why matlab take a lot of time in the symbolic calculation

I wrote a code to calculate the fundamental frequencies of a piezoelectric nanotube, I used the state space analytical method to solve my differential equation which based on the computation of the exponential of a matrix but the problem takes a lot for the execution and I did not know for I will be very happy if you clarify the problem for me

the code is

clear all

clc

%introduire des variable

h=1e-9;

L=10e-9;

C11=132e9;

e31=-4.1;

mu=0;

E11=5.841e-9;

E33=7.124e-9;

ro=7500;

B=(pi/h);

r1=49.5e-9;

r2=50.5e-9;

A=pi*((r2^2)-(r1^2));

I=pi*(((r2^4)/4)-((r1^4/4)));

fun1=@(x,r) e31.*(r.^2).*sin(x).*B.*sin(B.*r.*sin(x));

fun2=@(x,r) E11.*r.*(cos(B.*r.*sin(x))).^2;

fun3=@(x,r) E33.*(B^2).*r.*(sin(B.*r.*sin(x))).^2;

F31 = integral2(fun1,0,2*pi,49.5,50.5);

X11 = integral2(fun2,0,2*pi,49.5,50.5);

X33 = integral2(fun3,15,2*pi,49.5,50.5);

NE=0;

%introduire la matrice A

syms w

p1=((ro*A*w^2)/(C11*I));

p2=((-(mu^2*ro*A*w^2)+(F31^2/X11))/(C11*I));

p3=((F31*X33)/(X11*C11*I));

p4=(F31/X11);

p5=(X33/X11);

a=[0 1 0 0 0 0;0 0 1 0 0 0;0 0 0 1 0 0;...

(p1) 0 (p2) 0 (p3) 0;0 0 0 0 0 1;0 0 (p4) 0 (p5) 0];

[V,D]=eig(a);

H=eye(6);

M=V*diag(exp(diag(D*L)))/V;

%condition aux limite simplement appuyé

l1=H(1,:);

l3=H(3,:);

l5=H(5,:);

m1=M(1,:);

m3=M(3,:);

m5=M(5,:);

%la matrice final

K=[l1;l3;l5;m1;m3;m5];

DA=det(K);

%fréquence fondamental

for n = 1:5

F = vpasolve(DA,w,[0.4 3],'Random',true)

end

Answer 1

[Sulaymon Eshkabilov]

The problem with integral2 calculation part not with vpasolve(). You need to check the limit values for xmin, xmax, ymin, ymax in interal2(). This below given syntax controls tolerances and finds the solutions, BUT

...
F31 = integral2(fun1,0,2*pi,49.5,50.5, 'Method','iterated','AbsTol',0,'RelTol',1e-10);  % Solves the problem partially
X11 = integral2(fun2,0,2*pi,49.5,50.5, 'Method','iterated','AbsTol',0,'RelTol',1e-10);
X33 = integral2(fun3,15,2*pi,49.5,50.5, 'Method','iterated','AbsTol',0,'RelTol',1e-10);
NE=0;

As it is given, the polynomial coeffs p2, p3 .. p5 = nan.

Therefore, it does not produce any solutions.

[merwan behar]

yes I made a mistake in the meantime it's [49.9e-9 50.5e-9]

I made his modifications

fun1=@(x,r) e31.*(r.^2).*sin(x).*B.*sin(B.*r.*sin(x));

fun2=@(x,r) E11.*r.*(cos(B.*r.*sin(x))).^2;

fun3=@(x,r) E33.*(B^2).*r.*(sin(B.*r.*sin(x))).^2;

F31 = integral2(fun1,0,2*pi,49.5e-9,50.5e-9,'method','iterated');

X11 = integral2(fun2,0,2*pi,49.5e-9,50.5e-9,'method','iterated');

X33 = integral2(fun3,15,2*pi,49.5e-9,50.5e-9,'method','iterated');