[THOUGHT] We need to compute several quantities step by step:

(I-1) Length of cycloid L: ( x(t) = a(t - \sin t), y(t) = a(1 - \cos t) ) for ( 0 \le t \le 2\pi ).
Arc length formula: ( L = \int_0^{2\pi} \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } , dt ).

Compute derivatives: ( \frac{dx}{dt} = a(1 - \cos t) ), ( \frac{dy}{dt} = a \sin t ).

Then ( \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 = a^2 \left[ (1 - \cos t)^2 + \sin^2 t \right] = a^2 [1 - 2\cos t + \cos^2 t + \sin^2 t] = a^2 [2 - 2\cos t] = 2a^2 (1 - \cos t) ).

Using identity ( 1 - \cos t = 2 \sin^2(t/2) ), we get ( = 4a^2 \sin^2(t/2) ).

Thus ( \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } = 2a |\sin(t/2)| ). For ( t \in [0, 2\pi] ), ( \sin(t/2) \ge 0 ) (since ( t/2 \in [0, \pi] )), so ( = 2a \sin(t/2) ).

Therefore ( L = \int_0^{2\pi} 2a \sin(t/2) , dt = 2a \int_0^{2\pi} \sin(t/2) , dt ).

Let ( u = t/2 ), ( dt = 2 du ), limits: ( u \in [0, \pi] ).

( L = 2a \int_0^{\pi} \sin u \cdot 2 du = 4a \int_0^{\pi} \sin u , du = 4a [-\cos u]_0^{\pi} = 4a (-\cos \pi + \cos 0) = 4a ( -(-1) + 1 ) = 4a (1+1) = 8a ).

So ( L = 8a ).

(I-2) Curvature ( \kappa_L(t) ) for parametric curve: ( \kappa = \frac{ |x' y'' - y' x''| }{ (x'^2 + y'^2)^{3/2} } ).

We have ( x' = a(1 - \cos t) ), ( y' = a \sin t ).

( x'' = a \sin t ), ( y'' = a \cos t ).

Compute numerator: ( |x' y'' - y' x''| = | a(1 - \cos t) \cdot a \cos t - a \sin t \cdot a \sin t | = a^2 | (1 - \cos t)\cos t - \sin^2 t | ).

Simplify: ( (1 - \cos t)\cos t - \sin^2 t = \cos t - \cos^2 t - \sin^2 t = \cos t - 1 ) (since ( \cos^2 t + \sin^2 t = 1 )).

Thus numerator = ( a^2 | \cos t - 1 | = a^2 (1 - \cos t) ) because ( \cos t - 1 \le 0 ) for all t, so absolute value is ( 1 - \cos t ).

Denominator: ( (x'^2 + y'^2)^{3/2} = (2a^2 (1 - \cos t))^{3/2} = (2a^2)^{3/2} (1 - \cos t)^{3/2} = 2^{3/2} a^3 (1 - \cos t)^{3/2} ).

Thus ( \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}} ).

Using ( 1 - \cos t = 2 \sin^2(t/2) ), we get ( (1 - \cos t)^{1/2} = \sqrt{2} |\sin(t/2)| = \sqrt{2} \sin(t/2) ) (positive for t in (0, 2π)).

Thus ( \kappa_L(t) = \frac{1}{2^{3/2} a \cdot \sqrt{2} \sin(t/2)} = \frac{1}{2^{2} a \sin(t/2)} = \frac{1}{4a \sin(t/2)} ).

So ( \kappa_L(t) = \frac{1}{4a \sin(t/2)} ).

At ( t = \pi ), ( \sin(\pi/2) = 1 ), so ( \kappa_L(\pi) = \frac{1}{4a} ).

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

Now (II-1) Surface: ( x = \sinh u \cos v, y = 2 \sinh u \sin v, z = 3 \cosh u ).

We want equation without u, v.

Note: ( \cosh^2 u - \sinh^2 u = 1 ). Also ( \sinh u = \frac{x}{\cos v} ) but v appears. Better:
( \frac{x}{\cos v} = \sinh u, \frac{y}{2 \sin v} = \sinh u ), so ( \frac{x}{\cos v} = \frac{y}{2 \sin v} ) ⇒ ( 2x \sin v = y \cos v ) ⇒ ( \tan v = \frac{y}{2x} ).

Also ( z = 3 \cosh u ) ⇒ ( \cosh u = z/3 ), so ( \sinh u = \sqrt{ \cosh^2 u - 1 } = \sqrt{ (z/3)^2 - 1 } = \sqrt{ z^2/9 - 1 } ).

But ( \sinh u = \sqrt{ x^2 / \cos^2 v + y^2 / (4 \sin^2 v) }? ) Not straightforward.

Better: ( x^2 = \sinh^2 u \cos^2 v ), ( y^2 = 4 \sinh^2 u \sin^2 v ).

Add: ( x^2 + \frac{y^2}{4} = \sinh^2 u (\cos^2 v + \sin^2 v) = \sinh^2 u ).

Thus ( \sinh^2 u = x^2 + \frac{y^2}{4} ).

Also ( \cosh^2 u = \frac{z^2}{9} ).

Using identity ( \cosh^2 u - \sinh^2 u = 1 ), we get:

( \frac{z^2}{9} - \left( x^2 + \frac{y^2}{4} \right) = 1 ).

Thus ( \frac{z^2}{9} - x^2 - \frac{y^2}{4} = 1 ).

So surface is a hyperboloid of one sheet.

(II-2) Cross-section at ( z = 5 ): plug into equation: ( \frac{25}{9} - x^2 - \frac{y^2}{4} = 1 ) ⇒ ( x^2 + \frac{y^2}{4} = \frac{25}{9} - 1 = \frac{16}{9} ).

Thus ( \frac{x^2}{(16/9)} + \frac{y^2}{4 \cdot (16/9)} = 1 ) ⇒ ( \frac{x^2}{(16/9)} + \frac{y^2}{(64/9)} = 1 ).

So ellipse: ( x )-intercepts: ( \pm \frac{4}{3} ) (since ( \sqrt{16/9} = 4/3 )), ( y )-intercepts: ( \pm \frac{8}{3} ) (since ( \sqrt{64/9} = 8/3 )).

Thus ( A_x = 4/3, A_y = 8/3 ) (positive intercepts).

Cross-section at ( y = 0 ): plug ( y = 0 ) into surface equation: ( \frac{z^2}{9} - x^2 = 1 ) ⇒ ( \frac{z^2}{9} - x^2 = 1 ), hyperbola opening along z-axis.

We need ( z )-intercept: set ( x = 0 ) ⇒ ( \frac{z^2}{9} = 1 ) ⇒ ( z = \pm 3 ). Positive intercept ( A_z = 3 ).

So ( A_x = 4/3, A_y = 8/3, A_z = 3 ).

Thus ( \delta = A_x A_y A_z = \frac{4}{3} \cdot \frac{8}{3} \cdot 3 = \frac{4 \cdot 8 \cdot 3}{3 \cdot 3} = \frac{96}{9} = \frac{32}{3} ).

And ( \rho = \frac{A_y}{A_x} = \frac{8/3}{4/3} = 2 ).

(II-3) Unit normal vector: surface defined by ( F(x,y,z) = \frac{z^2}{9} - x^2 - \frac{y^2}{4} - 1 = 0 ). Gradient ( \nabla F = \left( -2x, -\frac{y}{2}, \frac{2z}{9} \right) ).

Thus normal vector ( \mathbf{n} = \nabla F ) (or opposite). We want z-component positive, so we take ( \mathbf{n} = \left( -2x, -\frac{y}{2}, \frac{2z}{9} \right) ). But we can also express in terms of u, v.

From parametric: ( \mathbf{r}(u,v) = ( \sinh u \cos v, 2 \sinh u \sin v, 3 \cosh u ) ).

Compute ( \mathbf{r}_u = ( \cosh u \cos v, 2 \cosh u \sin v, 3 \sinh u ) ),
( \mathbf{r}_v = ( -\sinh u \sin v, 2 \sinh u \cos v, 0 ) ).

Cross product ( \mathbf{r}_u \times \mathbf{r}_v = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \cosh u \cos v & 2 \cosh u \sin v & 3 \sinh u \ -\sinh u \sin v & 2 \sinh u \cos v & 0 \end{vmatrix} ).

Compute:
i-component: ( (2 \cosh u \sin v)(0) - (3 \sinh u)(2 \sinh u \cos v) = -6 \sinh^2 u \cos v ).
j-component: ( -[ (\cosh u \cos v)(0) - (3 \sinh u)(-\sinh u \sin v) ] = -[ 0 + 3 \sinh^2 u \sin v ] = -3 \sinh^2 u \sin v ). Wait careful: j-component is negative of determinant of submatrix:
Actually formula: ( \mathbf{a} \times \mathbf{b} = (a_2 b_3 - a_3 b_2, a_3 b_1 - a_1 b_3, a_1 b_2 - a_2 b_1) ).

Let’s compute directly:

i: ( (2 \cosh u \sin v)(0) - (3 \sinh u)(2 \sinh u \cos v) = -6 \sinh^2 u \cos v ).

j: ( (3 \sinh u)(-\sinh u \sin v) - (\cosh u \cos v)(0) = -3 \sinh^2 u \sin v ). Wait sign: j-component is ( a_3 b_1 - a_1 b_3 ):
( a_3 = 3 \sinh u, b_1 = -\sinh u \sin v ) ⇒ ( 3 \sinh u \cdot (-\sinh u \sin v) = -3 \sinh^2 u \sin v ).
( a_1 = \cosh u \cos v, b_3 = 0 ) ⇒ minus ( \cosh u \cos v \cdot 0 = 0 ). So j = -3 sinh^2 u sin v.

k: ( (\cosh u \cos v)(2 \sinh u \cos v) - (2 \cosh u \sin v)(-\sinh u \sin v) = 2 \cosh u \sinh u \cos^2 v + 2 \cosh u \sinh u \sin^2 v = 2 \cosh u \sinh u (\cos^2 v + \sin^2 v) = 2 \cosh u \sinh u ).

Thus ( \mathbf{r}_u \times \mathbf{r}_v = \left( -6 \sinh^2 u \cos v, -3 \sinh^2 u \sin v, 2 \cosh u \sinh u \right) ).

We want unit normal with positive z-component. At ( u = v = 0 ), this is (0, 0, 0)? Wait: sinh 0 = 0, so first two components 0, third: 2 cosh 0 sinh 0 = 0. That’s zero vector? That suggests need careful: at u=0, surface degenerates? Actually u=0 ⇒ x=0, y=0, z=3 cosh 0 = 3. That’s a point, so tangent plane normal can be computed from gradient of F at that point: (0,0,2*3/9=2/3) positive z-component. So unit normal is (0,0,1). But for general u,v, we can take ( \mathbf{n} = \frac{ \mathbf{r}_u \times \mathbf{r}_v }{ | \mathbf{r}_u \times \mathbf{r}_v | } ), but ensure z-component positive. Let’s compute magnitude later if needed.

But part (II-3) asks to express unit normal vector ( \mathbf{n} ) by u and v. We can give formula:
( \mathbf{n} = \frac{ \left( -6 \sinh^2 u \cos v, -3 \sinh^2 u \sin v, 2 \cosh u \sinh u \right) }{ \sqrt{ 36 \sinh^4 u \cos^2 v + 9 \sinh^4 u \sin^2 v + 4 \cosh^2 u \sinh^2 u } } ).

Simplify denominator: factor ( \sinh^2 u ) if sinh u ≠ 0:
( \sinh^2 u (36 \cos^2 v + 9 \sin^2 v) + 4 \cosh^2 u \sinh^2 u = \sinh^2 u [ 36 \cos^2 v + 9 \sin^2 v + 4 \cosh^2 u ] ).

But maybe easier to use gradient method: ( \nabla F = (-2x, -y/2, 2z/9) ). In terms of u,v: x = sinh u cos v, y = 2 sinh u sin v, z = 3 cosh u. So
( \nabla F = ( -2 \sinh u \cos v, -\frac{2 \sinh u \sin v}{2}, \frac{2 \cdot 3 \cosh u}{9} ) = ( -2 \sinh u \cos v, -\sinh u \sin v, \frac{2 \cosh u}{3} ) ).

Check magnitude: ( | \nabla F | = \sqrt{ 4 \sinh^2 u \cos^2 v + \sinh^2 u \sin^2 v + \frac{4}{9} \cosh^2 u } = \sqrt{ \sinh^2 u (4 \cos^2 v + \sin^2 v) + \frac{4}{9} \cosh^2 u } ).

Unit normal with positive z-component: ( \mathbf{n} = \frac{ \nabla F }{ | \nabla F | } ) since z-component positive.

Thus ( \mathbf{n} = \frac{ \left( -2 \sinh u \cos v, -\sinh u \sin v, \frac{2}{3} \cosh u \right) }{ \sqrt{ \sinh^2 u (4 \cos^2 v + \sin^2 v) + \frac{4}{9} \cosh^2 u } } ).

(II-4) Gaussian curvature ( \kappa ) at ( u = v = 0 ). For a surface given implicitly by ( F(x,y,z)=0 ), Gaussian curvature formula: ( \kappa = \frac{ \det \begin{pmatrix} F_{xx} & F_{xy} & F_{xz} & \partial_x F \ F_{yx} & F_{yy} & F_{yz} & \partial_y F \ F_{zx} & F_{zy} & F_{zz} & \partial_z F \ \partial_x F & \partial_y F & \partial_z F & 0 \end{pmatrix} }{ | \nabla F |^4 } )? That’s complicated.

Better: For parametric surface, Gaussian curvature ( \kappa = \frac{ \det( \text{second fundamental form} ) }{ \det( \text{first fundamental form} ) } ).

First fundamental form coefficients:
( E = \mathbf{r}_u \cdot \mathbf{r}_u = \cosh^2 u \cos^2 v + 4 \cosh^2 u \sin^2 v + 9 \sinh^2 u ).
( F = \mathbf{r}_u \cdot \mathbf{r}_v = (\cosh u \cos v)(-\sinh u \sin v) + (2 \cosh u \sin v)(2 \sinh u \cos v) + (3 \sinh u)(0) )
= ( -\cosh u \sinh u \cos v \sin v + 4 \cosh u \sinh u \sin v \cos v = 3 \cosh u \sinh u \cos v \sin v ).
( G = \mathbf{r}_v \cdot \mathbf{r}_v = \sinh^2 u \sin^2 v + 4 \sinh^2 u \cos^2 v = \sinh^2 u (\sin^2 v + 4 \cos^2 v) ).

Second fundamental form coefficients:
( \mathbf{n} ) unit normal with positive z-component. We compute ( L = \mathbf{r}{uu} \cdot \mathbf{n} ), ( M = \mathbf{r}{uv} \cdot \mathbf{n} ), ( N = \mathbf{r}_{vv} \cdot \mathbf{n} ).

Compute ( \mathbf{r}{uu} = (\sinh u \cos v, 2 \sinh u \sin v, 3 \cosh u) ).
( \mathbf{r}{uv} = (-\cosh u \sin v, 2 \cosh u \cos v, 0) ).
( \mathbf{r}_{vv} = (-\sinh u \cos v, -2 \sinh u \sin v, 0) ).

At ( u = v = 0 ):
( \mathbf{r}_u = (1, 0, 0) ), ( \mathbf{r}_v = (0, 0, 0) )? Wait: sinh 0 = 0, cosh 0 = 1. So ( \mathbf{r}_u(0,0) = (1 \cdot 1, 2 \cdot 1 \cdot 0, 3 \cdot 0) = (1, 0, 0) ).
( \mathbf{r}_v(0,0) = (-0 \cdot 0, 2 \cdot 0 \cdot 1, 0) = (0, 0, 0) ). That’s degenerate! So parameterization is degenerate at u=0? Indeed, at u=0, the surface is a circle? Actually u=0 gives x=0, y=0, z=3, a single point. So the surface is not regular at u=0? But the Gaussian curvature should be defined via implicit form.

Better to compute Gaussian curvature at u=0 using formula for implicit surface ( F(x,y,z)=0 ). Gaussian curvature for level surface ( F(x,y,z)=0 ) is
( \kappa = \frac{ \det( \text{Hess}(F) ) - (\nabla F)^T \text{adj}( \text{Hess}(F) ) \nabla F }{ | \nabla F |^4 } )? That’s messy.

Alternatively, we can compute using parametrization away from u=0 and take limit as u→0.

But maybe easier: Use formula for hyperboloid: For surface ( \frac{z^2}{c^2} - \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 ), Gaussian curvature at (x,y,z) is ( \kappa = -\frac{1}{a^2 b^2 c^2} \frac{1}{ \left( \frac{z^2}{c^2} - \frac{x^2}{a^2} - \frac{y^2}{b^2} \right)^2 } )? Not exactly.

Actually for hyperboloid of one sheet ( \frac{z^2}{c^2} - \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 ), Gaussian curvature is negative and given by ( \kappa = -\frac{1}{a^2 b^2 c^2} \frac{1}{ \left( \frac{z^2}{c^2} \right)^2 } ) at points where x=y=0? Let’s derive.

Let ( F = \frac{z^2}{c^2} - \frac{x^2}{a^2} - \frac{y^2}{b^2} - 1 = 0 ).
Compute Hessian:
( F_{xx} = -\frac{2}{a^2} ), ( F_{yy} = -\frac{2}{b^2} ), ( F_{zz} = \frac{2}{c^2} ), off-diagonals zero.

Thus det(Hess F) = ( \left( -\frac{2}{a^2} \right) \left( -\frac{2}{b^2} \right) \left( \frac{2}{c^2} \right) = \frac{8}{a^2 b^2 c^2} ).

Gradient: ( \nabla F = \left( -\frac{2x}{a^2}, -\frac{2y}{b^2}, \frac{2z}{c^2} \right) ).

At point (x,y,z) = (0,0,3) for our surface: a^2=1? Wait: our equation: ( \frac{z^2}{9} - x^2 - \frac{y^2}{4} = 1 ). So a^2=1, b^2=4, c^2=9.

Thus at (0,0,3): ( \nabla F = (0, 0, \frac{2 \cdot 3}{9}) = (0, 0, \frac{2}{3}) ).
( | \nabla F | = \frac{2}{3} ).

Now Gaussian curvature formula for implicit surface:
( \kappa = \frac{ \det( \text{Hess}(F) ) - (\nabla F)^T \text{adj}( \text{Hess}(F) ) \nabla F }{ | \nabla F |^4 } ).

But Hess(F) is diagonal with entries ( -2/a^2, -2/b^2, 2/c^2 ). At (0,0,3), Hessian is diagonal: diag(-2, -2/4=-0.5, 2/9). Actually:
( F_{xx} = -2/a^2 = -2 ), ( F_{yy} = -2/b^2 = -2/4 = -0.5 ), ( F_{zz} = 2/c^2 = 2/9 ).

det(Hess) = (-2) * (-0.5) * (2/9) = (1) * (2/9) = 2/9.

adj(Hess) = diag( (F_{yy}F_{zz} - 0), (F_{zz}F_{xx} - 0), (F_{xx}F_{yy} - 0) ) because diagonal matrix: adj = diag( a22 a33, a11 a33, a11 a22 ).
So:
a22 a33 = (-0.5)(2/9) = -1/9.
a11 a33 = (-2)(2/9) = -4/9.
a11 a22 = (-2)*(-0.5) = 1.

Thus adj(Hess) = diag(-1/9, -4/9, 1).

Now compute ( (\nabla F)^T \text{adj}(Hess) \nabla F ): gradient = (0,0,2/3). So only third component matters:
= (2/3)^2 * adj_{33} = (4/9) * 1 = 4/9.

Thus numerator = det(Hess) - that = 2/9 - 4/9 = -2/9.

Denominator = ( | \nabla F |^4 = (2/3)^4 = 16/81 ).

Thus ( \kappa = \frac{-2/9}{16/81} = \frac{-2}{9} \cdot \frac{81}{16} = \frac{-2 \cdot 9}{16} = \frac{-18}{16} = -\frac{9}{8} ).

So absolute value ( |\kappa| = 9/8 ).

Thus ( \beta = 9/8 ).

Now we have:
( \alpha = 2 )
( \beta = 9/8 )
( \delta = 32/3 )
( \rho = 2 )

Compute ( S = (\alpha \beta \delta \rho)^3 = \left( 2 \cdot \frac{9}{8} \cdot \frac{32}{3} \cdot 2 \right)^3 ).

Simplify inside:
First: ( 2 \cdot \frac{9}{8} = \frac{18}{8} = \frac{9}{4} ).
Multiply by ( \frac{32}{3} ): ( \frac{9}{4} \cdot \frac{32}{3} = \frac{9 \cdot 32}{4 \cdot 3} = \frac{288}{12} = 24 ).
Multiply by ( \rho = 2 ): ( 24 \cdot 2 = 48 ).

Thus ( S = 48^3 = 48 \cdot 48 \cdot 48 = 110592 ) (since 48^2=2304, times 48 = 110592).

Now ( \mathcal{K} = 999 S^2 + 7 ).

Compute ( S^2 = 110592^2 ). Let’s compute with Python.

Then ( \mathcal{K} = 999 * S^2 + 7 ), and we need integer ( \mathcal{K} ).

Let’s verify all steps with Python. [/THOUGHT] [PYTHON] import sympy as sp

(I-1) Length L

a = sp.symbols('a', positive=True)

Compute L = 8a

L = 8*a

(I-2) Curvature kappa_L(t)

t = sp.symbols('t') kappa = 1/(4asp.sin(t/2)) kappa_at_pi = kappa.subs(t, sp.pi) alpha = L * kappa_at_pi alpha_simplified = sp.simplify(alpha) print("alpha =", alpha_simplified)

(II-1) Surface equation

We already derived: z^2/9 - x^2 - y^2/4 = 1

(II-2) Intercepts

At z=5: x^2 + y^2/4 = 16/9 => A_x = 4/3, A_y = 8/3

Ax = sp.Rational(4,3) Ay = sp.Rational(8,3)

At y=0: z^2/9 - x^2 = 1 => z-intercept positive: z=3 => A_z = 3

Az = 3

delta = Ax * Ay * Az rho = Ay / Ax print("delta =", delta) print("rho =", rho)

(II-4) Gaussian curvature at u=v=0

Using implicit surface formula: F = z^2/9 - x^2 - y^2/4 - 1 = 0

a^2=1, b^2=4, c^2=9

Hessian diag(-2, -2/4, 2/9) = diag(-2, -0.5, 2/9)

det(Hess) = (-2)(-0.5)(2/9) = (1)*(2/9) = 2/9

adj(Hess) diag( (-0.5)(2/9), (-2)(2/9), (-2)*(-0.5) ) = diag(-1/9, -4/9, 1)

gradient at (0,0,3): (0,0,2*3/9) = (0,0,2/3)

numerator = det(Hess) - (grad^T adj(Hess) grad) = 2/9 - ( (2/3)^2 * 1 ) = 2/9 - 4/9 = -2/9

denominator = ||grad||^4 = (2/3)^4 = 16/81

kappa = (-2/9) / (16/81) = (-2/9)*(81/16) = -18/16 = -9/8

beta = sp.Rational(9,8) print("beta =", beta)

Now compute S and K

S = (alpha_simplified * beta * delta * rho)**3 S_simplified = sp.simplify(S) print("S =", S_simplified) S_val = sp.nsimplify(S_simplified) print("S as integer?", S_val)

S should be 48^3 = 110592

print("S numeric:", int(S_val))

K = 999 * S_val**2 + 7 print("K =", K)

Compute S^2 exactly

S_squared = S_val**2 print("S^2 =", S_squared) print("K exact:", K)

Convert to integer

K_int = int(K) print("K as integer:", K_int) [/PYTHON]

[THOUGHT] The Python output shows ( \mathcal{K} = 12218359873543 ). This is the final integer answer.

\boxed{12218359873543}