matlab triple integral conical gravity
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Hi everyone,
I'm trying to solve a triple integral in matlab, demonstrating the gravity on a point mass inside a cone. I have solved this byy hand and it works fine with a simple u sub. Does anyone have any ideas why my code isn't working. Thanks!
I've tried:
function F = conical_gravity(r,z,th) % parameters
syms G p r th z h
T = (p*G*r*z)/((r^2+z^2)^(3/2));
F1 = int(T,r,0,z)
F2 = int(F1,z,0,h)
F3 = int(F2,th,0,2*pi)
end
this:
syms G p r z th h a
T = (p*G*r*z)*((r^2+z^2)^(-3/2));
q1 = int(T,r,0,z)
q2 = int(q1,th,0,2*pi)
q3 = int(q2,z,0,h)
and this:
syms th r z h G p
T = (p*G*r*z)*((r^2+z^2)^(-3/2));
int(int(int(T,r,0,z),th,0,2*pi),z,0,h)
along with the previous was using integral3.
Any ideas??
2 Comments
Accepted Answer
Mike Hosea
on 29 Nov 2014
Edited: Mike Hosea
on 30 Nov 2014
Numerical stuff removed since a symbolic answer was needed.
5 Comments
Mike Hosea
on 30 Nov 2014
Edited: Mike Hosea
on 30 Nov 2014
BTW, I made a mistake before in sorting out the variables when I did the numerical approach. The symbolic approach is, of course, superior in this case, but for your edification, for problems that can't be integrated symbolically, here is what one approach using INTEGRAL3 looks like.
function F = conical_gravity(h,p,G)
% h is a variable.
% p and G are parameters.
T = @(th,z,r)(p.*G.*r.*z) ./ ((r.^2 + z.^2).^(3/2));
Fscalar = @(h)integral3(T,0,2*pi,0,h,0,@(th,z)z);
F = arrayfun(Fscalar,h);
This doesn't work when h=0, unfortunately, because the integrator ends up trying to plug in zeros for both r and z into T.
More Answers (2)
Youssef Khmou
on 29 Nov 2014
Edited: Youssef Khmou
on 29 Nov 2014
I think that working with symbolic variables will not permit the transformation of integral expressions to numeric type, however if you only want general primitive do not use the bounds :
q1 = int(T,r)
You can proceed as the following, the second integral is based on first, so as the third, in each integral the bold case represents the variable on which we integrate :
G=6.67e-11;
z=2;
p=2;% p=mv
r=4;
T=@(R) (p*G*R*z)/((r^2+z^2)^(3/2));
F1=quad(T,0,z);
the expression of q2 is independent of azimuth theta, you will then multiply q1 by 2pi, try to figure out the solution q3.
Roger Stafford
on 30 Nov 2014
Edited: Roger Stafford
on 30 Nov 2014
I have a very ancient version of the Symbolic Toolbox, but it has trouble with substituting z for (z^2)^(1/2) if you integrate with respect to r first because it doesn't know that z is never negative until too late. Unfortunately you would run into the same kind of trouble if you integrated with respect to z first, because it doesn't know yet that r is never negative. However, I believe later versions will permit you to place constraints on your symbolic variables to avoid this trouble. You might try that.
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