Course: Calculus & Linear Algebra (TU Delft)
Topic: Lecture 1 - Surface Parametrization, Tangents, and Area

1. Finding Parameters $(u,v)$ from a Point $P$

Before calculating vectors, you must know which input values $(u,v)$ produced the point $(x,y,z)$.

2. Finding the Normal Vector ($\mathbf{n}$)

The normal vector is perpendicular to the surface at a specific point.

Steps:

1. Calculate partial derivatives ${\mathbf{r}}_u$ and ${\mathbf{r}}_v$.
2. Plug in the numerical values for $u$ and $v$.
3. Perform the Cross Product:$$\mathbf{n}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ x_u& y_u& z_u\\ x_v& y_v& z_v\end{array}\right|$$
4. Verification: Ensure $\mathbf{n}\cdot {\mathbf{r}}_u=0$ and $\mathbf{n}\cdot {\mathbf{r}}_v=0$.

3. Parametrizing a Sphere

Using spherical coordinates is the standard way to describe a sphere of radius $R$.

Finding Domain Limits ($a\le u\le b$):
If the sphere is cut by planes (e.g., between $z=5$ and $z=-5$):

1. Use the $z$ formula: $z=R\cos (u)$.
2. Solve for $u$: $u=\arccos (z/R)$.
3. Remember: $u$ (or $\varphi$) always stays between $0$ and $\pi$.

4. Equation of the Tangent Plane

A flat plane touching the surface at point $P(x_0,y_0,z_0)$.

Geometric Shortcut:
If the surface is a sphere centered at the origin, the normal vector $\mathbf{n}$ at point $P$ is simply the vector from the origin to $P$:

5. Surface Area Element ($dS$)

Used to calculate the area of a small "patch" on the surface.

Steps:

1. Find the normal vector $\mathbf{n}={\mathbf{r}}_u\times {\mathbf{r}}_v$.
2. Calculate the Magnitude (Scalar):
   $|\mathbf{n}|=\sqrt{x^2+y^2+z^2}$.
3. Multiply by the given change in parameters ($du$ and $dv$).
4. Result: Always a positive scalar (number), not a vector.

6. Total Surface Area for $z=f(x,y)$

When the surface is a "topography" defined by a function $z=f(x,y)$ over a region $D$ in the $xy$-plane.

The "Circular Boundary" Workflow

If the region $D$ is defined by a cylinder or circle (e.g., $x^2+y^2=R^2$), follow these steps:

1. Calculate Partial Derivatives: Find $f_x$ and $f_y$.
2. Set up the Integral: Plug them into $\sqrt{1+f_x^2+f_y^2}$.
3. Convert to Polar Coordinates:
   • Replace $x^2+y^2$ with $r^2$.
   • Replace $dA$ with $rdrd\theta$ (Don't forget the $r$!).
   • Set limits: $0\le \theta \le 2\pi$ and $0\le r\le R$.
4. Solve using u-substitution: Usually, the "inside" of the square root becomes your $u$.

Example Case ($z=5xy$):

• $f_x=5y$, $f_y=5x$
• Integrand: $\sqrt{1+25y^2+25x^2}=\sqrt{1+25(x^2+y^2)}$
• Polar: ${\int }_0^{2\pi }{\int }_0^R\sqrt{1+25r^2}\cdot rdrd\theta$

Why this works:

This is actually the same as the parametric method. If you treat $x$ and $y$ as your parameters ($u$ and $v$), the parametrization is $\mathbf{r}(x,y)=⟨x,y,f(x,y)⟩$. The magnitude of the normal vector $|{\mathbf{r}}_x\times {\mathbf{r}}_y|$ always simplifies down to $\sqrt{1+f_x^2+f_y^2}$.

7. The Torus (The "Donut" Shape) - Pappus's Theorem

A torus is created by taking a circle of radius $r$ and rotating it around an axis at a distance $R$ from its center.

The Standard Parametrization

For a torus centered on the $z$-axis:

• $x=(R+r\cos v)\cos u$
• $y=(R+r\cos v)\sin u$
• $z=r\sin v$
  (Where $R$ = major radius to center of tube, $r$ = minor radius of the tube itself).

The "Product of Circumferences" Shortcut

The surface area of a torus is simply the circumference of the small circle multiplied by the circumference of the path it travels.

From your exercise:

• $R=8$ (distance from origin)
• $r=5$ (radius of the vertical circle)
• Area $=4\cdot {\pi }^2\cdot 8\cdot 5=\mathbf{160}{\pi }^{\mathbf{2}}$

Integration Detail (The "Why")

If you are asked to show the work, the magnitude of the cross product (the surface area element) for a torus always simplifies to:

When you integrate this from $0\to 2\pi$ for both $u$ and $v$, the $\cos v$ term integrates to $0$ over a full circle, leaving you with just the $2\pi \cdot 2\pi \cdot R\cdot r$ result.

Tags: #calculus #linear-algebra #parametric-surfaces #surface-area #tudelft