Calculating the surface area, mass, mass center of gravity and flux of a paraboloid

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I'm an engineering student who is this year taking the calculus-classes for the first time to apply vector analysis. All the concepts are a completely new and thus, I tried solving an exam question. However, there are no solutions given with this question.

The question is explained underneath: Given the following paraboloid: $$x^2+y^2=36-4z$$ Extra parameters: $$x\geqslant0 $$ $$z\geqslant0 $$

The paraboloid is surface S and the projection on the xy-plane is called S1.

Therefore, the normal vector on S is asked, the parameterization, de mass of the object (given a density function), the mass center of gravity and the flux trough S1 (given a vector field).

$$f(x,y,z)=2*({x^2+y^2+4})^{1/2}$$

At this point, explaining all the math is a bit difficult. But I solved it trough using MATLAB:

clear all

syms u v x y z

Sigma = [ucos(v), usin(v), 9-u^2/4] %parametrization of x, y and z

f(x,y,z) = 2*sqrt(x^2+y^2+4) %mass function per surface area

Calculating normal vector

n = cross(diff(Sigma, u), diff(Sigma, v))

Here the mass of the surface is calculated.

assume(u, 'real')

assume(v, 'real')

norm_n(u, v) = simplify(norm(n))

f_na_Sigma(u, v) = f(Sigma(1), Sigma(2), Sigma(3))

int_f = int(int(f_na_Sigma(u,v) * norm_n(u, v), v, [-pi/2, pi/2]), u, [0, 6])

mass=int_f


integrand=norm_n

surface = int(int(integrand, v, [-pi/2, pi/2]), u, [0, 6])

%The center of gravity is calculated

center_x= int(int(u*cos(v)*f_na_Sigma(u,v) * norm_n(u, v), v, [-pi/2, pi/2]), u, [0, 6]) / mass

center_y= int(int(u*sin(v)*f_na_Sigma(u,v) * norm_n(u, v), v, [-pi/2, pi/2]), u, [0, 6]) / mass

center_z= int(int((9-u^2/4)*f_na_Sigma(u,v) * norm_n(u, v), v, [-pi/2, pi/2]), u, [0, 6]) / mass


syms x y u v

Sigma=[6ucos(v), 6usin(v), 0]

n = cross(diff(Sigma, u), diff(Sigma, v))

F=[x*z,y^2,-5]

F=subs(F,x,Sigma(1))

F=subs(F,y,Sigma(2))

F=subs(F,z,Sigma(3))

integrand=dot(F,n)

flux2=int(int(integrand,u,[0,1]),v,[-pi/2, pi/2])


%Flux trough projection of top parabola on xy-plane

%conclusion: there goes more 'virtual fluid' out of the surface rather than goes inwards

After bringing the code of MATLAB in this thread. I realize this doesn't look too clear. If anyone would have tips for this, it would be greatly appreciated. The same goes for the calculations. Are they correct? How could I do better?

Thanks by advance. Best regards.