How do I determine if the flux is positive or negative?

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SUMMARY

This discussion focuses on determining the sign of flux in vector calculus, particularly in relation to surface integrals and the orientation of surfaces like paraboloids. Participants clarify that the sign of the flux depends on the direction of the normal vector obtained from the cross product of the parameterization vectors, Ru and Rv. A negative flux result is acceptable if the orientation is correctly defined, especially when dealing with nonnegative integrands in triple integrals. The importance of checking the limits and direction of the normal vector is emphasized to ensure accurate calculations.

PREREQUISITES
  • Understanding of vector calculus concepts, specifically flux and surface integrals.
  • Familiarity with parameterization of surfaces using vectors Ru and Rv.
  • Knowledge of cross product operations and their geometric interpretations.
  • Ability to evaluate triple integrals and recognize the significance of integrand signs.
NEXT STEPS
  • Study the properties of the cross product in vector calculus.
  • Learn about surface parameterization techniques for different geometric shapes.
  • Explore the implications of orientation in surface integrals and flux calculations.
  • Review examples of triple integrals with nonnegative integrands to solidify understanding.
USEFUL FOR

Students and educators in mathematics, particularly those studying vector calculus, as well as professionals involved in fields requiring knowledge of surface integrals and flux analysis.

mmmboh
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Hi I have an assignment due tomorrow, and have answered all the questions, I am just unsure about positive or negative flux
dcby91.jpg


I know 5b) is positive, I think 6 is positive, and I'm guessing the other two are positive, but can anyone confirm?

And if I did get the sign wrong, would I have to redo the cross product, or can I just say at the end for instance that the flux is positive and change the sign from a negative to positive?

Also
2u7ui5s.jpg


I got a negative answer, is that ok? or should I just say it's positive and change the sign to a positive one?

Thanks!
 
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Help please it's for tomorrow
 
anyone?
 
If you make the graphs, I will work on it but I don't want to make them.
 
33df6nl.jpg


Those are my graphs, I know they suck...

This is due tomorrow morning, help would be amazing really! and for the last one I posted, is it just a volume so it should be positive? I am a little confused. I solved them all though.
 
When you calculate Ru X Rv to get the normal to the surface, you should check at that step whether it agrees with the direction specified as positive. That's where you should adjust the sign.

For the triple integral in your last problem it is not OK to get a negative answer since your integrand is nonnegative. Check that your limits are correct and in the positive direction through the volume.

I agree your graphs are awful, the "disk" and "cone" don't pass the laugh test. :smile:
 
Ok thanks :) Are you able to confirm whether the others are in fact positive values?
 
Or how do I check at Ru X Rv whether I have the right direction?
 
mmmboh said:
Or how do I check at Ru X Rv whether I have the right direction?

Depends on the problem. For example, if your flux is directed generally upwards you would look for Ru X Rv to have a positive z component and, if it didn't, use the opposite cross product.

As to whether the other answers should be positive, I didn't work them. But if your flux is in the radial direction like it is, you can probably tell whether it is going through your surface in the positive orientation direction or not. At least you can if your graphs bear some semblance to reality.
 
  • #10
So for number 5a after I do Ru X Rv I get -2x-2y+1...so that is in the positive z direction right...and my final answer is -pi/2...assuming my work is correct this makes sense?
 
  • #11
mmmboh said:
So for number 5a after I do Ru X Rv I get -2x-2y+1

That isn't a vector. Do you mean <-2, -2, 1>?

...so that is in the positive z direction right...and my final answer is -pi/2...assuming my work is correct this makes sense?

Given that the orientation is upward, which is towards the inside of the paraboloid, and that the radial direction is towards the outside of the paraboloid, negative makes sense to me. Does it to you?
 
  • #12
yes i meant -2i-2j+k...

Um well if the paraboloid looks like how I drew it, wouldn't an upward flux be leaving the paraboloid...unless I drew it wrong.
 
  • #13
For number 6, after I do the cross, I get (-ucosv)i+(-usinv)j+uk...does this mean it is in the positive y-direction? because u can be negative right...my final answer ends up being -13pi/8 for that one..
 
  • #14
mmmboh said:
yes i meant -2i-2j+k...

Um well if the paraboloid looks like how I drew it, wouldn't an upward flux be leaving the paraboloid...unless I drew it wrong.

Yes, it is leaving the paraboloid. But what about the orientation of the paraboloid? The paraboloid is oriented upwards. The "top" or up side of the paraboloid is the inside. The flux is not going towards the positive orientation of the surface, even though it is heading generally upwards.
 
  • #15
Gotta run. Good luck.
 

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