Is This Surface Integral Correct?

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SUMMARY

The forum discussion centers on the verification of a surface integral related to vector identities in fluid dynamics. The user questions the correctness of the expression derived from the equation A4, specifically regarding the term involving the peculiar velocity vector, u. The identity in question is confirmed as valid, with the user realizing a minor error in their interpretation of the signs in the equation. The discussion concludes with the clarification that the last term should indeed be -2u instead of +2u, resolving the confusion.

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Apashanka
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Homework Statement
Requesting for re check
Relevant Equations
Requesting for re check
Problem Statement: Requesting for re check
Relevant Equations: Requesting for re check

IMG_20190527_112941.jpg

In this eq.A4 putting ##v=Hr+u## the first integrand in eq.A5 is coming as ##H(r(\nabla•u)-(r•\nabla)u+2u)\ne\nabla×(r×u)##
Am I right??
Can I request anyone to please recheck it...
using this the author has put the term ##\int_s(da•\nabla×(r×u))=0##(as closed surface) how then it is coming??
Will anyone please help...
##r## is the position vector and ##u ## is the peculiar velocity
 
Last edited:
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Apashanka said:
In this eq.A4 putting ##v=Hr+u## the first integrand in eq.A5 is coming as ##H(r(\nabla•u)-(r•\nabla)u+2u)\ne\nabla×(r×u)##
Am I right??
Shouldn't the last term on the left be ##-2u## instead of ##+2u##?

I'm not sure of the full meaning and context of the symbols. However, as a vector identity, I believe it is true that
$${\bf r} (\nabla \cdot {\bf u})-({\bf r} \cdot \nabla) {\bf u} -2 {\bf u}=\nabla×({\bf r} × {\bf u})$$
 
Last edited:
TSny said:
Shouldn't the last term on the left be ##-2u## instead of ##+2u##?

I'm not sure of the full meaning and context of the symbols. However, as a vector identity, I believe it is true that
$$r(\nabla•u)-(r•\nabla)u-2u=\nabla×(r×u)$$

Sir the last term is coming as ##2u## instead of ##-2u## e.g putting ##v=Hr+u## in eqA4 we get ##v(\nabla•v)=(Hr+u)(3H+\nabla•u)=3H^2r+Hr(\nabla•u)+3Hu+u(\nabla•u)##
Similarly ##-(v•\nabla)v=-[(Hr+u)•\nabla](Hr+u)=-(H^2r+H(r•\nabla)u+Hu+(u•\nabla)u)##
Similarly ##-\frac{2}{3V^2}[\int_s(da•v)]^2=-\frac{2}{3V^2}[\int_s(da•Hr)]^2-\frac{2}{3V^2}[\int(da•u)]^2=-6H^2-\frac{2}{3V^2}[\int(da•u)]^2##
Now adding these terms of the integrand we rest all term same except ##2Hu## instead of ##-2Hu## for which the author has simplified this to ##\nabla×(r×u)## and put it's surface integral to 0 (assuming closed surface) which is not coming true here...
That's what I want to clarify
 
Apashanka said:
Similarly ##-\frac{2}{3V^2}[\int_s(da•v)]^2=-\frac{2}{3V^2}[\int_s(da•Hr)]^2-\frac{2}{3V^2}[\int(da•u)]^2##
There is a third term that should be included on the right side. Including this missing term should make everything OK.
 
TSny said:
There is a third term that should be included on the right side. Including this missing term should make everything OK.
Sorry I didn't get you
 
Apashanka said:
Sorry I didn't get you
Oh yes sir got it ...silly mistake
...thanks
 

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