matlab语言编程w0=1u0=0.4u=[0.7 0.4 0.1]v0=0.2v=[0.1 0.3 0.2]i=3j

matlab语言编程
w0=1
u0=0.4
u=[0.7 0.4 0.1]
v0=0.2
v=[0.1 0.3 0.2]
i=3
j=mod(i,3)+1
syms ux vy
R=sqrt(((u0-ux)^2)+((v0-vy)^2)+(w0)^2)
f=(3*((u0-ux)^2)/(R^5))-1/(R^3)
I=int(int(f,vy,((v(3)-v(2))/(u(3)-u(2)))*ux+v(2)-((v(3)-v(2))/(u(3)-u(2)))*u(2),((v(1)-v(2))/(u(1)-u(2)))*ux+v(1)-((v(1)-v(2))/(u(1)-u(2)))*u(2)),ux,u(2),u(1))+int(int(f,vy,((v(3)-v(2))/(u(3)-u(2)))*ux+v(2)-((v(3)-v(2))/(u(3)-u(2)))*u(2),((v(1)-v(3))/(u(1)-u(3)))*ux+v(1)-((v(1)-v(3))/(u(1)-u(3)))*u(1)),ux,u(1),u(3))
w0=0
u0=0.4
u=[0.7 0.4 0.1]
v0=0.2
v=[0.1 0.3 0.2]
i=3
j=mod(i,3)+1
a(i)=(u(i)-u0)*(v(j)-v(i))-(v(i)-v0)*(u(j)-u(i))
b(i)=(u(j)-u(i))*(u(i)-u0)+(v(j)-v(i))*(v(i)-v0)
c(i)=(u(j)-u(i))*(u(j)-u0)+(v(j)-v(i))*(v(j)-v0)
l(i)=sqrt((u(j)-u(i))^2+(v(j)-v(i))^2)
r(i)=sqrt((u0-u(i))^2+(v0-v(i))^2+w0^2)
R(j)=sqrt((u0-u(j))^2+(v0-v(j))^2+w0^2)
%g(j)=sqrt((l(i))^2+2*((b(i))^2)+(r(i))^2)
s1(i)=(u(j)-u(i))*(v(j)-v(i))*(r(i)-R(j))/(r(i)*R(j)*((l(i))^2))
s2(i)=a(i)*(c(i)*r(i)-b(i)*R(j))/(r(i)*R(j)*(((a(i))^2)+((l(i))^2)*(w0^2)))
t1(i)=(-(u(i)-u0)*a(i)-(v(j)-v(i))*w0^2)/((u(j)-u(i))*r(i)*w0)
t2(i)=(-(u(j)-u0)*a(i)-(v(j)-v(i))*w0^2)/((u(j)-u(i))*R(j)*w0)
t3(i)=(-(v(i)-v0)*a(i)+(u(j)-u(i))*w0^2)/((v(j)-v(i))*r(i)*w0)
t4(i)=(-(v(j)-v0)*a(i)+(u(j)-u(i))*w0^2)/((v(j)-v(i))*R(j)*w0)
d(i)=sqrt((u0-u(i))^2+(v0-v(i))^2)
h1(i)=((u(j)-u(i))*(v(j)-v(i))*(d(i)-d(j)))/(d(i)*d(j)*(l(i))^2)
h2(i)=(c(i)*r(i)-b(i)*r(j))/(a(i)*d(i)*d(j))
%f1(i)=(1/w0)*(atan(t1(i))-atan(t2(i)))+(2/(3*w0))*(-atan(t3(i))+atan(t4(i)))+(1/3)*(s1(i)-(s2(i)/l(i))*((v(j)-v(i))^2))
%f_1=f1(1)+f1(2)+f1(3)
%f9(i)=(1/(2*w0))*(atan(t1(i))-atan(t2(i))+atan(t3(i))-atan(t4(i)))
%f_9=f9(1)+f9(2)+f9(3)
IL1(i)=h1(i)-(h2(i)/l(i))*((v(j)-v(i))^2)
I_L1=IL1(1)+IL1(2)+IL1(3)
%I_1=3*f_1-f_9
在这上面还要再添加上上面东西可以实现这个功能 就是对这个程序还要怎么对其进行处理
麻烦知道的速度解决下还好.
这是两种积分方法的比较问题

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