Is This Surface Integral Correct?

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Homework Help Overview

The discussion revolves around verifying the correctness of a surface integral related to vector calculus, specifically involving the manipulation of vector identities and integrals in the context of fluid dynamics or similar fields.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are examining the validity of a vector identity and its implications in the context of a surface integral. There are questions about the signs of terms in the equations and whether certain terms have been omitted or misrepresented.

Discussion Status

Some participants are actively questioning the setup and the terms involved in the equations, with attempts to clarify the relationships between the variables. There is an acknowledgment of potential errors in the original formulation, and some guidance has been offered regarding the inclusion of additional terms.

Contextual Notes

Participants note the importance of understanding the vector identities and the implications of assuming a closed surface for the integral. There are indications of confusion regarding the notation and the specific terms used in the equations, which may affect the interpretation of the problem.

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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