restart;
f:=(x,y)->evalf(2*x*Int(sqrt(1+y^2*(t*x-1)^2/(1-(t*x-1)^2)), t = 0 .. 1)):
m:=19;n:=9;
G:=Matrix(m*n,3,datatype=float);
k:=0;
for i to m  do   for j to n do
k:=k+1;
G[k,1]:=2*i/(m+1); G[k,2]:=j/(n+1); G[k,3]:=f(2*i/(m+1),j/(n+1))
od od:

f_fit:=a1*x*sqrt(1+y^2*((a2*x-a3)^2/(1-(a4*x-1)^2)));
Statistics[NonlinearFit](f_fit, G, [x, y], initialvalues = [a1 = 1, a2 = 0, a3 = 0, a4=1/2],
output = [leastsquaresfunction, residuals]);
f1:=unapply(%[1],x,y);

printf("x    y    f(x,y)    f_fit(x,y)  \n");    
for x_ from 0.1 to 1.9 by 0.1 do
for y_ from 0.1 to 0.9 by 0.1 do
printf("%2.1f  %2.1f  %6f  %6f\n",x_,y_,f(x_,y_), f1(x_,y_)      );
od; od;

You may replace arcsin by wrongarcsin where:

wrongarcsin := x -> piecewise(x<0, arcsin(x)+2*Pi, arcsin(x));

You may simply "cut" the polynomial, or use:

hor:= (f,x,i) -> numapprox[hornerform](select(u -> degree(u,x)<=i, f) ,x);

f:=add( (10*i+1)*x^i, i=0..6);
hor(f,x,3);

Your notations are not clear (at least for me).

Given the function  g = f(r,theta)*r
you can integrate it in the domain

A = { (r,theta) :  a<= r <= b,  c(r) <= theta <= d(r)}

or

 B = { (r,theta) :  a<= theta <= b,  c(theta) <= r <= d(theta)}.

In the A case, the integral is
int(int(g, theta=c(r)..d(r)), r = a..b);

and similarly for B.

Please rephrase your question in this context.

numtheory[cfrac](3/11+3*sqrt(3),'periodic', 'quotients');

The only possibility I see is to visualize.

E.g. the z-sections using

ee:=subs( [a[1]=x,a[2]=y,a[3]=z], %);
f:=unapply(rhs(ee)-lhs(ee),x,y,z);
Explore(  plots[implicitplot](f(x,y,z)>=0, x=-20..20,y=-20..20,filledregions,numpoints=15000),
parameters = [z=-20..20] );

Cf:=proc(x,n:=floor(Digits/log10(2)+2))
if not(type(x,realcons)) then
  if length(1-x)< length(x) then return 1-'Cf'(1-x) else return 'Cf'(x) fi;
fi;
if x=0 or x=1 then return x fi; if n<0 then return 0.0 fi;
piecewise(x<1/3, Cf(3*x,n-1)/2, x<=2/3, 1/2, 1/2+Cf(3*x-2,n-1)/2)
#  piecewise(x>1/2, 1-Cf(1-x,n-1), x<1/3, Cf(3*x,n-1)/2, 1/2)
end;

Cf(1-3*x)+Cf(3*x);
    1

Cf(2015/2222);
    13/16

plot(Cf,0..1);

The condition number has a standard definition, and cannot be changed. Probably there is a bug also in LU or Cholesky factorization which should be investigated.

Norm(B,infinity),Norm(B^(-1),infinity); ==> 3,2  hence OK.

Edit: for other matrices such as A:=HilbertMatrix(3,3) the results are correct.

asympt((x^2+3)/(x+2),x,8);

(there are mistakes in your computations)

convert(sin(x),FormalPowerSeries);

For two sets only, it's easier to use seq.

A:={a,b,c}; B:={x,y}:

{ seq(seq( u*v, u=A ), v=B) };

Proj:=proc(A::Matrix,b::Vector[column],v::Vector[column])
uses LinearAlgebra;
local u0,P,_xxx;
u0:=LinearSolve(A,b,free=_xxx);  #error if {x:Ax=b} is empty
u0:=eval(u0,_xxx=0);
P:=ProjectionMatrix(NullSpace(A));
#   u0+P.(v-u0)
u0+P.(Vector(v)-u0) # To work with Gradient without ConvertVector(...,free)
end;

Example

Proj(<<1|2|3>,<2|4|6>>, <1,2>, <0,0,0>);

G:=proc(k::posint, n::posint)
local f,g,x,y,u,v;
f:=(1/2)*(1-cos(k*Pi*x)*cos(n*Pi*y))/(sin((1/2)*Pi*x)^2+sin((1/2)*Pi*y)^2);
g:=simplify(subs([x=2/Pi*arctan(u),y=2/Pi*arctan(v)],f)*4/(Pi^2*(u^2+1)*(v^2+1)));
expand(g);
int(int(%,u=0..infinity),v=0..infinity)
end:

t:=time():
G(9,9);
time()-t;

   1.279

After some "reverse engineering", it is a group but for mod 70.

Simply construct the Cayley table:

G:=[5, 15, 25, 45, 55, 65];
Matrix(6, (i,j) -> G[i]*G[j] mod 70);

Alternatively, you may use Magma (see ?Magma) but the problem is too simple to need it..

phasor:=(r,a)->r*exp(I*a*Pi/180):   #simpler version
ec:=phasor(U__C, -90) = phasor(400, X)-phasor(150, 0)+phasor(1744/10, 90):
s:=solve(evalc([Re(ec),Im(ec)]),allsolutions);


uc:=eval(U__C,s):
[eval(uc,[_Z1=0,_B1=0]),eval(uc,[_Z1=0,_B1=1])];


evalf(%);


So, you will take the second (positive) solution.

Note that
fsolve(evalc([Re(ec),Im(ec)]), {U__C=0..1000, X=0..360});
does not provide a solution (it should!), but
fsolve(evalc([Re(ec),Im(ec)]), {U__C=200, X=270});