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1-Visitor
July 7, 2015
Question

Function dependent on two variables Help

  • July 7, 2015
  • 4 replies
  • 4060 views

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.

4 replies

23-Emerald V
July 7, 2015

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

24-Ruby III
July 9, 2015

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

19-Tanzanite
July 8, 2015

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

24-Ruby III
July 9, 2015

Alan,

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

1-Visitor
July 9, 2015

Whats the matter here?

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

LT

23-Emerald V
July 9, 2015

Leopold Turek wrote:

Whats the matter here?

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

LT

You are.  

arivera-21-VisitorAuthor
1-Visitor
July 9, 2015

That was it, thank you very much.