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1-Visitor
Jul 7, 2015 8:54 PM UTC
Question

Function dependent on two variables Help

I am trying to replicate something done in Mathcad 15.0 in Prime but I am having difficulties.

I would like to double integrate two functions, but can't get a function dependent on two variable to work.

I have attached my best try, and a pdf of how it was done in Mathcad 15.0. I can't get the n(x) to work in the gamma,h1(x) function. I am able to skirt this for a while but never get my integration to work. Please see attached and thank you in advance for your help. AlanStevens‌ I feel like you would have the answer looking at one of your other posts.

7 replies

23-Emerald V
Jul 7, 2015 11:23 PM UTC

Andres Rivera wrote:

I am trying to replicate something done in Mathcad 15.0 in Prime but I am having difficulties.

I would like to double integrate two functions, but can't get a function dependent on two variable to work.

I have attached my best try, and a pdf of how it was done in Mathcad 15.0. I can't get the n(x) to work in the gamma,h1(x) function. I am able to skirt this for a while but never get my integration to work. Please see attached and thank you in advance for your help. AlanStevens I feel like you would have the answer looking at one of your other posts.

It looks like you should have σ'h(n,θ) rather than just σ'h

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24-Ruby IV
Jul 9, 2015 9:12 AM UTC

Basically the attached archive has PDF file with the listing of the program, which is how it is written.

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19-Tanzanite
Jul 8, 2015 7:58 AM UTC

Andres Rivera wrote:

AlanStevens I feel like you would have the answer looking at one of your other posts.

Unfortunately, I don't have Prime 3, so I'm no use to you here.

Alan

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24-Ruby IV
Jul 9, 2015 9:06 AM UTC

Alan,

Please find in attachment PDF and XPS version of this Mathcad Prime worksheet.

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1-Visitor
Jul 9, 2015 2:25 PM UTC

Whats the matter here?

Is nobody listening to what Stuart said at the very beginning?

LT

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23-Emerald V
Jul 9, 2015 2:52 PM UTC

Leopold Turek wrote:

Whats the matter here?

Is nobody listening to what Stuart said at the very beginning?

LT

You are.  

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arivera-21-VisitorAuthor
1-Visitor
Jul 9, 2015 3:16 PM UTC

That was it, thank you very much.

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