1. **Problem:** Consider the surface in $\mathbb{R}^3$ parameterized by $\vec{\Phi}(r, \theta) = (r \cos \theta, r \sin \theta, \theta)$ with $0 \leq r \leq 1$ and $0 \leq \theta \leq 4\pi$. (a) **Sketch and describe the surface:** - The surface is formed by points where the $x$ and $y$ coordinates lie on a disk of radius $r$ in the plane, and the $z$ coordinate equals the angle $\theta$. - As $\theta$ increases from $0$ to $4\pi$, the surface makes two full turns around the $z$-axis, creating a spiral ramp or helical surface. - The radius $r$ varies from $0$ (center axis) to $1$ (outer edge), so the surface looks like a spiral disk ascending along $z$. (b) **Find a unit normal to the surface:** - Compute partial derivatives: $$\vec{\Phi}_r = \frac{\partial \vec{\Phi}}{\partial r} = (\cos \theta, \sin \theta, 0)$$ $$\vec{\Phi}_\theta = \frac{\partial \vec{\Phi}}{\partial \theta} = (-r \sin \theta, r \cos \theta, 1)$$ - The normal vector is the cross product: $$\vec{N} = \vec{\Phi}_r \times \vec{\Phi}_\theta = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \cos \theta & \sin \theta & 0 \\ -r \sin \theta & r \cos \theta & 1 \end{vmatrix} = (\sin \theta, -\cos \theta, r)$$ - The magnitude is: $$\|\vec{N}\| = \sqrt{\sin^2 \theta + \cos^2 \theta + r^2} = \sqrt{1 + r^2}$$ - Unit normal: $$\hat{\mathbf{n}} = \frac{1}{\sqrt{1 + r^2}} (\sin \theta, -\cos \theta, r)$$ (c) **Equation of the tangent plane at $(x_0, y_0, z_0)$:** - The tangent plane at $\vec{\Phi}(r_0, \theta_0)$ is given by: $$\vec{N} \cdot (\vec{x} - \vec{\Phi}(r_0, \theta_0)) = 0$$ - Using $\vec{N} = (\sin \theta_0, -\cos \theta_0, r_0)$ and $\vec{\Phi}(r_0, \theta_0) = (r_0 \cos \theta_0, r_0 \sin \theta_0, \theta_0)$: $$\sin \theta_0 (x - r_0 \cos \theta_0) - \cos \theta_0 (y - r_0 \sin \theta_0) + r_0 (z - \theta_0) = 0$$ (d) **Show the horizontal line segment of unit length from the $z$-axis through $(x_0, y_0, z_0)$ lies in the surface and tangent plane:** - The point $(x_0, y_0, z_0)$ corresponds to $(r_0, \theta_0)$ with $x_0 = r_0 \cos \theta_0$, $y_0 = r_0 \sin \theta_0$, $z_0 = \theta_0$. - The horizontal line segment from the $z$-axis through $(x_0, y_0, z_0)$ of unit length is: $$\vec{L}(t) = (t \cos \theta_0, t \sin \theta_0, \theta_0), \quad 0 \leq t \leq 1$$ - For each $t$, $\vec{L}(t) = \vec{\Phi}(t, \theta_0)$ lies on the surface. - Check if $\vec{L}(t)$ lies in the tangent plane: Substitute into tangent plane equation: $$\sin \theta_0 (t \cos \theta_0 - r_0 \cos \theta_0) - \cos \theta_0 (t \sin \theta_0 - r_0 \sin \theta_0) + r_0 (\theta_0 - \theta_0) = 0$$ Simplifies to zero, so the line segment lies in the tangent plane. 2. **Problem:** Parameterization of a torus: $$\vec{\Phi}(u,v) = ((R + r \cos u) \cos v, (R + r \cos u) \sin v, r \sin u)$$ with $0 \leq u,v \leq 2\pi$, $0 < r < 1$. (a) **Show points satisfy:** $$ (\sqrt{x^2 + y^2} - R)^2 + z^2 = r^2 $$ - Compute $\sqrt{x^2 + y^2} = R + r \cos u$. - Substitute: $$ (R + r \cos u - R)^2 + (r \sin u)^2 = (r \cos u)^2 + (r \sin u)^2 = r^2 (\cos^2 u + \sin^2 u) = r^2 $$ - Hence, the equation holds. (b) **Show surface is regular:** - Compute partial derivatives: $$\vec{\Phi}_u = (-r \sin u \cos v, -r \sin u \sin v, r \cos u)$$ $$\vec{\Phi}_v = (-(R + r \cos u) \sin v, (R + r \cos u) \cos v, 0)$$ - The cross product $\vec{\Phi}_u \times \vec{\Phi}_v$ is nonzero for all $u,v$ because $r > 0$ and $R + r \cos u > 0$. - Thus, the surface is regular everywhere. 3. **Problem:** Parametrize surface $x^2 - y^2 = 1$ with $x > 0$, $-1 \leq y \leq 1$, $0 \leq z \leq 1$. - Use hyperbolic functions: $$x = \cosh t, \quad y = \sinh t, \quad t \in [-\sinh^{-1}(1), \sinh^{-1}(1)]$$ - Parametrization: $$\vec{\Phi}(t,z) = (\cosh t, \sinh t, z), \quad z \in [0,1]$$ - Area element: $$\vec{\Phi}_t = (\sinh t, \cosh t, 0), \quad \vec{\Phi}_z = (0,0,1)$$ - Cross product magnitude: $$\|\vec{\Phi}_t \times \vec{\Phi}_z\| = \sqrt{\cosh^2 t + \sinh^2 t} = \sqrt{\cosh(2t)}$$ - Area integral: $$A = \int_0^1 \int_{-\sinh^{-1}(1)}^{\sinh^{-1}(1)} \sqrt{\cosh(2t)} \, dt \, dz = \int_{-\sinh^{-1}(1)}^{\sinh^{-1}(1)} \sqrt{\cosh(2t)} \, dt$$ 4. **Problem:** Verify in spherical coordinates on sphere radius $R$: $$\|\vec{T}_\phi \times \vec{T}_\theta\| d\phi d\theta = R^2 \sin \phi \, d\phi d\theta$$ - Spherical coordinates: $$\vec{r}(\phi, \theta) = (R \sin \phi \cos \theta, R \sin \phi \sin \theta, R \cos \phi)$$ - Compute partial derivatives: $$\vec{T}_\phi = \frac{\partial \vec{r}}{\partial \phi} = (R \cos \phi \cos \theta, R \cos \phi \sin \theta, -R \sin \phi)$$ $$\vec{T}_\theta = \frac{\partial \vec{r}}{\partial \theta} = (-R \sin \phi \sin \theta, R \sin \phi \cos \theta, 0)$$ - Cross product magnitude: $$\|\vec{T}_\phi \times \vec{T}_\theta\| = R^2 \sin \phi$$ - Hence: $$\|\vec{T}_\phi \times \vec{T}_\theta\| d\phi d\theta = R^2 \sin \phi \, d\phi d\theta$$ **Final answers:** - Unit normal: $\hat{\mathbf{n}} = \frac{1}{\sqrt{1 + r^2}} (\sin \theta, -\cos \theta, r)$ - Tangent plane: $\sin \theta_0 (x - r_0 \cos \theta_0) - \cos \theta_0 (y - r_0 \sin \theta_0) + r_0 (z - \theta_0) = 0$ - Torus satisfies $(\sqrt{x^2 + y^2} - R)^2 + z^2 = r^2$ - Torus is regular everywhere - Parametrization of hyperbolic surface: $\vec{\Phi}(t,z) = (\cosh t, \sinh t, z)$ - Area integral: $A = \int_{-\sinh^{-1}(1)}^{\sinh^{-1}(1)} \sqrt{\cosh(2t)} \, dt$ - Spherical coordinates area element verified: $\|\vec{T}_\phi \times \vec{T}_\theta\| d\phi d\theta = R^2 \sin \phi \, d\phi d\theta$