Electric Flux 8-23-2020 (Part 2)

First of all few steps to follow
1 Basically,in my below calculation I am trying to find flux through lower sphere $(z<0)$ ,so in that case basically i have evaluated the flux through $x^{2}+y^{2}=1$ .Becauss if you think carefully both are same.
2) Using Gauss law we can easily calculate the flux through total sphere $\phi=\frac{q_{inside}}{\epsilon_{0}}$
3) If we substract the flux of $(z<0)$ from total flux we will get the flux of upper half $(z>0)$

$\frac{\phi_{z>0}}{\phi_{Total}}=\frac{2.89435668}{3.76991118}=\boxed{0.767751955}$

A very nice solution has already been provided by @Lil Doug . I am sharing my numerical approach. I have resorted to evaluating a surface integral without having to deal with any of the algebraic expressions.

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clear all
clc

% Numerical resoloution:
dalpha = pi/1000;
dtheta = pi/100;
dphi   = pi/100;

% Ring radius:
R     = 3/5;

% Flux initialisation:
Phi    = 0;

% Nested loops for numerical integration:
% Surface integral over the top half of the sphere:
for alpha = 0:dalpha:2*pi

    for theta = 0:dtheta:pi/2

        for phi = 0:dphi:2*pi

            % Unit sphere centered at origin parameterisation:
            rs    = [sin(theta)*cos(phi);sin(theta)*sin(phi);cos(theta)];

            % Ring parameterisation:
            rr    = [R*cos(alpha);R*sin(alpha);1/2];

            % Coulomb's Law:
            dE    = ((1/(4*pi))*(R*dalpha)*(rs - rr))/(norm(rs - rr)^3);

            % Surface area element:
            dS    = sin(theta)*dtheta*dphi*rs;

            % Elementary flux computation (dot product):
            dPhi  = dE'*dS;

            % Numerical integration
            Phi   = Phi + dPhi;
        end

    end

end

% Ration of flux through top half to that through the whole sphere
Answer = Phi/3.77
% Answer = 0.7771