Problem Analysis #3

[THOUGHT] The surface is a torus with major radius 2, minor radius 1. For part I, at P corresponding to θ=π/4, φ=π, the inward unit normal is (1/√2, 1/√2, 0), so n1 + n2 = √2, (n1 + n2)^2 = 2. For II, A_S = 4π² - 4π. For III, V_ov = 2π² - 4π/3. Then V_ov/(2π) - A_S/(4π) = 1/3. For IV, verify κ_Q at t=π/2 for c(t) = ((2 + cos t) cos t, (2 + cos t) sin t, sin t) using sympy to compute ||c' × c''|| / ||c'||^3, expecting √41 / 5^{3/2}, so κ_Q^2 = 41/125. Then the expression simplifies to 999 * 250 + 7 = 249757. Now verify κ_Q with code. [/THOUGHT] [PYTHON] import sympy as sp

t = sp.symbols('t') alpha = 2 + sp.cos(t) cx = alpha * sp.cos(t) cy = alpha * sp.sin(t) cz = sp.sin(t)

c = sp.Matrix([cx, cy, cz]) c1 = sp.diff(c, t) c2 = sp.diff(c1, t)

t_val = sp.pi / 2 c1_val = c1.subs(t, t_val) c2_val = c2.subs(t, t_val)

cross = c1_val.cross(c2_val) norm_cross = sp.sqrt(cross.dot(cross)) norm_c1 = sp.sqrt(c1_val.dot(c1_val)) kappa = norm_cross / norm_c1**3

print(kappa) print(float(kappa2)) # to check 41/125=0.328 print(kappa2.simplify()) [/PYTHON]

[THOUGHT] Fix the SyntaxError by removing comments and the float line. Compute kappa exactly and simplify kappa**2. [/THOUGHT] [PYTHON] import sympy as sp

t = sp.symbols('t') alpha = 2 + sp.cos(t) cx = alpha * sp.cos(t) cy = alpha * sp.sin(t) cz = sp.sin(t)

c = sp.Matrix([cx, cy, cz]) c1 = c.diff(t) c2 = c1.diff(t)

t_val = sp.pi / 2 c1_val = c1.subs(t, t_val) c2_val = c2.subs(t, t_val)

cross = c1_val.cross(c2_val) norm_cross = sp.sqrt(cross.dot(cross)) norm_c1 = sp.sqrt(c1_val.dot(c1_val)) kappa = norm_cross / norm_c1**3

print(kappa) print(kappa.simplify()) print((kappa**2).simplify()) [/PYTHON]

[THOUGHT] The calculations confirm: $\mathbf{n} = \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \right)$, so $(n_1 + n_2)^2 = 2$. $A_S = 4\pi^2 - 4\pi$. $V_{\mathrm{ov}} = 2\pi^2 - \frac{4\pi}{3}$. Thus $\frac{V_{\mathrm{ov}}}{2\pi} - \frac{A_S}{4\pi} = \frac{1}{3}$. $\kappa_Q^2 = \frac{41}{125}$. Then $\mathcal{K} = 999 \cdot 123 \cdot \frac{1}{3} \cdot \frac{2}{41/125} + 7 = 999 \cdot 123 \cdot \frac{1}{3} \cdot \frac{250}{41} + 7$. Since $123 = 3 \cdot 41$, this simplifies to $333 \cdot 3 \cdot 250 + 7 = 999 \cdot 250 + 7 = 249750 + 7 = 249757$. [/THOUGHT] $\boxed{249757}$

Explanation

The reasoning derives the values for each component of $\mathcal{K}$ systematically from the surface parametrization and problem constraints.

For part I, the point $P = \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \right)$ corresponds to $\theta = \pi/4$, $\phi = \pi$. Compute partial derivatives: $\mathbf{r}_\theta = \left( -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \right)$, $\mathbf{r}_\phi = (0, 0, -1)$. The cross product $\mathbf{r}_\theta \times \mathbf{r}_\phi = \left( -\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0 \right)$ has unit length and points outward for the solid torus $V$. Thus, the inward unit normal is $\mathbf{n} = \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0 \right)$, so $n_1 + n_2 = \sqrt{2}$ and $(n_1 + n_2)^2 = 2$.

For part II, the area element is $\|( \mathbf{r}_\theta \times \mathbf{r}_\phi )\| = 2 + \cos \phi$. Region $W$ requires $\sqrt{x^2 + y^2} = 2 + \cos \phi \leq 2$, so $\cos \phi \leq 0$ or $\phi \in [\pi/2, 3\pi/2]$. Thus, $A_S = \int_0^{2\pi} \int_{\pi/2}^{3\pi/2} (2 + \cos \phi) \, d\phi \, d\theta = 2\pi \left[ 2\phi + \sin \phi \right]_{\pi/2}^{3\pi/2} = 2\pi (2\pi - 2) = 4\pi^2 - 4\pi$.

For part III, volume $V_{\mathrm{ov}}$ uses cylindrical coordinates: $V_{\mathrm{ov}} = 4\pi \int_1^2 \rho \sqrt{1 - (\rho - 2)^2} \, d\rho$. Substitute $u = \rho - 2$, yielding $\int_{-1}^0 (u + 2) \sqrt{1 - u^2} \, du = -\frac{1}{3} + \frac{\pi}{2}$. Thus, $V_{\mathrm{ov}} = 4\pi \left( \frac{\pi}{2} - \frac{1}{3} \right) = 2\pi^2 - \frac{4\pi}{3}$.

Then, $\frac{V_{\mathrm{ov}}}{2\pi} = \pi - \frac{2}{3}$ and $\frac{A_S}{4\pi} = \pi - 1$, so $\frac{V_{\mathrm{ov}}}{2\pi} - \frac{A_S}{4\pi} = \frac{1}{3}$.

For part IV, curve $C$: $\mathbf{c}(t) = \left( (2 + \cos t) \cos t, (2 + \cos t) \sin t, \sin t \right)$. At $Q$ ($t = \pi/2$), Python computation gives $\kappa_Q = \sqrt{205}/25$, so $\kappa_Q^2 = 41/125$.

Substitute into $\mathcal{K}$: $999 \cdot 123 \cdot \frac{1}{3} \cdot \frac{2}{41/125} + 7 = 999 \cdot 123 \cdot \frac{1}{3} \cdot \frac{250}{41}$. Since $123 = 3 \cdot 41$, simplify to $333 \cdot 3 \cdot 250 + 7 = 999 \cdot 250 + 7 = 249750 + 7 = 249757$.