Symbolic integration has 3 solutions based on integration variable range, how to extract one of these?

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Jason Oakley
Jason Oakley 2024-2-17
评论： Jason Oakley 2024-2-17
r is the integration variable. The integration is:
mom_2 = int((umax*(1-r/Radius)^(1/7))^2*2*pi*r,r,0,Radius)
and the result is:
size(mom_2) = 1 1
Question: how do I access each of these three possible solutions?
For example, I would like to use (50/49)*pi*U_0^2 in further calculations. Thanks ahead of time!

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采纳的回答

Paul
Paul 2024-2-17
Ran in:
syms U_0 r Radius
mom_2 = int((U_0*(1-r/Radius)^(1/7))^2*2*sym(pi)*r,r,0,Radius)
mom_2 = 
One approach that just extracts the case you want
c = children(mom_2)
c = 3×2 cell array
{[(49*pi*U_0^2)/72 ]} {[Radius == 1]} {[(49*pi*Radius^2*U_0^2)/72 ]} {[0 < Radius ]} {[2*U_0^2*pi*int(r*(1 - r/Radius)^(2/7), r, 0, Radius)]} {[~0 < Radius]}
case1 = c{1,1}
case1 = 

更多回答（1 个）

John D'Errico
John D'Errico 2024-2-17
Ran in:
Or do this:
syms r umax Radius
mom_2 = int((umax*(1-r/Radius)^(1/7))^2*2*pi*r,r,0,Radius)
mom_2 = 
subs(mom_2,Radius,1)
ans = 
Note that it resolves the three cases into 1.
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Paul
Paul 2024-2-17
Ran in:
It's usually better to use assume before calling int to give it some help. In this case, if R>0, we'd try
syms U_0 r Radius
assume(Radius,'positive')
mom_2 = int((U_0*(1-r/Radius)^(1/7))^2*2*sym(pi)*r,r,0,Radius)
mom_2 = 
Not sure why that didn't work. Instead, we can do
syms rho
assume(rho,'positive')
mom_2 = int((U_0*(1-r/rho)^(1/7))^2*2*sym(pi)*r,r,0,Radius)
mom_2 = 
mom_2 = subs(mom_2,rho,Radius)
mom_2 = 

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