# Surface Integrals - what do they mean and how do I choose my normal vector?

QUESTION I AM STUCK ON:

I have a surface, $S$, defined by $8x=y^2$, bounded by the planes $y=4, z=6$. We define a vector field by $\textbf{F} = 2y\textbf{i} - z\textbf{j} + x^2 \textbf{k}$.

We want to work out: $\int_{S} (\textbf{F} \cdot \hat{\textbf{n}} ) dS$ by projecting onto the plane $x=0$, where $\hat{\textbf{n}}$ points in the direction of increasing $x$.

MY CONFUSION:

Firstly, what exactly are we working out? It's not the area of the surface is it? (Out of interest, what would that be?)

Secondly, I've been having trouble finding the vector in the direction of increasing [insert variable here]. How would I go about doing this?

EDIT: Does "where $\hat{\textbf{n}}$ points in the direction of increasing $x$" mean that you're taking the normal vector to the surface $8x=y^2$, and choosing the direction such that it has a non-negative x-coordinate?

• Surface area would be $\int_S 1 dS$. – mathematician Oct 28 '16 at 20:59
• There is a bound missing. – Faraad Armwood Oct 28 '16 at 21:08

## 1 Answer

$\bullet$ No this is not surface area and here is why:

Consider a small patch $S_0$ on your surface $S$. Now consider a parametrization $G: D \to S$ where $D$ has coordinate functions $u,v$. Let $(u_0,v_0) \in D$ then we have:

$$G(u_0,v_0) - G(u_0+\Delta u,y_0) \approx G_u(u_0,v_0)\Delta u$$

$$G(u_0,v_0) - G(u_0,v_0+\Delta v) \approx G_v(u_0,v_0)\ \Delta v$$

Hence, the normal vector for $S_0$ is approximately $G_u(u_0,v_0) \Delta u \times G_v(u_0,v_0) \Delta v := \vec{n}(u_0,v_0) \Delta u \Delta v$. The magnitude of this vector greatly approximates the area of the patch $S_0$ i.e we have:

$$\textbf{Area}(S_0) \approx \|\vec{n}(u_0,v_0) \Delta u \Delta v\| = \|\vec{n}(u_0,v_0)\| \Delta u \Delta v$$

Thus, if we take a partition of $D$ say $\{(x_i,y_j): i \in [0,N], j \in [0,M]\}$ which also has the property that:

$$\lim_{N,M \to \infty}\sum_{i,j} \|\vec{n}(u_i,v_j)\|\ \Delta u \Delta v< \infty$$

then we know:

$$\textbf{Surface Area of S}=\lim_{N,M \to \infty} \sum_{j=0}^M \sum_{i=0}^N \|\vec{n}(u_i,v_j)\| \ \Delta u \Delta v = \iint_D \|\vec{n}(u,v)\| \ du \ dv$$

$\bullet$ In the case of vector fields, the integral you are trying to compute is flux. The flux of a surface measures how much fluid is flowing either from the inside of the surface to the outside or the outside of the surface to the inside. The reason you have to choose $\vec{n}$ i.e a normal for your surface is because you need a way to how the fluid is flowing (where to the two flows are the ones I've mentioned).

$\bullet$ To show that this integral is calculating exactly that, again we restrict ourselves to looking at what is going on at a small patch. Let $S_0$ be as above and consider $\vec{n}(u_0,v_0) \Delta v \Delta$ defined above as well.

$\bullet$ What may really be confusing you is your integral. I think it is off. By definition the surface element $dS=\|\vec{n}(u,v)\| \ du dv$ and so the integral should read:

$$\int_S \vec{F} \cdot \textbf{e}_{\vec{n}} \ dS = \int_s \vec{F} \cdot \frac{\vec{n}(u,v)}{\|\vec{n}(u,v)\|} \ dS=\iint_D \vec{F}(G(u,v)) \cdot \vec{n}(u,v) \ du \ dv$$

Using the fact that $u \cdot v = \|u\| \|v\| \cos \theta_{\vec{u},\vec{v}}$ then it is from here, the integral makes sense because:

$$\vec{F} \cdot \textbf{e}_{\vec{n}} = \|\vec{F}\| \cos \theta_{\vec{F}, \vec{n}} = \|\textrm{proj} \vec{F}_{\vec{n}}\|$$

i.e the above quantity multiplied by $\|\vec{n}(u,v)\| \Delta u \Delta v$ gives the volume of fluid passing through the patch $S_0$.