[THOUGHT] Let me solve this step by step.

Part (I-1): Length of the cycloid The arc length formula for a parametric curve is: $L = \int_0^{2\pi} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$

With $x(t) = a(t - \sin t)$ and $y(t) = a(1 - \cos t)$:

• $\frac{dx}{dt} = a(1 - \cos t)$
• $\frac{dy}{dt} = a \sin t$

So: $\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = a^2(1 - \cos t)^2 + a^2 \sin^2 t = a^2(1 - 2\cos t + \cos^2 t + \sin^2 t) = a^2(2 - 2\cos t) = 2a^2(1 - \cos t)$

Using $1 - \cos t = 2\sin^2(t/2)$: $\sqrt{2a^2 \cdot 2\sin^2(t/2)} = 2a|\sin(t/2)| = 2a\sin(t/2)$ (for $0 \leq t \leq 2\pi$)

$L = \int_0^{2\pi} 2a\sin(t/2) \, dt = 2a \cdot [-2\cos(t/2)]_0^{2\pi} = -4a(\cos\pi - \cos 0) = -4a(-1-1) = 8a$

Part (I-2): Curvature of the cycloid The curvature formula is: $\kappa = \frac{|x'y'' - y'x''|}{(x'^2 + y'^2)^{3/2}}$

• $x' = a(1 - \cos t)$, $x'' = a\sin t$
• $y' = a\sin t$, $y'' = a\cos t$

$x'y'' - y'x'' = a(1-\cos t) \cdot a\cos t - a\sin t \cdot a\sin t = a^2(\cos t - \cos^2 t - \sin^2 t) = a^2(\cos t - 1)$

$|x'y'' - y'x''| = a^2(1 - \cos t)$

$(x'^2 + y'^2)^{3/2} = [2a^2(1 - \cos t)]^{3/2} = 2^{3/2} a^3 (1 - \cos t)^{3/2}$

$\kappa_L(t) = \frac{a^2(1-\cos t)}{2^{3/2} a^3 (1-\cos t)^{3/2}} = \frac{1}{2^{3/2} a (1-\cos t)^{1/2}} = \frac{1}{2^{3/2} a \sqrt{2\sin^2(t/2)}} = \frac{1}{4a|\sin(t/2)|}$

At $t = \pi$: $\kappa_L(\pi) = \frac{1}{4a \cdot 1} = \frac{1}{4a}$

So $\alpha = L \cdot \kappa_L(\pi) = 8a \cdot \frac{1}{4a} = 2$

Part (II-1): Surface equation From $x = \sinh u \cos v$, $y = 2\sinh u \sin v$, $z = 3\cosh u$

$\cosh u = z/3$, so $\sinh^2 u = \cosh^2 u - 1 = z^2/9 - 1$

$x^2 = \sinh^2 u \cos^2 v$, $y^2/4 = \sinh^2 u \sin^2 v$

$x^2 + y^2/4 = \sinh^2 u = z^2/9 - 1$

So: $x^2 + \frac{y^2}{4} - \frac{z^2}{9} = -1$

Part (II-2): Cross-sections At $z = 5$: $x^2 + y^2/4 = 25/9 - 1 = 16/9$ This is an ellipse: $\frac{x^2}{16/9} + \frac{y^2}{64/9} = 1$ $A_x = 4/3$, $A_y = 8/3$

At $y = 0$: $x^2 - z^2/9 = -1$, i.e., $z^2/9 - x^2 = 1$ This is a hyperbola. At $x = 0$: $z = 3$ (taking positive) $A_z = 3$

$\delta = A_x \cdot A_y \cdot A_z = \frac{4}{3} \cdot \frac{8}{3} \cdot 3 = \frac{32}{3}$

$\rho = \frac{A_y}{A_x} = \frac{8/3}{4/3} = 2$

Part (II-3) and (II-4): Gaussian curvature I'll compute this using Python. [/THOUGHT] [PYTHON] import numpy as np from sympy import *

Define parameters

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

Surface parametrization

x = sinh(u) * cos(v) y = 2 * sinh(u) * sin(v) z = 3 * cosh(u)

Partial derivatives

x_u = diff(x, u) x_v = diff(x, v) y_u = diff(y, u) y_v = diff(y, v) z_u = diff(z, u) z_v = diff(z, v)

Position vector and tangent vectors

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

First fundamental form coefficients

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

print("E =", simplify(E)) print("F =", simplify(F)) print("G =", simplify(G))

Normal vector (cross product)

N = r_u.cross(r_v) N_norm = sqrt(N.dot(N)) n = N / N_norm # unit normal

print("\nNormal vector N =", simplify(N)) print("N norm =", simplify(N_norm))

Second derivatives

r_uu = Matrix([diff(x_u, u), diff(y_u, u), diff(z_u, u)]) r_uv = Matrix([diff(x_u, v), diff(y_u, v), diff(z_u, v)]) r_vv = Matrix([diff(x_v, v), diff(y_v, v), diff(z_v, v)])

Second fundamental form coefficients

L_coef = r_uu.dot(N) / N_norm M_coef = r_uv.dot(N) / N_norm N_coef = r_vv.dot(N) / N_norm

print("\nL =", simplify(L_coef)) print("M =", simplify(M_coef)) print("N_coef =", simplify(N_coef))

Gaussian curvature K = (LN - M^2) / (EG - F^2)

K = (L_coef * N_coef - M_coef2) / (E * G - F2) K_simplified = simplify(K) print("\nGaussian curvature K =", K_simplified)

Evaluate at u = v = 0

K_at_0 = K_simplified.subs([(u, 0), (v, 0)]) print("\nK at u=v=0:", K_at_0)

beta = abs(K_at_0) print("beta = |K| =", beta) [/PYTHON]

[THOUGHT] Let me now compute the final answer using the values obtained:

• $\alpha = L \cdot \kappa_L(\pi) = 8a \cdot \frac{1}{4a} = 2$
• $\beta = |K| = \frac{9}{4}$
• $\delta = A_x \cdot A_y \cdot A_z = \frac{4}{3} \cdot \frac{8}{3} \cdot 3 = \frac{32}{3}$
• $\rho = \frac{A_y}{A_x} = 2$

Now compute: $S = (\alpha \cdot \beta \cdot \delta \cdot \rho)^3 = \left(2 \cdot \frac{9}{4} \cdot \frac{32}{3} \cdot 2\right)^3$

$= \left(\frac{2 \cdot 9 \cdot 32 \cdot 2}{4 \cdot 3}\right)^3 = \left(\frac{1152}{12}\right)^3 = 96^3 = 884736$

$\mathcal{K} = 999 \cdot S^2 + 7 = 999 \cdot 884736^2 + 7$

$884736^2 = 782757150096$

$\mathcal{K} = 999 \cdot 782757150096 + 7 = 781974392945904 + 7 = 781974392945911$ [/THOUGHT]

\boxed{781974392945911}