?
 
mitchturb said:
What does it mean defined outside the cylinder?

I am actually confused my self with that... as far as my understanding goes it just means the direction points outwards...

someone care to correct me and give me more help to solve this problem?
 
Usually it mean that inside the cylinder and on its boundaries the field is null. Think about electrostatics ;)

MM
 

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