matlab实现曲线积分(path_integral)
MATLAB语言并未直接提供曲线积分和曲面积分的现成函数。以下是曲线积分函数:
function I = path_integral(F,vars,t,a,b) %path_integral %第一类曲线积分 % I = path_integral(f, [x,y], t, t_m, t_M) % I = path_integral(f, [x,y,z], t, t_m, t_M) % Examples: % 计算int_l(z^2/(x^2+y^2))ds, l是如下定义的螺线 % x=acost, y=asint, z=at, 0<=t<=2*pi, a>0 % MATLAB求解语句 % syms t; syms a positive; % x=a*cos(t); y=a*sin(t); z=a*t; % f=z^2/(x^2+y^2); % I=path_integral(f,[x,y,z],t,0,2*pi) % %第二类曲线积分 % I = path_integral([P,Q], [x,y], t, a, b) % I = path_integral([P,Q,R], [x,y,z], t, a, b) % I = path_integral(F, v, t, a, b) % Examples: % 曲线积分int_l( (x+y)/(x^2+y^2)*dx - (x-y)/(x^2+y^2)*dy ), % l为正向圆周x^2+y^2=a^2 % 正向圆周的参数函数描述: x=acost, y=asint, (0<=t<=2pi) % MATLAB求解语句 % syms t; syms a positive; % x=a*cos(t); y=a*sin(t); % F=[ (x+y)/(x^2+y^2), -(x-y)/(x^2+y^2) ]; % I=path_integral(F,[x,y],t,2*pi,0) if length(F)==1 I = int(F*sqrt(sum(diff(vars,t).^2)),t,a,b); else F = F(:).'; vars = vars(:); I = int(F*diff(vars,t),t,a,b); end
posted on 2020-08-06 13:21 last_point 阅读(3193) 评论(0) 编辑 收藏 举报