Triple integers (cylindrical method)

I am trying to write a code for a triple integral with function of r (radius) t (theta) z

xy/2 <= z <= (x^2+2y^2) 1/2

x = rcos theta

y = r sin theta

r^2sin(theta)*cos(theta) <= z <= r^2 cos^2 theta + 2*r^2 sin^2 theta

0<= r <= 1

0 <= theta <= pi/2

I tried this to get my answer

%%% computing the integral to find the actual volume of the region

% we know our theta (t) is bounded by 0 and pi/2

% we know the radius is bounded by 0 and 1

syms r t z

theta = t;

x = @(r,t) r*cos(t);

y = @(r,t) r*sin(t);

zmin = @(x,y) x*y/2;

zmax = @(x,y) sqrt (x.^2+2*y.^2);

rmin = 0;

rmax = x.^2 + y.^2 ==1;

% rmax is 1

% since we are finding the volume with triple integral the function would

% be 1

fun = @(z,r,t) r

volumee = integral3(fun,zmin,zmax,0,1,0,pi/2)

I am trying to get this

[Ibraheem]

No matter what i do i keep geting an error

Answer 1

[Torsten]

lower_z = @(r,theta) 0.5*r.^2.*sin(theta).*cos(theta);
upper_z = @(r,theta) r.*sqrt(cos(theta).^2+2*sin(theta).^2);
fun = @(r,theta,z) r;
value = integral3(fun,0,1,0,pi/2,lower_z,upper_z)

value = 0.5742