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