# The coordonates for the center of B is (x, r+d).
x^2 + (d+r)^2 = (R+r)^2;
solve(%,x);
u := factor(%[1]);

# The coordinates of the contact are:
R/(R+r) * <u, d+r>;

a:=1; b:=2; c:=3;  # semi axes
plot3d([a*cos(theta)*sin(phi),b*sin(theta)*sin(phi),c*cos(phi)],
theta=0..2*Pi,phi=0..Pi,scaling=constrained,axes=boxed);

For a=b=c=1 (as in your code) you have a sphere.

restart;   # Continuous roots w1(k)..w5(k) of a polynomial (wrt parameter k in 0..1)
with(plots):
f := -135/4*w^5+369/16*w^3*k^2+47/4*I*w^4-93/16*I*w^2*k^2+w^3-2/3*w^3*k*B-27/16*k^4*w+3/16*I*k^4-1/3*w*k^2+2/9*k^3*w*B:
B0:=1:
ini:=fsolve(eval(f,[B=B0,k=1/2]),w):
ode:=diff(eval(f,[w=W(k),B=B0]),k):
solode:=seq(dsolve({ode,W(1/2)=ii},numeric),ii=[ini]):
display(Matrix(2,5,
        [ [seq( odeplot(solode[i], [k,Re(W(k))], k=0..1,title=w[i]), i=1..5)],
          [seq( odeplot(solode[i], [k,Im(W(k))], k=0..1), i=1..5)] ]
),thickness=3);

Download PolynomialFit-simplex.mw

I seems that a polynomial of degree <=13 satisfying the "marked" conditions in the provided graph and having also a similar "shape" (i.e. same sign for the derivative) does not exist.

N:=13:
t:= unapply(add(x^k*u[k],k=0..N),x):
sol:=solve([
        t(-7/2)=-5,
        t(-2)  = 2, 
        t(0)   = -2, 
        t(1)   = 0, 
        t(5/2) = -3, 
        t(4)=3, 
        D(t)(-2)=0,
        D(t)(0)=0,
        D(t)(1)=0,
        D(t)(5/2)=0,
        D(t)(-7/2)=a,  # a>0
        D(t)(4)=b,     # b>0
        t(-1)=c,       # -1<c<1
        t(7/2)=d       # -1<d<1
      ], [seq(u[i],i=0..N)] 
     
)[]:
P:=eval(t(x),sol):
#indets(P);
Explore(plot(P, x=-3.5..4), parameters=[[a=0. .. 100],[b=0. .. 100], [c=-1. .. 1],[d=-1. .. 1]]);

residue connot compute the residue at a pole with symbolic order.
It's easy to write a procedure in this case provided that you know the order of the pole. It is based on Maple's ability to compute the derivative of a symbolic order, but it will fail for complex expressions.

Res:=(f,z,a,p) -> limit(diff(f*(z-a)^p,[z$(p-1)]), z=a)/(p-1)!:
# p = the order of the pole z=a

f := exp(-z/A)/z^(2 + n):
Res(f,z,0,n+2);
        

Try to increase Digits. For example,

restart;
Digits:=10;
n:=10;
x:=ln(1+0.1^n);
x^x;

==> Float(undefined). Increasing Digits ==> OK.

You must use

Explore('a'[i], i = 1 .. 3);

otherwise accessing a[i]  with symbolic i, produces an error.

(Explore has special evaluation rules, the first argument is passed unevaluated).

pde:={diff(u(x,t),t) =K*diff(u(x,t),x, x), D[1](u)(0,t)=0}: 
# u(x,infinity) = infinity,  diff(u(x,t),x) =0 when x=0 for all t 
sol:=pdsolve(pde, generalsolution);

# ==> (changing the constants)

u(x,t) = A*( exp(c*(K*c*t+x))+exp(c*(K*c*t-x)) );

# A>0, c>0 [K>0]
# pdsolve says it's the general solution, but with pdsolve you can never know

SolveTools:-SemiAlgebraic([x^2+y^2+z^2-4*x+6*y-2*z-11 = 0, 2*x+2*y-z = -18], [x,y,z]);

       [[x = -4/3, y = -19/3, z = 8/3]]
              

You should post text, or preferably Maple code instead of low resolution pictures.

Anyway, you do not need Change if your f is smooth:
int(diff(f(s),s), s = a..b, continuous);
      - f(a) + f(b)

Chebyshev already had the smart idea in 1853, see here.

After a change of variables ( x+1 --> x), it's a binomial integral

Int( x^m * (A + B*x^n)^p, x);

where m=-3/2, n=3*b, p=-1/2

Using Chebyshev's result, assuming that b is rational (and a<>0, a<>1 of course), the integral is an elementary function iff
b = 1/(6*k), or b = 1/(6*k + 3),  where k is an integer.

Yes, the computation are very slow in the general case.

You can speed up considerably the computations if you use hardware floats.

If A is the 0-1 matrix, define B:=Matrix(A, datatype=float)   and compute C := B^k.
The result will be exact provided that the entries of C are < 2^53 =~ 9.0*10^15.

Another posibility is to use LinearAlgebra:-Modular.

Here I have assumed the usual multiplication. For a custom multiplication you may use LinearAlgebra:-Generic.

The level sets BS_Price = B (B constant) are 4-dimensional manifolds.
The equation can be solved e.g. wrt K. (It is convenient to use IERF the inverse erf function, see the code below).

To visualize the 3d sections, one can use Explore.

restart;
BS_Price=exp(-r*T)*(-(1/2)*erf((1/4)*sqrt(2)*(sigma^2*T-2*r*T+2*ln(K)-2*ln(S__0))/(sigma*sqrt(T)))+1/2);
IERF(2*B*exp(r*T)-1) = (-(1/2)*sigma^2*T+r*T-ln(K)+ln(S__0))*sqrt(2)/(2*sigma*sqrt(T));
EQK:=solve(%, K);
IERF:= z -> RootOf(erf(_Z)-z):
Explore(plot3d(EQK, r = 0 .. -ln(B)/T, T=0.5 .. 2, labels=[r,T,K]), 
parameters=[sigma=0.01 ..1, S__0=0.01 ..2, B=0.01  ..0.95]);