Lecture 2 Surface Integrals & Flux

Course: Calculus & Linear Algebra (TU Delft)
Topic: Lecture 2 - Surface Integrals & Flux (The "How-To" Guide)

1. Scalar Surface Integrals (Mass, Charge, Area)

Use this when you are integrating a "flat" number (a scalar function $f$ ) over a "curvy" surface.

The "Stretch Factor" Logic

To integrate over a curve, we must account for how much the surface is "stretched" compared to the flat $u v$ -plane. This is the $d S$ factor.

\iint_{S} f d S = \iint_{D} f (r (u, v)) \cdot \underset{The Stretch Factor}{\underset{⏟}{| r_{u} \times r_{v} |}} d u d v

The Workflow (Example: Charge on a Cylinder)

If you have $λ = x$ on a cylinder $y^{2} + z^{2} = 9$ (Radius 3) from $x = 0$ to $6$ :

Parametrize: $x = x, y = 3 \cos θ, z = 3 \sin θ$ .
Find $d S$ : For a cylinder, $d S = R d x d θ = 3 dx d θ$ .
Set up: $\int_{0}^{2 π} \int_{0}^{6} (x) \cdot (3) d x d θ$ .
Solve: $3 \cdot [\frac{1}{2} x^{2}]_{0}^{6} \cdot [θ]_{0}^{2 π} = 108 π$ .

2. Flux Integrals (Vector Flow)

Use this to measure how much of a vector field $F$ is "piercing" through a surface.

The Orientation Rule

Flux requires a direction.

Outward: Default for closed solids (spheres, boxes).
Upward: Positive $z$ -component.
Rightwards: Positive $y$ -component (like our square $y = 4$ example).

The Workflow (Example: The "Bowl" $z = 4 - x^{2} - y^{2}$ )

Find the Normal Vector ( $d S$ ): For $z = g (x, y)$ , the upward vector is $⟨ - g_{x}, - g_{y}, 1 ⟩$ .
- $n = ⟨ 2 x, 2 y, 1 ⟩$ .
Dot Product: $F \cdot n$ .
- If $F = ⟨ y, - x, z + 1 ⟩$ , then $F \cdot n = 2 x y - 2 x y + (z + 1) = z + 1$ .
Substitution: You must replace $z$ with the surface equation $4 - x^{2} - y^{2}$ .
- Integrand: $(4 - r^{2}) + 1 = 5 - r^{2}$ .
Integrate: Use Polar! $\int_{0}^{2 π} \int_{0}^{2} (5 - r^{2}) r d r d θ$ .

3. Dealing with Piecewise Surfaces (Closed Solids)

When a surface has a "top" and a "bottom" (like a pill or a capped bowl), you must calculate them separately.

The Consistency Check

Make sure both normals point OUT of the solid.

Top: Usually $n = ⟨ \dots, \dots, 1 ⟩$ (Up).
Bottom: Usually $n = ⟨ 0, 0, - 1 ⟩$ (Down).

Example (The Floor $z = 0$ ):

If $n = ⟨ 0, 0, - 1 ⟩$ and $F = ⟨ \dots, z + 1 ⟩$ :
Dot product $= - (z + 1)$ .
At the floor ( $z = 0$ ), this is $- 1$ .
Total Flux through floor $= (- 1) \cdot (Area of the base)$ .

4. Calculus Shortcuts & Integration Tricks

Save time during the exam with these three patterns.

1. Separable Variables

If your integral looks like $\int \int g (u) h (v) d u d v$ , you can split it:

(\int g (u) d u) \cdot (\int h (v) d v)

Used in: The cylinder charge problem ( $x$ and $θ$ separated).

2. Symmetry Cancellations

If you integrate an odd function (like $x, y, \sin θ$ ) over a symmetric domain (like a full circle), the result is 0.
Look for this in flux dot products to delete terms early!

3. The $z = f (x, y)$ Area Shortcut

If the problem asks for Surface Area (Scalar integral of $f = 1$ ):

d S = \sqrt{1 + (f_{x})^{2} + (f_{y})^{2}} d A

Used in: The cone problem $z = 25 \sqrt{x^{2} + y^{2}}$ where $d S = \sqrt{1 + 25^{2}} d A$ .

5. Summary Table for $d S$ and $d S$

Surface	Parametrization	Scalar $d S$ (Mass)	Vector $d S$ (Flux)
Sphere ( $R$ )	Spherical $(ϕ, θ)$	$R^{2} \sin ϕ d ϕ d θ$	$⟨ x, y, z ⟩ R \sin ϕ d ϕ d θ$
Cylinder ( $R$ )	$(z, θ)$	$R d z d θ$	$⟨ \cos θ, \sin θ, 0 ⟩ R d z d θ$
Graph $z = g$	$(x, y)$	$\sqrt{1 + g_{x}^{2} + g_{y}^{2}} d A$	$⟨ - g_{x}, - g_{y}, 1 ⟩ d A$