Compute the downwards flux of F

  • Context: MHB 
  • Thread starter Thread starter Uniman
  • Start date Start date
  • Tags Tags
    Flux
Click For Summary
SUMMARY

The discussion focuses on computing the downward flux of the vector field \(\mathbf{F}(x,y,z)=\mathbf{i} \cos\left( \frac{y}{z}\right)-\mathbf{j} \frac{x}{z} \sin \left(\frac{y}{z}\right)+\mathbf{k} \frac{xy}{z^2} \sin \left( \frac{y}{z} \right)\) over the surface \(S\) where \(z=1\). The downward flux is calculated using the integral \(-\iint_{S}xy\sin y\,dS\), leading to the double integral \(-\int_{x=0}^{1}\int_{y=0}^{\pi}xy\sin y\,dy\,dx\). The user is encouraged to continue the calculations based on this setup.

PREREQUISITES
  • Understanding of vector fields and flux integrals
  • Familiarity with double integrals in multivariable calculus
  • Knowledge of surface integrals and their applications
  • Proficiency in using LaTeX for mathematical expressions
NEXT STEPS
  • Study the properties of vector fields in calculus
  • Learn about surface integrals and their applications in physics
  • Explore techniques for evaluating double integrals
  • Investigate the divergence theorem and its relation to flux
USEFUL FOR

Students and professionals in mathematics, physics, and engineering who are working with vector fields and flux calculations, particularly those involved in multivariable calculus and surface integrals.

Uniman
Messages
11
Reaction score
0

Attachments

  • Screen Shot 2012-10-19 at 6.19.57 PM.png
    Screen Shot 2012-10-19 at 6.19.57 PM.png
    11.3 KB · Views: 90
Physics news on Phys.org
Uniman said:
https://www.physicsforums.com/attachments/418
The work done so far

https://www.dropbox.com/s/qmfidqot3nttlw5/phpQd8GGH.png
https://www.dropbox.com/s/xn4t5m2aleopwy7/phpldZ0KX.png

Is the answer correct...

Hi Univman, :)

The links to your work on the problem seem not to work. Anyway, you have been given the vector field,

\[\mathbf{F}(x,y,z)=\mathbf{i} \cos\left( \frac{y}{z}\right)-\mathbf{j} \frac{x}{z} \sin \left(\frac{y}{z}\right)+\mathbf{k} \frac{xy}{z^2} \sin \left( \frac{y}{z} \right)\]

The downward flux of \(\mathbf{F}\) over \(S\) is given by,

\[\iint_{S}\mathbf{F}.(-\mathbf{k})\,dS=-\iint_{S}\mathbf{F}.\mathbf{k}\,dS=-\iint_{S}\frac{xy}{z^2}\sin\left(\frac{y}{z}\right)\,dS\]

On the surface \(S\); \(z=1\). Therefore we have,

\[\iint_{S}\mathbf{F}.(-\mathbf{k})\,dS=-\iint_{S}xy\sin y\,dS=-\int_{x=0}^{1}\int_{y=0}^{\pi}xy\sin y\,dy\,dx\]

Hope you can continue. :)

Kind Regards,
Sudharaka.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
7K
  • · Replies 236 ·
8
Replies
236
Views
12K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K