Problem Analysis #3

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Problem Statement

In the three-dimensional orthogonal coordinate system $xyz$ , consider the surface $S$ defined by $\begin{pmatrix}x(\theta, \phi) \\y(\theta, \phi) \\z(\theta, \phi)\end{pmatrix}=\begin{pmatrix}\cos\theta & -\sin\theta & 0 \\\sin\theta & \cos\theta & 0 \\0 & 0 & 1\end{pmatrix}\begin{pmatrix}\cos\phi + 2 \\0 \\\sin\phi\end{pmatrix},$ where $\theta$ and $\phi$ are parameters of the surface $S$ , and $0 \leq \theta < 2\pi,\qquad 0 \leq \phi < 2\pi.$ Let $V$ be the region surrounded by the surface $S$ , and let $W$ be the region satisfying the inequality $x^2 + y^2 \leq 4$ . Answer the following questions for the surface $S$ .

[I.] Find the unit normal vector oriented inward the region $V$ at the point $P = \begin{pmatrix} \dfrac{1}{\sqrt{2}} \\ \dfrac{1}{\sqrt{2}} \\ 0 \end{pmatrix}$ on the surface $S$ .
[II.] Find the area of the portion of the surface $S$ included in the region $W$ .
[III.] Find the volume of the overlapping region $V \cap W$ .
[IV.] Consider the curve $C$ on the surface $S$ defined by setting $\theta = \phi$ in the parametrization of $S$ . Find the curvature of the curve $C$ at the point $Q = \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}$ on the curve $C$ . Note that, in general, given a three-dimensional curve defined by $c(t) =\begin{pmatrix}x(t) \\y(t) \\z(t)\end{pmatrix}$ with parameter $t$ , the curvature $\kappa(t)$ of the curve at the point $c(t)$ is given by $\kappa(t)= \frac{\left\lVert \dfrac{dc(t)}{dt} \times \dfrac{d^2c(t)}{dt^2}\right\rVert} {\left\lVert \dfrac{dc(t)}{dt}\right\rVert^3}.$ Let $\mathbf{n} = (n_1,n_2,n_3)$ be the inward unit normal vector at $P$ obtained in {\rm I}. Let $A_S$ be the area obtained in {\rm II}, and let $V_{\mathrm{ov}}$ be the overlapping volume $V\cap W$ obtained in {\rm III}. Let $\kappa_Q$ be the curvature at $Q$ obtained in {\rm IV}.Define $\mathcal{K}= 999 \cdot 123 \cdot\left(\frac{V_{\mathrm{ov}}}{2\pi}- \frac{A_S}{4\pi}\right)\cdot\frac{(n_1 + n_2)^2}{\kappa_Q^2}\;+\; 7.$ Compute the integer $\mathcal{K}$ .\

Ground Truth

249757

Execution Trace

Reasoning Process

[THOUGHT] Let's break down the problem step-by-step.

Part I: Find the unit normal vector oriented inward at point (P)

Parametrization of Surface (S): [ \begin{pmatrix} x(\theta, \phi) \ y(\theta, \phi) \ z(\theta, \phi) \end{pmatrix}

\begin{pmatrix} \cos\theta & -\sin\theta & 0 \ \sin\theta & \cos\theta & 0 \ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \cos\phi + 2 \ 0 \ \sin\phi \end{pmatrix}

\begin{pmatrix} \cos\theta (\cos\phi + 2) \ \sin\theta (\cos\phi + 2) \ \sin\phi \end{pmatrix} ]
Compute Tangent Vectors: [ \frac{\partial \mathbf{r}}{\partial \theta} = \begin{pmatrix} -\sin\theta (\cos\phi + 2) \ \cos\theta (\cos\phi + 2) \ 0 \end{pmatrix} ] [ \frac{\partial \mathbf{r}}{\partial \phi} = \begin{pmatrix} -\cos\theta \sin\phi \ -\sin\theta \sin\phi \ \cos\phi \end{pmatrix} ]
Find Cross Product: [ \frac{\partial \mathbf{r}}{\partial \theta} \times \frac{\partial \mathbf{r}}{\partial \phi} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -\sin\theta (\cos\phi + 2) & \cos\theta (\cos\phi + 2) & 0 \ -\cos\theta \sin\phi & -\sin\theta \sin\phi & \cos\phi \end{vmatrix}

\begin{pmatrix} \cos\theta (\cos\phi + 2) \cos\phi \ \sin\theta (\cos\phi + 2) \cos\phi \ (\cos\phi + 2) \sin^2\theta + (\cos\phi + 2) \cos^2\theta \end{pmatrix}

\begin{pmatrix} \cos\theta (\cos\phi + 2) \cos\phi \ \sin\theta (\cos\phi + 2) \cos\phi \ (\cos\phi + 2) \end{pmatrix} ]
Normalize the Cross Product: [ \left| \frac{\partial \mathbf{r}}{\partial \theta} \times \frac{\partial \mathbf{r}}{\partial \phi} \right| = \sqrt{\cos^2\theta (\cos\phi + 2)^2 \cos^2\phi + \sin^2\theta (\cos\phi + 2)^2 \cos^2\phi + (\cos\phi + 2)^2}

(\cos\phi + 2) \sqrt{\cos^2\phi + 1} ] [ \mathbf{n} = \frac{1}{(\cos\phi + 2) \sqrt{\cos^2\phi + 1}} \begin{pmatrix} \cos\theta (\cos\phi + 2) \cos\phi \ \sin\theta (\cos\phi + 2) \cos\phi \ (\cos\phi + 2) \end{pmatrix}

\frac{1}{\sqrt{\cos^2\phi + 1}} \begin{pmatrix} \cos\theta \cos\phi \ \sin\theta \cos\phi \ 1 \end{pmatrix} ]
Determine (\theta) and (\phi) at Point (P): [ P = \begin{pmatrix} \frac{1}{\sqrt{2}} \ \frac{1}{\sqrt{2}} \ 0 \end{pmatrix} ] [ \cos\theta (\cos\phi + 2) = \frac{1}{\sqrt{2}}, \quad \sin\theta (\cos\phi + 2) = \frac{1}{\sqrt{2}}, \quad \sin\phi = 0 ] This implies (\cos\phi = -1) or (\cos\phi = 0). Since (\sin\phi = 0), (\cos\phi = 1). Hence, (\theta = \frac{\pi}{4}).
Calculate ( \mathbf{n} ) at (P): [ \mathbf{n} = \frac{1}{\sqrt{1^2 + 1}} \begin{pmatrix} \cos\frac{\pi}{4} \cdot 1 \ \sin\frac{\pi}{4} \cdot 1 \ 1 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} \frac{1}{\sqrt{2}} \ \frac{1}{\sqrt{2}} \ 1 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} \frac{1}{\sqrt{2}} \ \frac{1}{\sqrt{2}} \ 1 \end{pmatrix}

\begin{pmatrix} \frac{1}{2} \ \frac{1}{2} \ \frac{1}{\sqrt{2}} \end{pmatrix} ] Since we need the inward normal, we take the negative: [ \mathbf{n} = \begin{pmatrix} -\frac{1}{2} \ -\frac{1}{2} \ -\frac{1}{\sqrt{2}} \end{pmatrix} ]

Part II: Find the area of the portion of the surface (S) included in the region (W)

Region (W) Condition: [ x^2 + y^2 \leq 4 ] From parametrization: [ x^2 + y^2 = (\cos\theta (\cos\phi + 2))^2 + (\sin\theta (\cos\phi + 2))^2 = (\cos\phi + 2)^2 ] Hence, [ (\cos\phi + 2)^2 \leq 4 \implies \cos\phi + 2 \leq 2 \implies \cos\phi \leq 0 \implies \phi \in \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] ]
Surface Area Integral: [ A_S = \int_{0}^{2\pi} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \left| \frac{\partial \mathbf{r}}{\partial \theta} \times \frac{\partial \mathbf{r}}{\partial \phi} \right| , d\phi , d\theta = \int_{0}^{2\pi} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} (\cos\phi + 2) \sqrt{\cos^2\phi + 1} , d\phi , d\theta ] Let's compute this integral using Python. [/THOUGHT] [PYTHON] import numpy as np from scipy.integrate import dblquad

Define the integrand

def integrand(phi, theta): return (np.cos(phi) + 2) * np.sqrt(np.cos(phi)**2 + 1)

Integrate over the specified ranges

area_s, _ = dblquad(integrand, 0, 2 * np.pi, lambda theta: np.pi / 2, lambda theta: 3 * np.pi / 2) print(area_s)

Tool Output

[PYTHON OUTPUT] 31.85323152880882 [/PYTHON OUTPUT]

Reasoning Process

[THOUGHT] Given the Python output for the area, ( A_S \approx 31.85323152880882 ). This output should be treated as accurate. Let’s continue with the next steps using this value. [/THOUGHT] [THOUGHT]

Part III: Find the volume of the overlapping region (V \cap W)

Region (W) Condition: [ x^2 + y^2 \leq 4 ] From parametrization: [ x^2 + y^2 = (\cos\theta (\cos\phi + 2))^2 + (\sin\theta (\cos\phi + 2))^2 = (\cos\phi + 2)^2 \leq 4 \implies \cos\phi \leq 0 \implies \phi \in \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] ]
Volume Integral: [ V_{\text{ov}} = \int_{0}^{2\pi} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \int_{0}^{\sin\phi} r , dz , d\phi , d\theta ] where ( r = \cos\phi + 2 ).

The integral simplifies to: [ V_{\text{ov}} = \int_{0}^{2\pi} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} (\cos\phi + 2) \sin\phi , d\phi , d\theta ] Let’s compute this integral using Python. [/THOUGHT] [PYTHON] from scipy.integrate import dblquad

Define the integrand for the volume

def integrand_volume(phi, theta): return (np.cos(phi) + 2) * np.sin(phi)

Integrate over the specified ranges

volume_ov, _ = dblquad(integrand_volume, 0, 2 * np.pi, lambda theta: np.pi / 2, lambda theta: 3 * np.pi / 2) print(volume_ov)

Tool Output

[PYTHON OUTPUT] 2.205547862066261e-15 [/PYTHON OUTPUT]

Reasoning Process

[THOUGHT] The volume of the overlapping region ( V \cap W ) is approximately ( 2.205547862066261 \times 10^{-15} ), which is extremely close to zero. This implies that the volume of the overlapping region is effectively zero. We will use ( V_{\text{ov}} = 0 ) for further calculations.

Next, we proceed with Part IV: Find the curvature of the curve ( C ) at point ( Q ).

Part IV: Find the curvature of the curve ( C ) at point ( Q )

Parametrization of Curve ( C ): Set ( \theta = \phi ): [ c(\theta) = \begin{pmatrix} \cos\theta (\cos\theta + 2) \ \sin\theta (\cos\theta + 2) \ \sin\theta \end{pmatrix} ]
First Derivative: [ c'(\theta) = \begin{pmatrix} -\sin\theta (\cos\theta + 2) + \cos^2\theta \ \cos\theta (\cos\theta + 2) - \sin^2\theta \ \cos\theta \end{pmatrix}

\begin{pmatrix} -\sin\theta \cos\theta - 2\sin\theta + \cos^2\theta \ \cos^2\theta + 2\cos\theta - \sin^2\theta \ \cos\theta \end{pmatrix} ]
Second Derivative: [ c''(\theta) = \begin{pmatrix} -\cos^2\theta + \sin^2\theta - 2\cos\theta \ -2\sin\theta\cos\theta + 2\sin\theta - 2\cos\theta \ -\sin\theta \end{pmatrix}

\begin{pmatrix} -\cos^2\theta + \sin^2\theta - 2\cos\theta \ -\sin(2\theta) + 2\sin\theta - 2\cos\theta \ -\sin\theta \end{pmatrix} ]
Determine ( \theta ) at Point ( Q ): [ Q = \begin{pmatrix} 0 \ 2 \ 1 \end{pmatrix} ] [ \cos\theta (\cos\theta + 2) = 0, \quad \sin\theta (\cos\theta + 2) = 2, \quad \sin\theta = 1 ] [ \theta = \frac{\pi}{2} ]
Evaluate ( c'(\theta) ) and ( c''(\theta) ) at ( \theta = \frac{\pi}{2} ): [ c'\left(\frac{\pi}{2}\right) = \begin{pmatrix} -1 \cdot 2 + 0 \ 0 + 2 - 1 \ 0 \end{pmatrix}

\begin{pmatrix} -2 \ 1 \ 0 \end{pmatrix} ] [ c''\left(\frac{\pi}{2}\right) = \begin{pmatrix} -0 + 1 - 0 \ -0 + 2 - 0 \ -1 \end{pmatrix}

\begin{pmatrix} 1 \ 2 \ -1 \end{pmatrix} ]
Cross Product: [ c'(\theta) \times c''(\theta) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -2 & 1 & 0 \ 1 & 2 & -1 \end{vmatrix}

\begin{pmatrix} -1 \ -2 \ -5 \end{pmatrix} ]
Magnitudes: [ \left| c'(\theta) \times c''(\theta) \right| = \sqrt{(-1)^2 + (-2)^2 + (-5)^2} = \sqrt{30} ] [ \left| c'(\theta) \right| = \sqrt{(-2)^2 + 1^2 + 0^2} = \sqrt{5} ]
Curvature: [ \kappa_Q = \frac{\left| c'(\theta) \times c''(\theta) \right|}{\left| c'(\theta) \right|^3} = \frac{\sqrt{30}}{(\sqrt{5})^3} = \frac{\sqrt{30}}{5\sqrt{5}} = \frac{\sqrt{6}}{5} ]

Calculation of (\mathcal{K})

Given: [ \mathbf{n} = \begin{pmatrix} -0.5 \ -0.5 \ -\frac{1}{\sqrt{2}} \end{pmatrix} ] [ n_1 + n_2 = -0.5 - 0.5 = -1 ] [ (n_1 + n_2)^2 = (-1)^2 = 1 ] [ \kappa_Q = \frac{\sqrt{6}}{5} ] [ \kappa_Q^2 = \left(\frac{\sqrt{6}}{5}\right)^2 = \frac{6}{25} ] [ \frac{(n_1 + n_2)^2}{\kappa_Q^2} = \frac{1}{\frac{6}{25}} = \frac{25}{6} ] [ V_{\text{ov}} = 0, \quad A_S \approx 31.85323152880882 ] [ \frac{V_{\text{ov}}}{2\pi} - \frac{A_S}{4\pi} = 0 - \frac{31.85323152880882}{4\pi} = -\frac{31.85323152880882}{4\pi} ] [ \mathcal{K} = 999 \cdot 123 \cdot \left(-\frac{31.85323152880882}{4\pi}\right) \cdot \frac{25}{6} + 7 ] [ \mathcal{K} = -999 \cdot 123 \cdot \frac{31.85323152880882}{4\pi} \cdot \frac{25}{6} + 7 ] Let's compute this value using Python. [/THOUGHT] [PYTHON] import math

Given values

V_ov = 0 A_S = 31.85323152880882 k_Q = math.sqrt(6) / 5

Calculate the components of the final expression

part1 = (V_ov / (2 * math.pi)) - (A_S / (4 * math.pi)) part2 = (25 / 6) final_value = 999 * 123 * part1 * part2 + 7 print(final_value)

Tool Output

[PYTHON OUTPUT] -1297778.7233273766 [/PYTHON OUTPUT]

Reasoning Process

[THOUGHT] The Python output for (\mathcal{K}) is approximately (-1297778.7233273766). Since we are required to provide the integer value, we round this to the nearest integer. [/THOUGHT] \boxed{-1297779}

Final Model Output

-1297779

Comparative Analysis (P3) (Average across runs)

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Grok-4.1-fast

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Llama-VARCO-8B-Instruct

6884.0

Problem Analysis #3

Part I: Find the unit normal vector oriented inward at point (P)

Parametrization of Surface (S): [ \begin{pmatrix} x(\theta, \phi) \ y(\theta, \phi) \ z(\theta, \phi) \end{pmatrix}

\begin{pmatrix} \cos\theta & -\sin\theta & 0 \ \sin\theta & \cos\theta & 0 \ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \cos\phi + 2 \ 0 \ \sin\phi \end{pmatrix}

\begin{pmatrix} \cos\theta (\cos\phi + 2) \cos\phi \ \sin\theta (\cos\phi + 2) \cos\phi \ (\cos\phi + 2) \sin^2\theta + (\cos\phi + 2) \cos^2\theta \end{pmatrix}

Normalize the Cross Product: [ \left| \frac{\partial \mathbf{r}}{\partial \theta} \times \frac{\partial \mathbf{r}}{\partial \phi} \right| = \sqrt{\cos^2\theta (\cos\phi + 2)^2 \cos^2\phi + \sin^2\theta (\cos\phi + 2)^2 \cos^2\phi + (\cos\phi + 2)^2}

(\cos\phi + 2) \sqrt{\cos^2\phi + 1} ] [ \mathbf{n} = \frac{1}{(\cos\phi + 2) \sqrt{\cos^2\phi + 1}} \begin{pmatrix} \cos\theta (\cos\phi + 2) \cos\phi \ \sin\theta (\cos\phi + 2) \cos\phi \ (\cos\phi + 2) \end{pmatrix}

Part II: Find the area of the portion of the surface (S) included in the region (W)

Define the integrand

Integrate over the specified ranges

Part III: Find the volume of the overlapping region (V \cap W)

Define the integrand for the volume

Integrate over the specified ranges

Part IV: Find the curvature of the curve ( C ) at point ( Q )

First Derivative: [ c'(\theta) = \begin{pmatrix} -\sin\theta (\cos\theta + 2) + \cos^2\theta \ \cos\theta (\cos\theta + 2) - \sin^2\theta \ \cos\theta \end{pmatrix}

Second Derivative: [ c''(\theta) = \begin{pmatrix} -\cos^2\theta + \sin^2\theta - 2\cos\theta \ -2\sin\theta\cos\theta + 2\sin\theta - 2\cos\theta \ -\sin\theta \end{pmatrix}

Evaluate ( c'(\theta) ) and ( c''(\theta) ) at ( \theta = \frac{\pi}{2} ): [ c'\left(\frac{\pi}{2}\right) = \begin{pmatrix} -1 \cdot 2 + 0 \ 0 + 2 - 1 \ 0 \end{pmatrix}

\begin{pmatrix} -2 \ 1 \ 0 \end{pmatrix} ] [ c''\left(\frac{\pi}{2}\right) = \begin{pmatrix} -0 + 1 - 0 \ -0 + 2 - 0 \ -1 \end{pmatrix}

Cross Product: [ c'(\theta) \times c''(\theta) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -2 & 1 & 0 \ 1 & 2 & -1 \end{vmatrix}

Calculation of (\mathcal{K})

Given values

Calculate the components of the final expression

Comparative Analysis (P3) (Average across runs)

Avg Time Taken

Avg Token Usage