## 85 Reputation

8 years, 209 days

## Anybody can help me...

Anybody can help me?

Best

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.

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

## My example...

Hello,

My example was useful for you?

Thanks

## a clear example....

Thank you so much.

I provided a clear example.

If you have other question let me know.

Thanks so much

example2.pdf

## inert form.....

How I can convert it to the inert format?

I used:  Format>Convert To>Inert Form

but I could not gain inert form..!!

Thanks

## First my problem...

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

## other code is working!!!...

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

## Differentiation...

TEMPThanks

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

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

TEMPThanks

subs.mw

Diff.pdf

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

Thanks

## loop [for]...

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

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

## Errors have been removed....

Thank

Errors have been removed.

I want to calculated the parameter L.

I provide an example. Please see it.

Thanks

## Error...

May You help me how I can remove this error?

Thanks

## I understand...

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

## Error, a row and column Vector cannot be...

Thank you. please see again attached file.

Best.

newmar_(3).mw

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