One Each Day Till They Go Away (Part 3)

Parameterising the paraboloid:

$x = 1+ r\cos{t}$ $y = r\sin{t}$ $z = r^2$

Plugging this parameterization into the equation of the sphere leads to the equation (after simplifying):

$r^4 + r^2 -2r \cos{t} -3=0$

This is a 4th degree equation in $r$ which is to be numerically solved for each value of $t$ where $0 \le t \le 2 \pi$ . Finally, the parameterised curve is:

$\vec{R}(t) = ( 1+ r\cos{t}) \ \hat{i} + r\sin{t} \ \hat{j} + r^2 \ \hat{k}$

Where $r$ is the real root of the 4th degree polynomial. Now there may be more than one real root of this equation, so finding that is something I am yet to figure out. To be very honest, I got lucky in finding the answer. The simulation code is attached below. I look forward to receiving feedback on this solution, as I am sure there is a better way of solving this.

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clear all
clc

% Resolving the angle in space:
dt  = 1e-4;
t   = 0:dt:2*pi;

% Computing r(t) by solving the 4th degree polynomial:
for k = 1:length(t)

Roots    = roots([1,0,1,-2*cos(t(k)),-3]);

for i = 1:4

    if imag(Roots(i))==0 
        rd = abs(Roots(i));
        if abs((rd*cos(t(k)) + 1)^2+ (rd*sin(t(k)))^2 + rd^4 - 4) < 1e-3
            r(k) = rd;
        end
    end

end

end

% Parameterised curve of intersection:
x    = r.*cos(t) + 1;
y    = r.*sin(t);
z    = r.^2;

% Derivatives wrt t of the parameterised curve components (Finite differences)
for k = 1:length(t)-1

    dx(k)  = (x(k+1)-x(k))/dt;
    dy(k)  = (y(k+1)-y(k))/dt;
    dz(k)  = (z(k+1)-z(k))/dt;

end

% Integral initialisation:
I = 0;

for k = 1:length(t)-1

    % Electric field:
    E  = [z(k);x(k);y(k)];

    % Line element vector:
    dR = [dx(k);dy(k);dz(k)]*dt;

    % Integrand:
    dI = E'*dR;

    % Numerical integration:
    I  = I + dI;
end

ANSWER = abs(I) - 2*pi
% ANSWER = 0.104