Divergence theorem for vector functions

In summary, the conversation discusses the satisfaction of divergence theorem conditions for both Surface S and 3D space E, as well as the properties of a scalar function with continuous partials. The individual must prove a unique integral involving a double integral and a triple integral, but is struggling with the concept of the surface integral and its relation to the volume integral. They are advised to use the divergence theorem and move the constant vector out of the integrals. The conversation also touches on the regular divergence theorem and its connection to Archimedes' principle. The conversation concludes with a discussion
  • #1
laplacianZero
23
1
Thread moved from technical forum, hence the Homework Template is missing.
Surface S and 3D space E both satisfy divergence theorem conditions.

Function f is scalar with continuous partials.

I must prove

Double integral of f DS in normal direction = triple integral gradient f times dV

Surface S is not defined by a picture nor with an equation.

Help me. I don't know how to start this unique integral.
 
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  • #2
Consider the vector field [itex]\mathbf{c}f[/itex] where [itex]\mathbf{c}[/itex] is an arbitrary constant vector.
 
  • #3
Yes. I tried it. I don't get anywhere after considering vector c is a constant vector.
 
  • #4
laplacianZero said:
Yes. I tried it. I don't get anywhere after considering vector c is a constant vector.
Pull it inside of the derivatives and integrals and use the divergence theorem.
 
  • #5
Divergence of constant vector is obviously zero.

The surface integral is ambiguous due to no picture or function that defines the surface.
 
  • #6
laplacianZero said:
Divergence of constant vector is obviously zero.

The surface integral is ambiguous due to no picture or function that defines the surface.
You do not need a picture. It is the boundary surface of the volume that the volume integral is over.
 
  • #7
How can the infinite sum of scalar times the surface areas equal the infinite sum of scalar times their volumes?
 
  • #8
Also,

Closed double integral of f times constant vector equation C dot dS in normal direction

=

Triple integral of divergence of f times constant vector C dV

Produces triple integral of gradient f dot constant vector C dV on the right side only.

I can't seem to do the surface integral.
 
  • #9
laplacianZero said:
How can the infinite sum of scalar times the surface areas equal the infinite sum of scalar times their volumes?
It is not what you have, you have
$$
\oint_{\partial V} f\, d\vec S = \int_V \nabla f \, dV.
$$
This is the integral of ##f## over the surface that is equal to the integral of the gradient of ##f## over the volume.

laplacianZero said:
I can't seem to do the surface integral.
You don't need to. You just need to show that they are equal.

laplacianZero said:
Triple integral of divergence of f times constant vector C dV
Please use the LaTeX feature. It is very difficult to read mathematical formulae in text.

laplacianZero said:
Closed double integral of f times constant vector equation C dot dS in normal direction
laplacianZero said:
Produces triple integral of gradient f dot constant vector C dV on the right side only.

Yes, so move the constant vector out of the integrals.
 
  • #10
Orodruin said:
It is not what you have, you have
$$
\oint_{\partial V} f\, d\vec S = \int_V \nabla f \, dV.
$$
This is the integral of ##f## over the surface that is equal to the integral of the gradient of ##f## over the volume.You don't need to. You just need to show that they are equal.Please use the LaTeX feature. It is very difficult to read mathematical formulae in text.

Yes, so move the constant vector out of the integrals.

Ok. So how can the infinite sum of scalar times surface areas equal the infinite sum of gradient of the scalar times the volumes?
 
  • #11
laplacianZero said:
Ok. So how can the infinite sum of scalar times surface areas equal the infinite sum of gradient of the scalar times the volumes?
This is what you are supposed to show!

Do you have the same problem with the regular divergence theorem?

It is really no stranger than an integral of a function being equal to the difference between the values of the primitive functions at the end-points.
 
  • #12
I understand the divergence theorem quite well. From 2 closed line integrals to the closed surface integral of the curl of the function to the triple integral of the divergence of the curl of that original function while putting emphasis on zero... I understand divergence theorem very well when I can rid the vectors from the dot product and the divergence Operator.

Here is my problem.

Let's start off with vector function equals scalar f times constant vector.

You told me to put this vector function into the divergence theorem.

Is this correct?
 
  • #13
laplacianZero said:
Let's start off with vector function equals scalar f times constant vector.

You told me to put this vector function into the divergence theorem.

Is this correct?
Please start using the math markup available, see LaTeX Primer. There really is no way for me to tell whether you have understood my meaning or not when you use English instead of math.
 
  • #14
Orodruin said:
This is what you are supposed to show!

Do you have the same problem with the regular divergence theorem?

It is really no stranger than an integral of a function being equal to the difference between the values of the primitive functions at the end-points.

The divergence theorem states that the flux of the vector field through the surface is equal to the divergence of the vector field throughout the volume. So, no I do not have the same problem with the divergence theorem
 
  • #15
laplacianZero said:
The divergence theorem states that the flux of the vector field through the surface is equal to the divergence of the vector field throughout the volume. So, no I do not have the same problem with the divergence theorem

And this one says that a scalar quantity integrated with the surface element is equal to the volume integral of the gradient. It is basically Archimedes' principle. Also, it follows directly from the divergence theorem if you go through the steps that have been outlined above.
 
  • #16
If we don't place Archimedes principle on this equality... there really is no meaning.
 
  • #17
laplacianZero said:
If we don't place Archimedes principle on this equality... there really is no meaning.
What do you mean "no meaning"? It is a mathematical identity. What you do with it and what it can be used to model is a different thing - but it definitely has mathematical meaning.
 
  • #18
The surface integral

Scalar function times vector dS

Does NOT make sense.

Furthermore, Volume integral of gradient of the scalar function times dV makes no sense either.

Equating these two integrals to each other just does not produce meaning as there is clear meaning of the divergence theorem.

"Total Flux of vector field through the total closed surface is equal to the divergence of the vector field times the volume of enclosed Space"
 
  • #19
How do I derive this vector integral from the simple divergence theorem?

I seem to lose the vector if I start off with

Scalar function times arbitrary constant Vector V as my starting vector field.
 
  • #20
laplacianZero said:
Scalar function times vector dS

Does NOT make sense.
This is just wrong. It is a mathematical statement. Whether it makes physical sense to describe something is a different matter. I have already provided you with cases where integrals of this type does make physical sense.
laplacianZero said:
Furthermore, Volume integral of gradient of the scalar function times dV makes no sense either.
Of course it does, it is just an integral of a vector field - the result is a vector. You cannot just blatantly proclaim that it does not make sense.

Also, please use mathematical notation instead of words. See:
Orodruin said:
Please start using the math markup available, see LaTeX Primer. There really is no way for me to tell whether you have understood my meaning or not when you use English instead of math.
 
  • #21
This thread leads nowhere and participants seem to get optionally angry, bored or desperate.
Thread closed.
 
  • Like
Likes jedishrfu

1. What is the divergence theorem for vector functions?

The divergence theorem for vector functions, also known as Gauss's theorem, states that the flux of a vector field through a closed surface is equal to the volume integral of the divergence of that field over the enclosed volume.

2. Why is the divergence theorem important?

The divergence theorem is important because it provides a way to calculate the flux of a vector field over a closed surface without having to explicitly integrate the field over the surface. This makes it a powerful tool in many areas of physics and engineering, such as fluid dynamics and electromagnetism.

3. What are some real-world applications of the divergence theorem?

The divergence theorem has numerous real-world applications, including calculating the flow of fluids through pipes and determining the electric field around a charged object. It is also used in the study of weather patterns and in the analysis of fluid dynamics in engines and turbines.

4. How is the divergence theorem related to other theorems in vector calculus?

The divergence theorem is closely related to two other important theorems in vector calculus: Green's theorem and Stokes' theorem. Green's theorem relates the circulation of a vector field around a closed curve to the double integral of the curl of that field over the enclosed area. Stokes' theorem extends this concept to surfaces in three-dimensional space, relating the circulation of a vector field around a closed surface to the triple integral of the curl of that field over the enclosed volume.

5. Are there any limitations to the divergence theorem?

The divergence theorem is only applicable to continuous vector fields and closed surfaces. It also assumes that the vector field and surface are well-behaved, meaning that they do not have any singularities or sharp corners. In addition, the divergence theorem is only valid in three-dimensional space and cannot be extended to higher dimensions.

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