What Is the Role of Divergence in Vector Calculus?

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SUMMARY

The discussion focuses on the role of divergence in vector calculus, particularly in simplifying complex problems involving geometric shapes like spheres and cubes. Participants explore the implications of applying the divergence operation to vector fields, specifically questioning the conditions under which divergence can be calculated. The conversation highlights the importance of understanding divergence in relation to volume changes, with one participant noting that they derived the volume increase as 1 + div(h). This indicates a foundational understanding of how divergence relates to physical interpretations in vector calculus.

PREREQUISITES
  • Understanding of vector calculus concepts, particularly divergence
  • Familiarity with geometric shapes in three-dimensional space
  • Basic knowledge of mathematical notation used in calculus
  • Experience with volume calculations in relation to vector fields
NEXT STEPS
  • Study the definition and properties of divergence in vector fields
  • Learn about the application of divergence in physical contexts, such as fluid dynamics
  • Explore the relationship between divergence and volume changes in vector calculus
  • Investigate the implications of divergence in different geometric configurations, such as spheres and cubes
USEFUL FOR

This discussion is beneficial for students and professionals in mathematics, physics, and engineering who are looking to deepen their understanding of vector calculus, particularly the concept of divergence and its applications in various geometric contexts.

Physgeek64
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Got it. Thank you ;)
 
Last edited:
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Hi there,

Looks a bit like homework, for which we have a nice template !

Is it allowed to simplify this exercise to a sphere with the origin at the center ? Because if it is about a cube with the origin way outside of the cube, things seem complicated to me...

In the simple case ##\vec r ## goes to ## \vec r'= {\rm \ ?} ## so that ## \vec h = \vec r'- \vec r = {\rm \ ?} ## and that you can subject to the divergence operation (from its definition) , I should hope ?
 
BvU said:
Hi there,

Looks a bit like homework, for which we have a nice template !

Is it allowed to simplify this exercise to a sphere with the origin at the center ? Because if it is about a cube with the origin way outside of the cube, things seem complicated to me...

In the simple case ##\vec r ## goes to ## \vec r'= {\rm \ ?} ## so that ## \vec h = \vec r'- \vec r = {\rm \ ?} ## and that you can subject to the divergence operation (from its definition) , I should hope ?

Its okay, I have sorted out my queries. But thank you. I didn't know whether to post this in homework as it is not technically homework, as I am doing it a while in advance so I was hoping more for general background information than for specifics on how to do the question. :)
 
Please don't delete the original post -- it makes the thread incomprehensible !

Was your answer ##3\alpha## ?

What kind of background info did you have in mind ?
 
BvU said:
Please don't delete the original post -- it makes the thread incomprehensible !

Was your answer ##3\alpha## ?

What kind of background info did you have in mind ?
ahh I am sorry- I just don't know how to delete the post.
Kind of- I got the volume increase to be 1+divh, which is the same thing?
 
Physgeek64 said:
ahh I am sorry- I just don't know how to delete the post.
Well, effectively you did by overwriting it. Perhaps you can restore the original ?

Kind of- I got the volume increase to be 1+divh, which is the same thing?
Relative, I assume. Sounds good. Can you underpin it ?
 

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