Is this surface integral correct?

In summary, the conversation discusses a problem with a specific equation, where the author has made an error in the simplification of a term. The group discusses the correct form of the equation and confirms that the missing term was the cause of the error. The issue is resolved and the author thanks the group for clarifying the mistake.
  • #1
Apashanka
429
15
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
 
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  • #2
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
$${\bf r} (\nabla \cdot {\bf u})-({\bf r} \cdot \nabla) {\bf u} -2 {\bf u}=\nabla×({\bf r} × {\bf u})$$
 
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  • #3
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
 
  • #4
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.
 
  • #5
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
 
  • #6
Apashanka said:
Sorry I didn't get you
Oh yes sir got it ...silly mistake
...thanks
 

What is a surface integral?

A surface integral is a mathematical concept used in multivariable calculus to calculate the area of a surface in three-dimensional space. It involves integrating a function over a given surface to determine the total value of that function over the surface.

How do I know if a surface integral is correct?

To determine if a surface integral is correct, you can use various methods such as checking for consistency with known results, using symmetry or divergence theorems, or verifying the integral using other techniques such as parametrization or Green's theorem.

What are the common mistakes when calculating a surface integral?

Some common mistakes when calculating a surface integral include incorrect parametrization of the surface, using the wrong limits of integration, and not properly accounting for the direction of the surface normal vector. It is important to carefully check all steps of the calculation to avoid these errors.

Can I use a calculator to evaluate a surface integral?

Yes, you can use a calculator or computer software to evaluate a surface integral. However, it is important to understand the underlying mathematical concepts and principles in order to use these tools effectively and verify the accuracy of the results.

How can I apply surface integrals in real-world situations?

Surface integrals have many applications in physics, engineering, and other fields. They can be used to calculate the flux of a vector field, determine the mass or charge distribution over a surface, and solve various optimization problems. Understanding surface integrals can also help in visualizing and analyzing three-dimensional objects and phenomena.

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