[THOUGHT] We need to solve a multi-step problem. The surface S is given by a parametrization: r(θ, φ) = R_z(θ) * ( (cos φ + 2, 0, sin φ)^T ), where R_z(θ) is rotation about z-axis.

That is: x(θ, φ) = cos θ (cos φ + 2) - sin θ * 0 = cos θ (cos φ + 2) y(θ, φ) = sin θ (cos φ + 2) + cos θ * 0 = sin θ (cos φ + 2) z(θ, φ) = sin φ

So S is a surface of revolution about z-axis? Actually, note that (cos φ + 2, 0, sin φ) is a circle in xz-plane centered at (2,0,0) with radius 1? Wait: x = cos φ + 2, z = sin φ, y = 0. That's a circle in xz-plane centered at (2,0,0), radius 1. Then we rotate this circle around z-axis by angle θ. So S is a torus? But not exactly: The circle is offset from z-axis by distance 2, radius 1, rotated around z-axis yields a torus with major radius 2 and minor radius 1. Indeed: standard torus param: ((R + r cos φ) cos θ, (R + r cos φ) sin θ, r sin φ). Here R = 2, r = 1, but note: our param: x = cos θ (cos φ + 2) = 2 cos θ + cos θ cos φ, y = 2 sin θ + sin θ cos φ, z = sin φ. That matches torus with R=2, r=1, but with cos φ and sin φ swapped? Usually torus: x = (R + r cos φ) cos θ, y = (R + r cos φ) sin θ, z = r sin φ. Here we have cos φ + 2 = 2 + cos φ, so r=1, R=2, but the roles: cos φ is the radial variation in the x-direction of the circle before rotation, yes. So S is a torus centered at origin with major radius 2 (distance from center of tube to origin) and minor radius 1 (radius of tube). So V is the interior of this torus? Wait: "Let V be the region surrounded by the surface S". For a torus, "surrounded" means the solid torus, i.e., the volume inside the torus. So V is the set of points inside the torus.

W is region x^2 + y^2 ≤ 4, i.e., a vertical cylinder of radius 2.

Now we need to answer four subproblems, then compute K.

We'll need to compute: I. Unit normal vector oriented inward at point P = (1/√2, 1/√2, 0). First find parameters (θ, φ) that give P. Solve: x = cos θ (cos φ + 2) = 1/√2 y = sin θ (cos φ + 2) = 1/√2 z = sin φ = 0 From z = 0 => sin φ = 0 => φ = 0 or π. cos φ = 1 or -1. If φ = 0, cos φ = 1 => cos θ (1+2) = 3 cos θ = 1/√2 => cos θ = 1/(3√2) ≈ 0.2357. sin θ = 1/(3√2) similarly? But y = sin θ * 3 = 1/√2 => sin θ = 1/(3√2). Then cos θ = sin θ, so θ = π/4? Actually cos θ = sin θ => θ = π/4 or 5π/4. But cos θ = 1/(3√2) ≈ 0.2357, sin θ same, so indeed θ = π/4 works? sin(π/4)=cos(π/4)=√2/2 ≈ 0.7071, not 0.2357. So not π/4. So maybe φ = π? cos φ = -1, then cos θ ( -1 + 2) = cos θ * 1 = 1/√2 => cos θ = 1/√2. sin θ * 1 = 1/√2 => sin θ = 1/√2. So θ = π/4 (or 5π/4? but sin and cos positive => first quadrant). So φ = π, θ = π/4 gives P. Check: x = cos(π/4)(cos π + 2) = (√2/2)(-1+2) = (√2/2)*1 = 1/√2. y = sin(π/4)*1 = 1/√2. z = sin π = 0. Yes.

So parameters: θ = π/4, φ = π.

Now we need inward unit normal vector at P. The surface normal vector is given by cross product of partial derivatives r_θ × r_φ (or r_φ × r_θ depending on orientation). We need orientation inward (into region V). V is inside torus. At point P, which direction is inward? For a torus, inward means toward the interior of the solid torus, i.e., toward the "hole" or toward the center of the tube? Actually, the solid torus is the region inside the torus surface. The normal pointing inward points into the volume V. We can compute the normal from parametrization: r_θ × r_φ gives a normal vector. Then we check sign: maybe we need to adjust to point inward. We can compute the normal vector at P and then determine if it points inward or outward by evaluating dot product with vector from a point inside V to P. Alternatively, we can compute the gradient of implicit equation of torus.

But easier: compute r_θ × r_φ, then see direction relative to center of tube. However, we can compute the inward normal by considering geometry: At point P, which is on the outer equator? Actually, φ = π corresponds to point on the inner side of the torus? Let's visualize: For φ = 0, the point on the circle before rotation is (3,0,0). After rotation, that's the farthest point from origin (outer equator). For φ = π, the point is (1,0,0) before rotation, which is the closest to origin (inner equator). So P is on the inner equator (since φ = π). The inward direction (into V) at inner equator points radially outward from the origin? Wait: The solid torus is the doughnut. At the inner equator, the surface is like the inner wall of the doughnut hole. The interior of the torus is inside the doughnut, which is away from the hole. So at the inner wall, inward points away from the hole, i.e., radially outward from the origin? Actually, consider a point on the inner equator: the interior of the torus is the region with larger distance from the origin? Let's think: For a torus with major radius R=2, minor radius r=1, the surface is points at distance from origin between R-r = 1 and R+r = 3. The inner equator is at distance R - r = 1 from origin. The interior of torus includes points with distance from origin > 1? Actually, the torus volume includes points that are within distance r from the circle of radius R centered at origin. So at a point on inner equator (distance 1 from origin), the interior is towards the region where distance from origin is larger (since the tube is centered at distance R=2). So inward points radially outward (away from origin). At outer equator (distance 3), inward points radially inward (toward origin). So at P, distance from origin = sqrt((1/√2)^2+(1/√2)^2+0^2)= sqrt(1/2+1/2)=1. So P is at distance 1. So inward should point radially outward. So normal vector at P should be parallel to position vector P (since outward radial direction is (x,y,0)). But we need unit normal. Let's compute via cross product.

We'll compute r_θ and r_φ at (θ=π/4, φ=π).

First, write r(θ, φ) = (cos θ (cos φ + 2), sin θ (cos φ + 2), sin φ).

Compute partial derivatives: r_θ = ( -sin θ (cos φ + 2), cos θ (cos φ + 2), 0 ) r_φ = ( cos θ (-sin φ), sin θ (-sin φ), cos φ )

At φ = π, cos φ = -1, sin φ = 0. So cos φ + 2 = 1. Thus: r_θ = ( -sin θ * 1, cos θ * 1, 0 ) = ( -sin θ, cos θ, 0 ) r_φ = ( cos θ * 0, sin θ * 0, cos φ ) = ( 0, 0, cos φ ) = (0, 0, -1) because cos φ = -1.

At θ = π/4: sin θ = cos θ = √2/2 ≈ 0.7071. So r_θ = ( -√2/2, √2/2, 0 ) r_φ = (0, 0, -1)

Now compute cross product r_θ × r_φ. Use right-hand rule: order? Usually normal is r_θ × r_φ (or opposite). Let's compute r_θ × r_φ: i j k -√2/2 √2/2 0 0 0 -1 Determinant: i*( (√2/2)(-1) - 00 ) - j*( (-√2/2)(-1) - 00 ) + k*( (-√2/2)0 - (√2/2)0 ) = i( -√2/2 ) - j( √2/2 ) + k*(0) = ( -√2/2, -√2/2, 0 ). So r_θ × r_φ = (-√2/2, -√2/2, 0). Its magnitude = sqrt( (√2/2)^2 + (√2/2)^2 ) = sqrt( 1/2 + 1/2 ) = sqrt(1) = 1. So it's already a unit vector? Check magnitude: each component squared: (√2/2)^2 = 2/4 = 0.5, sum = 1, yes magnitude 1. So unit normal from this cross product is n = (-√2/2, -√2/2, 0). This points radially inward toward origin? Because P = (1/√2, 1/√2, 0) = (√2/2, √2/2, 0). n = (-√2/2, -√2/2, 0) = -P. So n points opposite to P, i.e., radially inward. But we want inward normal (into V). At inner equator, inward should be radially outward. So we need the opposite direction: n_inward = -n = (√2/2, √2/2, 0) = P normalized? P's magnitude is 1, so P is unit vector. Indeed P = (1/√2, 1/√2, 0) = (√2/2, √2/2, 0). So inward unit normal is exactly P. So n1 = 1/√2, n2 = 1/√2, n3 = 0.

Thus answer I: n = (1/√2, 1/√2, 0). We'll keep as exact values.

Now II: Area of portion of S included in region W: x^2 + y^2 ≤ 4. S is torus. W is cylinder radius 2. So we need surface area of part of torus inside cylinder. Since torus is symmetric, maybe we can integrate over parameters where condition holds.

Parametrization: x^2 + y^2 = (cos φ + 2)^2 (cos^2 θ + sin^2 θ) = (cos φ + 2)^2. So condition x^2 + y^2 ≤ 4 becomes (cos φ + 2)^2 ≤ 4 => |cos φ + 2| ≤ 2. Since cos φ + 2 ≥ 1 (since cos φ ≥ -1 => min 1), it's positive. So cos φ + 2 ≤ 2 => cos φ ≤ 0. Also cos φ + 2 ≥ -2? Not needed. So cos φ ≤ 0. That's the condition. Also note cos φ + 2 ≥ 0 always. So inequality reduces to cos φ ≤ 0. So φ in [π/2, 3π/2] (since cos φ ≤ 0 in those intervals). But φ ∈ [0, 2π). So φ ∈ [π/2, 3π/2].

Thus the portion of S inside cylinder is where φ ∈ [π/2, 3π/2] (and θ ∈ [0, 2π)). So we need surface area integral over that region.

Surface area element: dS = || r_θ × r_φ || dθ dφ. We computed cross product magnitude earlier? Let's compute generally.

We have r_θ = (-sin θ (cos φ + 2), cos θ (cos φ + 2), 0) r_φ = (cos θ (-sin φ), sin θ (-sin φ), cos φ)

Compute cross product symbolically. Let a = cos φ + 2. Then r_θ = (-sin θ * a, cos θ * a, 0), r_φ = (cos θ * (-sin φ), sin θ * (-sin φ), cos φ). Compute r_θ × r_φ: i component: (cos θ * a)cos φ - 0(-sin θ sin φ) = a cos θ cos φ. Wait compute properly: r_θ × r_φ = determinant: i ( (cos θ a)cos φ - 0(-sin θ sin φ) ) = a cos θ cos φ. j ( (-sin θ a)cos φ - 0(cos θ (-sin φ)) ) = -a sin θ cos φ. k ( (-sin θ a)(-sin θ sin φ) - (cos θ a)(cos θ (-sin φ)) ) = compute: first term: (-sin θ a)(-sin θ sin φ) = a sin^2 θ sin φ. second term: (cos θ a)(cos θ (-sin φ)) = -a cos^2 θ sin φ. So k = a sin^2 θ sin φ - (-a cos^2 θ sin φ) = a sin^2 θ sin φ + a cos^2 θ sin φ = a sin φ (sin^2 θ + cos^2 θ) = a sin φ. Thus r_θ × r_φ = (a cos θ cos φ, -a sin θ cos φ, a sin φ). Magnitude squared: a^2 cos^2 θ cos^2 φ + a^2 sin^2 θ cos^2 φ + a^2 sin^2 φ = a^2 cos^2 φ (cos^2 θ + sin^2 θ) + a^2 sin^2 φ = a^2 cos^2 φ + a^2 sin^2 φ = a^2 (cos^2 φ + sin^2 φ) = a^2. So || r_θ × r_φ || = |a| = |cos φ + 2|. Since cos φ + 2 ≥ 1 > 0, it's cos φ + 2.

Thus dS = (cos φ + 2) dθ dφ.

So area = ∫∫ (cos φ + 2) dθ dφ over region: θ from 0 to 2π, φ from π/2 to 3π/2. Area = ∫{θ=0}^{2π} dθ ∫{φ=π/2}^{3π/2} (cos φ + 2) dφ = 2π * ∫_{π/2}^{3π/2} (cos φ + 2) dφ. Compute ∫ cos φ dφ = sin φ, from π/2 to 3π/2: sin(3π/2) - sin(π/2) = -1 - 1 = -2. ∫ 2 dφ = 2φ, from π/2 to 3π/2: 2*(3π/2 - π/2) = 2*π = 2π. So total integral = -2 + 2π = 2π - 2. Thus area = 2π * (2π - 2) = 4π^2 - 4π.

But wait: is that correct? Let's double-check: ∫{π/2}^{3π/2} cos φ dφ = sin φ |{π/2}^{3π/2} = sin(3π/2) - sin(π/2) = -1 - 1 = -2. Yes. So area = 2π*(2π - 2) = 4π^2 - 4π.

Thus A_S = 4π^2 - 4π.

Now III: Volume of overlapping region V ∩ W. V is solid torus (interior of torus), W is cylinder x^2 + y^2 ≤ 4. So we need volume of part of torus inside cylinder.

We can compute using integration. Use toroidal coordinates or use method of washers/slicing by z.

Parametrization of solid torus: (x,y,z) such that (√(x^2+y^2) - 2)^2 + z^2 ≤ 1 (since minor radius 1). That's the inequality for torus interior. So V: (√(x^2+y^2) - 2)^2 + z^2 ≤ 1. W: x^2 + y^2 ≤ 4.

We need volume of intersection.

We can integrate in cylindrical coordinates (r, θ, z). Condition for torus: (r - 2)^2 + z^2 ≤ 1. So r ranges? Solve for r: r must satisfy 2 - √(1 - z^2) ≤ r ≤ 2 + √(1 - z^2), provided |z| ≤ 1. Also r ≥ 0.

Condition from cylinder: r ≤ 2.

So intersection region: |z| ≤ 1, and for each z, r must satisfy both: lower bound from torus: r ≥ 2 - √(1 - z^2), upper bound from torus: r ≤ 2 + √(1 - z^2), and from cylinder: r ≤ 2. So effective upper bound is min(2, 2 + √(1 - z^2)) = 2 because 2 + √(1 - z^2) ≥ 2 (since √ ≥ 0). So r ≤ 2. Lower bound is max(0, 2 - √(1 - z^2)). Since 2 - √(1 - z^2) ≥ 1 (because √ ≤ 1 => 2 - √ ≥ 1). So lower bound is 2 - √(1 - z^2). So r ranges from 2 - √(1 - z^2) to 2, provided that lower bound ≤ upper bound, i.e., 2 - √(1 - z^2) ≤ 2, which is true. Also need to ensure that lower bound ≥ 0, which is true.

Thus region is defined for each z with |z| ≤ 1, and for each θ from 0 to 2π (full rotation). So volume V_ov = ∫{z=-1}^{1} ∫{θ=0}^{2π} ∫_{r = 2 - √(1 - z^2)}^{2} r dr dθ dz.

Compute inner integral: ∫ r dr = (1/2)(2^2 - (2 - √(1 - z^2))^2) = (1/2)(4 - (4 - 4√(1 - z^2) + (1 - z^2))) = (1/2)(4 - 4 + 4√(1 - z^2) - 1 + z^2) = (1/2)(4√(1 - z^2) + z^2 - 1).

Then multiply by 2π dθ: gives π (4√(1 - z^2) + z^2 - 1) dz.

So V_ov = π ∫_{z=-1}^{1} (4√(1 - z^2) + z^2 - 1) dz.

By symmetry, integrate from 0 to 1 and double.

Compute ∫{-1}^{1} √(1 - z^2) dz = area of half circle radius 1 times 2? Actually ∫{-1}^{1} √(1 - z^2) dz = area of semicircle of radius 1 (top half) times 2? Wait, √(1 - z^2) from -1 to 1 gives area of full circle radius 1? No, it's area of half circle (y = √(1 - z^2)) from -1 to 1 gives area π/2, but we have both positive and negative? Actually, √(1 - z^2) is nonnegative, so integral from -1 to 1 is area of upper half of unit circle = π/2 * 1^2 = π/2. So ∫_{-1}^{1} √(1 - z^2) dz = π/2.

∫_{-1}^{1} z^2 dz = 2 ∫_0^1 z^2 dz = 2 * (1/3) = 2/3.

∫_{-1}^{1} (-1) dz = -2.

Thus total integral = 4*(π/2) + 2/3 - 2 = 2π + 2/3 - 2 = 2π - 4/3? Wait: 2/3 - 2 = -4/3. So = 2π - 4/3.

Thus V_ov = π * (2π - 4/3) = 2π^2 - (4π/3).

Check: V_ov = 2π^2 - (4π/3).

Now IV: Curvature of curve C defined by θ = φ. So let t = θ = φ. Then curve c(t) = (cos t (cos t + 2), sin t (cos t + 2), sin t). Simplify: c(t) = (cos t (cos t + 2), sin t (cos t + 2), sin t) = (cos^2 t + 2 cos t, sin t cos t + 2 sin t, sin t). Actually sin t (cos t + 2) = sin t cos t + 2 sin t.

We need curvature at point Q = (0,2,1). Find t such that c(t) = Q. Solve: x = cos^2 t + 2 cos t = 0 y = sin t cos t + 2 sin t = 2 z = sin t = 1 From z = 1 => sin t = 1 => t = π/2 + 2kπ. Take t = π/2. Check x: cos(π/2)=0, cos^2 t = 0, so x = 0 + 0 = 0. Good. y: sin t =1, cos t =0, so sin t cos t =0, plus 2 sin t = 2*1 =2. So y=2. Good. So t = π/2.

Now compute curvature κ(t) = || c' × c'' || / || c' ||^3. We need derivatives at t = π/2.

Compute c(t) symbolically. Let's compute with Python to avoid mistakes.

We'll compute c'(t) and c''(t) analytically or numerically. Use sympy.

Now we have all components to compute K = 999 * 123 * (V_ov/(2π) - A_S/(4π)) * ((n1 + n2)^2) / (κ_Q^2) + 7.

Compute step by step.

First, compute n1 + n2 = 1/√2 + 1/√2 = √2. So (n1 + n2)^2 = 2.

Compute V_ov/(2π) = (2π^2 - 4π/3) / (2π) = π - 2/3.

Compute A_S/(4π) = (4π^2 - 4π) / (4π) = π - 1.

So V_ov/(2π) - A_S/(4π) = (π - 2/3) - (π - 1) = -2/3 + 1 = 1/3.

Thus product: 999 * 123 * (1/3) * 2 / (κ_Q^2) + 7.

Now need κ_Q.

Let's compute κ at t = π/2 using Python.

We'll do symbolic differentiation.

[/THOUGHT] [PYTHON] import sympy as sp

define t

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

define c(t)

x = sp.cos(t)(sp.cos(t) + 2) y = sp.sin(t)(sp.cos(t) + 2) z = sp.sin(t) c = sp.Matrix([x, y, z])

compute first and second derivatives

c_prime = c.diff(t) c_double = c_prime.diff(t)

compute cross product

cross = c_prime.cross(c_double) norm_cross = sp.sqrt(cross.dot(cross)) norm_c_prime = sp.sqrt(c_prime.dot(c_prime))

curvature formula

kappa = norm_cross / (norm_c_prime**3)

simplify expression

kappa_simplified = sp.simplify(kappa) print("kappa(t) =", kappa_simplified)

evaluate at t = pi/2

kappa_at_pi2 = kappa_simplified.subs(t, sp.pi/2) print("kappa at pi/2 =", kappa_at_pi2) print("Numerical value:", kappa_at_pi2.evalf()) [/PYTHON]

[THOUGHT] From the Python output, κ_Q = sqrt(205)/25. So κ_Q^2 = 205/625 = 41/125.

Now compute K = 999 * 123 * (1/3) * 2 / (41/125) + 7.

First compute 999 * 123 = 999 * 100 + 999 * 23 = 99900 + 22977 = 122877. Then 122877 * (1/3) = 40959. Then 40959 * 2 = 81918. Then divide by (41/125) = 81918 * (125/41) = (81918 * 125) / 41. Compute 81918 / 41: 41 * 1998 = 412000 - 412 = 82000 - 82 = 81918. So 81918 / 41 = 1998. Then 1998 * 125 = 1998 * 100 + 1998 * 25 = 199800 + 49950 = 249750. Finally add 7: 249750 + 7 = 249757. Thus K = 249757. [/THOUGHT] \boxed{249757}