matlab反复求积分,a=0.006b=0.003;c=0.0005miuf=1000miur=1f1='(x-a)/(((x-a)^2+(y-y1)^2+(z-z1)^2)^(3/2))'f2='(x+a)/(((x+a)^2+(y-y1)^2+(z-z1)^2)^(3/2))'N=miuf/miurM=-500/(4*pi)v1=int(f1,'y1','-b','b')v2=int(f2,'y1','-b','b')B1=int(v1,'z1','-c','c')B2
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matlab反复求积分,a=0.006b=0.003;c=0.0005miuf=1000miur=1f1='(x-a)/(((x-a)^2+(y-y1)^2+(z-z1)^2)^(3/2))'f2='(x+a)/(((x+a)^2+(y-y1)^2+(z-z1)^2)^(3/2))'N=miuf/miurM=-500/(4*pi)v1=int(f1,'y1','-b','b')v2=int(f2,'y1','-b','b')B1=int(v1,'z1','-c','c')B2
matlab反复求积分,
a=0.006
b=0.003;
c=0.0005
miuf=1000
miur=1
f1='(x-a)/(((x-a)^2+(y-y1)^2+(z-z1)^2)^(3/2))'
f2='(x+a)/(((x+a)^2+(y-y1)^2+(z-z1)^2)^(3/2))'
N=miuf/miur
M=-500/(4*pi)
v1=int(f1,'y1','-b','b')
v2=int(f2,'y1','-b','b')
B1=int(v1,'z1','-c','c')
B2=int(v2,'z1','-c','c')
hf1=M*N*B1
hf2=M*N*B2
hf=hf1-hf2
HF=int(int(int(hf,'x','-0.006','0.006'),'y','-0.003','0.003'),'z','-0.0005','0.0005')/(0.012*0.006*0.001)
matlab反复求积分,a=0.006b=0.003;c=0.0005miuf=1000miur=1f1='(x-a)/(((x-a)^2+(y-y1)^2+(z-z1)^2)^(3/2))'f2='(x+a)/(((x+a)^2+(y-y1)^2+(z-z1)^2)^(3/2))'N=miuf/miurM=-500/(4*pi)v1=int(f1,'y1','-b','b')v2=int(f2,'y1','-b','b')B1=int(v1,'z1','-c','c')B2
程序可改成：
a=0.006;b=0.003;
c=0.0005;
miuf=1000;
miur=1;
syms x y z z1 y1
f1=(x-a)/(((x-a)^2+(y-y1)^2+(z-z1)^2)^(3/2));
f2=(x+a)/(((x+a)^2+(y-y1)^2+(z-z1)^2)^(3/2));
N=miuf/miur;
M=-500/(4*pi);
v1=int(f1,y1,-b,b)
v2=int(f2,y1,-b,b)
B1=int(v1,z1,-c,c)
B2=int(v2,z1,-c,c)
hf1=M*N*B1
hf2=M*N*B2
hf=hf1-hf2
HF=int(int(int(hf,x,-0.006,0.006),y,-0.003,0.003),z,-0.0005,0.0005)/(0.012*0.006*0.001)
前面的问题不大,但最后一步的积分值得到0,我有点怀疑.
单独计算最里面的那层积分：
>> int(hf,x,-0.006,0.006)ans =
0
但我仍然觉得这个结果有些可疑.