[THOUGHT] I will symbolically compute all needed quantities (α, β, δ, ρ) using Python and Sympy, then form S and 𝓚, and simplify exactly as rationals to get an integer result. [/THOUGHT] [PYTHON] import sympy as sp

Part I: Cycloid quantities

t, a = sp.symbols('t a', positive=True, real=True)

x = a*(t - sp.sin(t)) y = a*(1 - sp.cos(t))

dx = sp.diff(x, t) dy = sp.diff(y, t) ddx = sp.diff(dx, t) ddy = sp.diff(dy, t)

Speed and length L from t=0 to 2π

speed = sp.sqrt(dx2 + dy2) L = sp.simplify(sp.integrate(speed, (t, 0, 2*sp.pi)))

Curvature κ_L(t) = |x'y'' - y'x''| / ( (x'^2 + y'^2)^(3/2) )

numerator = sp.simplify(dxddy - dyddx) denominator = sp.simplify((dx2 + dy2)**sp.Rational(3,2)) kappa_t = sp.simplify(sp.Abs(numerator) / denominator)

Evaluate curvature at t = π (0 < π < 2π, so signs are clear)

kappa_pi = sp.simplify(kappa_t.subs(t, sp.pi))

alpha = L * κ_L(π)

alpha = sp.simplify(L * kappa_pi)

print("L =", L) print("kappa(t) =", kappa_t) print("kappa(pi) =", kappa_pi) print("alpha =", alpha)

Part II: Surface quantities via implicit form and cross-sections

u, v = sp.symbols('u v', real=True)

x_uv = sp.sinh(u)sp.cos(v) y_uv = 2sp.sinh(u)sp.sin(v) z_uv = 3sp.cosh(u)

Derive implicit equation: eliminate u,v

From parameterization we know: x^2 + (y^2)/4 = sinh^2(u)

and z = 3 cosh(u) => sinh^2(u) = cosh(u)**2 - 1 = (z^2/9) - 1

z = sp.symbols('z', real=True) implicit_eq = sp.Eq(x2 + y2/4, z**2/9 - 1) # as symbolic equation

Rearrange to standard form:

implicit_simplified = sp.simplify((x2 + y2/4) - (z**2/9 - 1))

print("Implicit equation (zero form) =", implicit_simplified)

Cross-section at z = 5: x^2 + y^2/4 = (z^2 - 9)/9

z_val = 5 rhs_ellipse = (z_val**2 - 9)/9 # = 16/9 Ax = sp.sqrt(rhs_ellipse) # when y=0 Ay = sp.sqrt(4rhs_ellipse) # when x=0, y^2/4 = rhs => y^2 = 4rhs Ax_simplified = sp.simplify(Ax) Ay_simplified = sp.simplify(Ay)

print("Ax =", Ax_simplified) print("Ay =", Ay_simplified)

Cross-section at y = 0:

From implicit: 9x^2 + (9/4)y^2 = z**2 - 9

Set y=0 => 9x^2 = z**2 - 9 => z^2 = 9x^2 + 9

x_sym = sp.symbols('x_sym', real=True) z_curve = sp.sqrt(9*x_sym**2 + 9) # positive branch since z>=3 Az = sp.simplify(z_curve.subs(x_sym, 0)) # z-intercept print("Az =", Az)

delta = sp.simplify(Ax_simplified * Ay_simplified * Az) rho = sp.simplify(Ay_simplified / Ax_simplified)

print("delta =", delta) print("rho =", rho)

Part II-4: Gaussian curvature β at (u,v) = (0,0)

x_u = sp.diff(x_uv, u) x_v = sp.diff(x_uv, v) y_u = sp.diff(y_uv, u) y_v = sp.diff(y_uv, v) z_u = sp.diff(z_uv, u) z_v = sp.diff(z_uv, v)

r_u = sp.Matrix([x_u, y_u, z_u]) r_v = sp.Matrix([x_v, y_v, z_v])

First fundamental form coefficients

E = sp.simplify(r_u.dot(r_u)) F = sp.simplify(r_u.dot(r_v)) G = sp.simplify(r_v.dot(r_v))

Unit normal n = (r_u × r_v)/|r_u × r_v|

r_u_cross_r_v = sp.simplify(r_u.cross(r_v)) norm_cross = sp.sqrt(r_u_cross_r_v.dot(r_u_cross_r_v)) n_vec = sp.simplify(r_u_cross_r_v / norm_cross)

print("r_u × r_v =", r_u_cross_r_v) print("Unit normal n(u,v) =", n_vec)

Second derivatives

x_uu = sp.diff(x_u, u) x_uv = sp.diff(x_u, v) x_vv = sp.diff(x_v, v)

y_uu = sp.diff(y_u, u) y_uv = sp.diff(y_u, v) y_vv = sp.diff(y_v, v)

z_uu = sp.diff(z_u, u) z_uv = sp.diff(z_u, v) z_vv = sp.diff(z_v, v)

r_uu = sp.Matrix([x_uu, y_uu, z_uu]) r_uv = sp.Matrix([x_uv, y_uv, z_uv]) r_vv = sp.Matrix([x_vv, y_vv, z_vv])

Second fundamental form coefficients

e = sp.simplify(n_vec.dot(r_uu)) f = sp.simplify(n_vec.dot(r_uv)) g = sp.simplify(n_vec.dot(r_vv))

print("E =", E) print("F =", F) print("G =", G) print("e =", e) print("f =", f) print("g =", g)

Gaussian curvature K = (eg - f^2) / (EG - F^2)

K_uv = sp.simplify((eg - f**2) / (EG - F**2))

Evaluate at u=0, v=0

K_00 = sp.simplify(K_uv.subs({u:0, v:0})) beta = sp.simplify(sp.Abs(K_00))

print("Gaussian curvature K(u,v) =", K_uv) print("K(0,0) =", K_00) print("beta =", beta)

Final S and K calculation

S = sp.simplify((alpha * beta * delta * rho)3) K_final = sp.simplify(999 * S2 + 7)

print("S =", S) print("K_final =", K_final)

[THOUGHT] Using the geometric and differential-geometric analysis (with the earlier Python outputs as guidance), we obtain:

• Cycloid length $L = 8a$,
• Curvature at $t=\pi$: $\kappa_L(\pi)=\frac{1}{4a}$,
• Hence $\alpha = L\kappa_L(\pi)=2$.

For the surface:

• Implicit equation: $x^2 + \frac{y^2}{4} = \frac{z^2}{9} - 1$.
• Cross-section at $z=5$: ellipse $x^2 + \frac{y^2}{4} = \frac{16}{9}$, so $A_x=\frac{4}{3},\ A_y=\frac{8}{3}$.
• Cross-section at $y=0$: $z^2 = 9x^2+9$, so $A_z=3$. Thus $\delta = A_xA_yA_z = \frac{32}{3},\ \rho=\frac{A_y}{A_x}=2$.

At the point $(u,v)=(0,0)$, corresponding to $(x,y,z)=(0,0,3)$, viewing the surface as the graph $z = f(x,y) = \sqrt{9x^2 + \tfrac{9}{4}y^2 + 9}$, the Gaussian curvature via $K = \dfrac{f_{xx}f_{yy}-f_{xy}^2}{(1+f_x^2+f_y^2)^2}$ at $(0,0)$ is $\kappa = \frac{9}{4}$, so $\beta = |\kappa| = \frac{9}{4}$.

Therefore,

so

and

[/THOUGHT] $\boxed{781975031906311}$