quarter to full region in matlab

i have a some problem to make a full region

in my code that is just the thing only the qurter region

but i wanna make a full region so how can i get to there ? i have no special idea about how to do

in my code this quarter region is shown (+x,+y) region

this is the what i wanna figure that shown full region plus to this this picture that i make 3 dimension picture like first picture

%**************************************************************

% Using the finite difference method

% This program calculates the characteristic impedance of the transmission

% line shown in Figure 3.14

%*************************************************************************

clear all;format compact;

%Output;

%

% H NT Zo

% -----------------------

% 0.25 700 69.77

% 0.1 500 65.75

% 0.05 500 70.53

% 0.05 700 67.36

% 0.05 1000 65.50

H=0.05;

NT= 100;

A=2.5; B=2.5; D=0.5; W=1.0;

ER=2.35;

EO=8.81E-12;

U=3.0E+8;

NX=A/H;

NY=B/H;

ND=D/H;

NW=W/H;

VD=100.0;

%CALCULATE CHARGE WITH AND WITHOUT DIELECTRIC

ERR=1.0;

for L=1:2

E1=EO;

E2=EO*ERR;

%INITIALIZATION

V=zeros(NX+2,NY+2);

%SET POTENTIAL ON INNER CONDUCTOR(FIXED NODES) EQUAL TO VD

V(2:NW+1,ND+2)=VD;

%CALCULATE POTENTIAL AT FREE NODES

P1=E1/(2*(E1+E2));

P2=E2/(2*(E1+E2));

for K=1:NT

for I=-0:NY-1

for J=0:NY-1

if( (J==ND)&(I<=NW))

%do nothing

elseif(J==ND)

%IMPOSE BOUNDARY CONDITION AT THE INTERFACE

V(I+2,J+2)=0.25*(V(I+3,J+2)+V(I+1,J+2))+....

P1*V(I+2,J+3)+P2*V(I+2,J+1);

elseif(I==0)

%IMPOSE SYMMETRY CONDITION ALONG Y-AXIS

V(I+2,J+2)=(2*V(I+3,J+2)+V(I+2,J+3)+V(I+2,J+1))/4.0;

elseif(J==0)

%IMPOSE SYMMETRY CONDITION ALONG X-AXIS

V(I+2,J+2)=(V(I+3,J+2)+V(I+1,J+2)+2*V(I+2,J+3))/4.0

else

V(I+2,J+2)=(V(I+3,J+2)+V(I+1,J+2)+V(I+2,J+3)+V(I+2,J+1))/4.0;

end

end

end

%Animation of calculation

figure(1),surf(V),colorbar,title([num2str(K),'/',num2str(NT)])

%drawnow

end

-----------------------------------------------------------------------this is the code that can figure

%NOW,CALCULATE THE TOTAL CHARGE ENCLOSED IN A

%RECTANGULAR PATH SURROUNDING THE INNER CONDUCTOR

IOUT=round((NX+NW)/2);

JOUT=round((NY+ND)/2);

%SUM POTENTIAL ON INNER AND OUTER LOOPS

for K=1:2

SUM= E1*SUM(V(3:IOUT+1,JOUT+2))....

+E1*V(2,JOUT+2)/2+E2*V(IOUT+2,2)/2;

for J=1:JOUT-1

if(J<ND)

SUM=SUM+E2*V(IOUT+2,J+2);

elseif(J==ND)

SUM=SUM+(E1+E2)*V(IOUT+2,J+2)/2;

else

SUM=SUM+E1*V(IOUT+2,J+2);

end

end

if K==1

SV(1)=SUM;

end

IOUT=IOUT-1;

JOUT=JOUT-1;

end

SUM=SUM+2.0*E1*V(IOUT+2,JOUT+2);

SV(2)=SUM;

Q(L)=abs(SV(1)-SV(2));

ERR=ER;

end

% FINALLY,CALCULATE Zo

Co=4.0*Q(1)/VD;

C1=4.0*Q(2)/VD;

ZO=1.0/(U*sqrt(C0*C1));

disp([H,NT,ZO])

--------------------this is just the code that after the figure to calculate to some question