# Surface Integral, flux. Boundary and orientation

• Liferider
In summary, the boundary of a flat surface does not influence the flux integral through its normal vector. The normal vector only plays a role in determining the direction of the flux, not its magnitude. The magnitude of the normal vector should be unit length, and the projection of the flux component with respect to the normal is used to calculate the flux. The unit normal can be defined by dividing the vector function by the length of the gradient, or by using the vector differential of surface area. Both methods are equivalent in calculating the flux integral.
Liferider
In solving a flux integral over a flat surface, inclined above the xy-plane, does the boundary of the surface influence the flux only through the integral limits? (and not through its normal vector)

Let's say that there is an elliptic surface inclined above the xy-plane. The orientation is given by the plane: z=3-y. Now, when I am supposed to solve this surface integral, it seems I can easily use the normal vector of the plane z=3-y in the dS conversion:

dS = [-(∂f/∂x)$^{2}$, -(∂f/∂y)$^{2}$, 1]dA, where f(x,y)=3-y

This means that I do not need to think about the parametrization of the elliptic surface...
This would ofcourse only be true for flat surfaces. So, for clarification, am I doing something wrong?

To clarify, because this might seem like a strange question... Does the magnitude of the Normal vector play a role, or just its direction?

Liferider said:
To clarify, because this might seem like a strange question... Does the magnitude of the Normal vector play a role, or just its direction?

As a disclaimer you should probably get some further input since I haven't taken vector calculus in a long time, but as far as I understand it, the normal vector should be a unit normal and the magnitude should be contained in the information for the actual flux component itself.

If the magnitude was not unit (i.e. 1) then you would have to just normalize the integral in the appropriate way. You have to remember that you are calculating a projection of the flux with respect to the surface normal and if you review the definition of such a projection, you should find that the normal needs to be unit length.

This implies that only the direction matters and not the magnitude.

Thanks, so the reason why one projects the surface onto a plane like the xy-plane is to find how much of the different vector field components go trough the surface.

Liferider said:
Thanks, so the reason why one projects the surface onto a plane like the xy-plane is to find how much of the different vector field components go trough the surface.

In flux integrals its best to think of each point on your surface having its own normal vector at a point. On a plane, this doesn't change but on a curved surface it does.

What you are doing is just doing a projection of the 'flux component' with respect to the normal.

If the flux is parallel to the normal then the magnitude of the flux is at a maximum since it is going through the surface with maximum intensity. If the normal is perpendicular, then the flux will not even touch the surface resulting in a zero contribution since the flux is parallel to the surface itself.

The answer could be either "yes" or "no" depending upon exactly how you define the "normal vector". If you are given some vector function, $\vec{v}(s,t)$, defined over a surface F(s, t)= constant, to integrate the vector function over the surface by finding the gradient, $\nabla F$ which is, of course, normal to the surface.

Many texts form the unit normal, $\vec{v}$, by dividing that by the length $\left|\nabla F\right|$ but then form the "differential of surface area" $dS= \left|\nabla F\right|dsdt$. But I have never seen much sense in dividing by $\left|\nabla\right|$ to get the unit normal, then multiplying by it to get the differential of surface area! Instead, I prefer to define the "vector differential of surface area", as done in many other texts, $\vec{dS}= \nabla F dsdt$.

On the one hand you can calculate the flux as
$$\int \vec{v}\cdot\vec{n}dS= \int \vec{v}\cdot\vec{n} \left|\nabla F\right| dsdt$$
or as
$$\int \vec{v}\cdot\vec{dS}= \int \vec{v}\cdot\nabla F dsdt$$

I prefer the second but they are equivalent.

## 1. What is a surface integral?

A surface integral is a mathematical tool used in multivariable calculus to find the total flux (i.e. flow) of a vector field across a given surface. It involves breaking the surface into small pieces and calculating the flux through each piece, then summing up all the individual fluxes to find the total flux.

## 2. What is flux?

Flux, in the context of surface integrals, refers to the amount of flow (e.g. heat, liquid, etc.) passing through a given surface. It is a measure of the quantity of a vector field that is crossing the surface per unit time.

## 3. How is orientation determined in a surface integral?

The orientation of a surface is typically determined by the direction of its normal vector, which is a vector perpendicular to the surface at each point. The orientation of the surface is important in calculating the surface integral, as it affects the direction of the flux through the surface.

## 4. Can a surface integral be negative?

Yes, a surface integral can be negative. This can occur if the vector field is pointing in the opposite direction of the surface's orientation, resulting in a negative flux. It is important to pay attention to the orientation of the surface when calculating a surface integral to ensure the correct sign.

## 5. How is the boundary of a surface related to a surface integral?

The boundary of a surface is important in determining the limits of integration for a surface integral. The surface integral is typically evaluated by integrating over a closed surface, so the boundary of the surface serves as the limits of integration for the integral. Additionally, the boundary may also affect the orientation of the surface and thus the direction of the flux through the surface.

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