# Surface Integrals and Gauss's Law

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1. Jan 18, 2015

### Brennan999

When I learned Integrals in Calc III, the formula looked like this
∫∫ F(r(s,t))⋅(rs x rt)*dA
but in physics for Gauss's law it is
∫∫E⋅nhat dA
How are these the same basic formula? I know that nhat is a unit vector, so it is n/|n|, but in the actual equation, it is a dot between the cross product and E.

The problem is I dont know how they are the same since Gauss' Law wants E dot a magnitude, when a surface integral is F dot the vector cross product? They dont seem to be following the same rules

2. Jan 18, 2015

### Staff: Mentor

Let me guess: rs and rt are perpendicular unit vectors within the surface, correct?

Chet

3. Jan 18, 2015

### Brennan999

Yeah, I get that you need to cross them to get the perpendicular vector of the surface.
Basically, the surface integral eq in full form is
∫∫S F ⋅ (rs x rt)/|(rs x rt)| *(rs x rt) dA
and for Gauss's Law it is
∫∫S E⋅n/|n| dA

so you can see in the first one it simplifies to the cross product as a vector,
but the second one simplifies to the unit vector (because of dividing in magnitude) in the direction normal to the surface.

I dont get how these are equal

4. Jan 19, 2015

### Staff: Mentor

Are you sure about that first one that you wrote. It looks to me like there is an extra (rs x rt) in there.

Chet

5. Jan 22, 2015

### vanhees71

The general notion is a surface integral over a vector field,
$$I=\int_{F} \mathrm{d}^2 \vec{f} \cdot \vec{V}(\vec{r}).$$
Here $\mathrm{d}^2 \vec{f}$ is a vector on a infinitesimal surface element which is perpendicular to the surface and its magnitude is the area of this infinitesimal surface element.

To evaluate such a surface integral, one possibility is to parametrize the surface with two generalized coordinates
$$F: \quad \vec{r}=\vec{r}(q_1,q_2).$$
In the most simple cases the $q_j$ run over a rectangular area in a plane, $(q_1,q_2) \in [a_1,b_1] \times [a_2,b_2]$. Then
$$\vec{T}_j=\frac{\partial}{\partial q_j} \vec{r}$$
defines two tangent vectors along the surface, and the infinitesimal vectors $$\mathrm{d} q_j \vec{T}_j$$ span an infitesimal parallelogram. Then from the geometric meaning of the cross product, it's clear that
$$\mathrm{d}^2 \vec{f}=\mathrm{d}q_1 \, \mathrm{d} q_2 \, \vec{T}_1 \times \vec{T}_2$$
are the surface-element vectors according to this parametrization of your surface. You get
$$I=\int_{a_1}^{b_1} \mathrm{d} q_1 \int_{a_2}^{b_2} \mathrm{d} q_2 [\vec{T}_1(q_1,q_2) \times \vec{T}_2(q_1,q_2)] \cdot \vec{V}[\vec{r}(q_1,q_2)].$$
It's pretty easy to show that the value of the integral is independent of the parametrization (up to a sign, which you have to carefully specify anyway by choosing the orientation of the surface by your choice of the order of the generalized coordinates).

6. Jan 22, 2015

### Staff: Mentor

This guy hasn't responded since Sunday. I'll give him an extra day or two, and then I'm closing this thread.

Chet