XMAX must be a floating point scalar in double integral. Kindly help.

CODE:

x=0
y=0
u0=4*pi*(10^-7);
f=5*10^6;
w=2*pi*f;
u=4*pi*(10^-7);
U=2;
a=0.5
N=5
for  d=0.1:0.1:1
    s=0.001
    N=5
                wd=0.011
        r0=0.0001608
e=7*(8.854*10^-12) %dry sand check again
sig=0.11 %variable so check again
alpha=3.9*10^-3 %temp coeff
T=26 %soil temp
T0=33 %room temp
dout=2*a
 d1=sqrt(2)*a
 aw=wd./2
  din=dout-(2*N*wd)-(2*(N-2)*s)
    davg=(dout-din)./(dout+din)
 ltc=4.*N.*(dout-(N-1).*(wd+s)) %ORIGINAL CORRECT
Aw=pi.*(aw.^2)
R=(r0.*ltc.*(1+alpha.*(T-T0)))./Aw
theta=90
  J=2.*sin(theta).*sin(theta)+cos(theta).*cos(theta)
  sd=(w.*(sqrt((u.*e./2).*(sqrt(1+(sig./(w.*e))^2)-1))))^(-1)
              G=exp(-d./sd)
      Mt1=0
      Bz=@(ai) (((u)/(2*pi*sqrt(((r0+ai)^2)+(d^2))))*((ellipticK(pi/2))+(((-r0^2-d^2+ai^2)/((r0-ai)^2+d^2))*ellipticE(pi/2))))
M=zeros(1,N*N)
for  i=1:1:N
    ai=a-2*(N-1)*wd-2*(N-1)*s:2*(wd+s):a-2*(N-N)*wd-2*(N-N)*s
for  j=1:1:N
    aj=a-2*(N-1)*wd-2*(N-1)*s:2*(wd+s):a-2*(N-N)*wd-2*(N-N)*s
M=integral2(Bz,0,ai,0,aj) 
end
end 
Mt1=Mt1+M
Mtotal=Mtl*J*G
end
        end

ERROR:

Error using integral2 (line 76)

XMAX must be a floating point scalar.

Error in FPSCCLM2 (line 93)

M=integral2(Bz,0,ai,0,aj)

Answer 1

[Walter Roberson]

    ai=a-2*(N-1)*wd-2*(N-1)*s:2*(wd+s):a-2*(N-N)*wd-2*(N-N)*s

Notice the colon operators. Your ai is going to be a vector. Likewise your aj is going to be a vector.

    M=integral2(Bz,0,ai,0,aj)

You are trying to use integral2() passing in a vector of maximum x values in the variable ai, and a vector of maximum y values in the variable aj . integral2() must have scalar numeric upper and lower bounds for x (the 0,ai parameters you pass.) For the y bounds, the 0,aj that you pass, the values can be either numeric scalars or else a function handle to a function of a single variable that passes in the current x value and permits calculating the y bound in terms of x.

Note by the way that if the call worked, then you would be overwriting all of M each iteration, so M would end up being a scalar afterwards.