Anybody can help me?

Best

@vv 

Yes lambda, mu, J2 etc are constants.
 the circled triple integral means is not important. It shows integral on the volume.

V indicates that integral is taken on the volume.

the change of variables is used for simplification only and it is not an important issue.

the quantities  Ji , miu, lambda, omega, rho, A,B,C are constants.

the assumption (4) is used and then substituting in the Eq. 1, but you consider only Eq

7 as a final equation and take differential with respect to the constants A,B,C.

Aijk and  Bijk and  Cijk are unique? Are

but Jijk for example is a function.see Eq. 4. They are functions of (sigma, theta, phi) and are differentiable.

Please see pages 2 and 3 of attached file.

Tnanks so much for your attention and your help.

%28ASCE%29EM%2E1943-7889%2E0001115.pdf

@vv 

Hello,

My example was useful for you?

I am waiting to receive your response.

Thanks

@vv 

Thank you so much.

I provided a clear example.

If you have other question let me know.

Thanks so much

example2.pdf

@Joe Riel 

How I can convert it to the inert format?

I used:  Format>Convert To>Inert Form 

but I could not gain inert form..!!

Thanks

@vv 

Thanks .

First my problem is about  not working [subs] rule, as I explained in previous  response.

Can you help me in this regard?

After i should use 

https://www.mapleprimes.com/questions/208952-How-To-Take-Derivative-Of--Sum#answer224355

@Joe Riel 

Thanks.

Please see two attached files below.

One of them is working. I need to get a result that shown in  Eq. 8 in maple file (name is work).

so, by considering the same procedure in maple file (subs 2) I can not gain same behavior in Eq. 5?

what is the problem. also, as you mentioned I used Int and Sum rule but they don not work as well

Thanks

subs2.mwWORK.mw

TEMPThanks

But yet subs do not work well and U functions are not replaced wif series form.

Please see equation 5.

For Differentiation I need a result as shown in teh attached figure.

TEMPThanks

subs.mw

Diff.pdf

@tomleslie 

Do you have another suggestion for calculating Gradient(V)?

Thanks

@Carl Love 

Thanks. It is working, but when n in the loop [for] increase, the error 'Error, Matrix index out of range
' is appear. This procedure should be repeated, but the size of matrix A is always is 1.

How I can solve this problem?

Thanks 

@tomleslie 

Thanks so much.

It is very interesting that this package is not able to calculate the gradient of the vector V!

Please note that for curvilinear coordinates the gradient is not exactly equal to transpose of  its Jacobian, due to this point that base vectors or  Scale factors are different from Cartesian one. 

see attached file. As is shown the base vectors (Scale factors) i.e., (h_r,h_sigma,h_p-hi) appear in gradient result.

see Scale factors at 

https://en.wikipedia.org/wiki/Bipolar_coordinates

gradient.pdf

@acer 

Thank

Errors have been removed.

I want to calculated the parameter L.

I provide an example. Please see it.

Thanks

FOR3.mwEXAMPLE.pdf

@acer 

May You help me how I can remove this error?

Thanks

@Carl Love 

I understand the meaning of the error message. But I use and write only Eq (18). So, I do not how I can remove this error.

newm-furmul.pdf

@Carl Love 

Thank you. please see again attached file.

Best.

newmar_(3).mw