Re: Applying the Integration Function to a List Of Regions
- To: mathgroup at smc.vnet.net
- Subject: [mg88758] Re: [mg88739] Applying the Integration Function to a List Of Regions
- From: János <janos.lobb at yale.edu>
- Date: Fri, 16 May 2008 05:29:01 -0400 (EDT)
- References: <200805151051.GAA21829@smc.vnet.net>
On May 15, 2008, at 6:51 AM, John Snyder wrote:
> Assume that I have already determined a list of 4 dimensional
> regions as
> follows:
>
> regions={{{x,0,a},{cx,0,a+x},{y,0,Sqrt[a^2-cx^2+2 cx
> x-x^2]},{cy,0,Sqrt[a^2-cx^2+2 cx
> x-x^2]+y}},{{x,0,a},{cx,0,a+x},{y,Sqrt[a^2-cx^2+2 cx x-x^2],2
> a-Sqrt[a^2-cx^2+2 cx x-x^2]},{cy,-Sqrt[a^2-cx^2+2 cx
> x-x^2]+y,Sqrt[a^2-cx^2+2 cx x-x^2]+y}},{{x,0,a},{cx,0,a+x},{y,2
> a-Sqrt[a^2-cx^2+2 cx x-x^2],2 a},{cy,-Sqrt[a^2-cx^2+2 cx x-x^2]+y,2
> a}},{{x,a,2 a},{cx,-a+x,2 a},{y,0,Sqrt[a^2-cx^2+2 cx
> x-x^2]},{cy,0,Sqrt[a^2-cx^2+2 cx x-x^2]+y}},{{x,a,2 a},{cx,-a+x,2
> a},{y,Sqrt[a^2-cx^2+2 cx x-x^2],2 a-Sqrt[a^2-cx^2+2 cx
> x-x^2]},{cy,-Sqrt[a^2-cx^2+2 cx x-x^2]+y,Sqrt[a^2-cx^2+2 cx
> x-x^2]+y}},{{x,a,2 a},{cx,-a+x,2 a},{y,2 a-Sqrt[a^2-cx^2+2 cx x-x^2],2
> a},{cy,-Sqrt[a^2-cx^2+2 cx x-x^2]+y,2 a}}};
>
> I want to integrate over each of these regions using an integrand
> of 1. I
> want my output to be as follows:
>
> {Integrate[1,{x,0,a},{cx,0,a+x},{y,0,Sqrt[a^2-cx^2+2 cx
> x-x^2]},{cy,0,Sqrt[a^2-cx^2+2 cx
> x-x^2]+y}],Integrate[1,{x,0,a},{cx,0,a+x},{y,Sqrt[a^2-cx^2+2 cx x-
> x^2],2
> a-Sqrt[a^2-cx^2+2 cx x-x^2]},{cy,-Sqrt[a^2-cx^2+2 cx
> x-x^2]+y,Sqrt[a^2-cx^2+2 cx x-x^2]+y}],Integrate[1,{x,0,a},{cx,0,a
> +x},{y,2
> a-Sqrt[a^2-cx^2+2 cx x-x^2],2 a},{cy,-Sqrt[a^2-cx^2+2 cx x-x^2]+y,2
> a}],Integrate[1,{x,a,2 a},{cx,-a+x,2 a},{y,0,Sqrt[a^2-cx^2+2 cx
> x-x^2]},{cy,0,Sqrt[a^2-cx^2+2 cx x-x^2]+y}],Integrate[1,{x,a,2 a},
> {cx,-a+x,2
> a},{y,Sqrt[a^2-cx^2+2 cx x-x^2],2 a-Sqrt[a^2-cx^2+2 cx
> x-x^2]},{cy,-Sqrt[a^2-cx^2+2 cx x-x^2]+y,Sqrt[a^2-cx^2+2 cx
> x-x^2]+y}],Integrate[1,{x,a,2 a},{cx,-a+x,2 a},{y,2 a-Sqrt[a^2-cx^2
> +2 cx
> x-x^2],2 a},{cy,-Sqrt[a^2-cx^2+2 cx x-x^2]+y,2 a}]}
>
> How can I do that without having to set up each of the integrals
> manually?
> I am looking for some way to do something like:
>
> Integrate @@ regions
>
> or
>
> Integrate @@@ regions
>
> But I can't figure out how to incorporate the 1 as the integrand
> when I try
> to set this up automatically.
>
> There must be a way?
>
> Thanks,
>
> John
>
Me as a newbie, I would try a variant of this to start with:
In[2]:=
(Integrate[Sequence[1],
#1] & ) /@ regions
But to tell the truth if I see an Integrate I duck for cover.
J=E1nos
------
"..because Annushka has already bought sunflower oil, and not only
bought it, but spilled it too."
Bulgakov: Master and Margarita
- References:
- Applying the Integration Function to a List Of Regions
- From: "John Snyder" <jsnyder@wi.rr.com>
- Applying the Integration Function to a List Of Regions