Flux through cylindrical wedge

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SUMMARY

The discussion centers on calculating the flux of the vector field \textbf{F} = x\textbf{i} + y\textbf{j} + z\textbf{k} through a cylindrical surface defined by x² + y² = 1 and bounded by the planes z = 0 and x + y + z = 2. By applying Gauss' Theorem, the divergence of \textbf{F} is determined to be 3, leading to a flux calculation of 6π based on the enclosed volume of 2π. Despite arriving at this result, the author expresses confusion over a multiple-choice test where the correct answer was not listed among the options provided.

PREREQUISITES
  • Understanding of vector calculus, specifically Gauss' Theorem
  • Familiarity with cylindrical coordinates
  • Knowledge of divergence and flux calculations
  • Ability to perform triple integrals
NEXT STEPS
  • Review the application of Gauss' Theorem in vector calculus
  • Study cylindrical coordinate transformations in detail
  • Practice calculating flux through various surfaces using divergence
  • Explore common pitfalls in multiple-choice tests related to vector fields
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Students studying vector calculus, particularly those tackling problems involving flux and divergence, as well as educators preparing multiple-choice assessments in this subject area.

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Homework Statement

Given \textbf{F} = x\textbf{i} + y\textbf{j} + z\textbf{k}, what is the flux of \textbf{F} through the cylinder x^2 + y^2 =1 bounded by the planes z=0, x+y+z=2.


The Attempt at a Solution


By Gauss' Theorem, \int\int_{S}\textbf{F}\cdot d\textbf{S} = \int\int\int_{V}(\nabla\cdot \textbf{F})dV

But \nabla\cdot \textbf{F}=3, so the flux through the surface equals 3 times the enclosed volume. Using cylindric coordinates to calculate the volume from the integrals or using the fact the the volume is half that of a cylinder of radius 4, the volume is 2\pi and that gives 6\pi as the flux.

However, on a multiple choice test I just took, the answers offered were 0, \pi, 2\pi, 4\pi, and 10\pi. Where did I make a mistake?
 
Last edited:
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Hello!

I did the volume integral with the boundaries:

<br /> \int\int\int_{V}(\nabla\cdot \textbf{F})dV<br />

z = [0, 2 - x - y]
r = [0, 1]
theta = [-pi, pi]

x = r cos(theta)
y = r sin(theta)

and I also got the answer of 6pi :)
 

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