matlab triple integral conical gravity
조회 수: 2 (최근 30일)
이전 댓글 표시
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??
댓글 수: 2
채택된 답변
Mike Hosea
2014년 11월 29일
편집: Mike Hosea
2014년 11월 30일
Numerical stuff removed since a symbolic answer was needed.
댓글 수: 5
Mike Hosea
2014년 11월 30일
편집: Mike Hosea
2014년 11월 30일
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.
추가 답변 (2개)
Youssef Khmou
2014년 11월 29일
편집: Youssef Khmou
2014년 11월 29일
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
2014년 11월 30일
편집: Roger Stafford
2014년 11월 30일
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.
댓글 수: 0
참고 항목
카테고리
Help Center 및 File Exchange에서 Assumptions에 대해 자세히 알아보기
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!