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 $\mathcal{K}$.\