# Definition for Surface Integral

1. Dec 17, 2013

### Jhenrique

Hello!

The definition of Line Integral can be this:
$$\int_s\vec{f}\cdot d\vec{r}=\int_s(f_1dx+f_2dy+f_3dz)$$

And the definition of Surface Integral can be this:
$$\int\int_S(f_1dydz+f_2dzdx+f_3dxdy)$$

However, in actually:
$$\\dx=dy\wedge dz \\dy=dz\wedge dx \\dz=dx\wedge dy$$
What do the Surface Integral be equal to:
$$\int\int_S(f_1dy\wedge dz+f_2dz\wedge dx+f_3dx\wedge dy)=\int\int_S(f_1dx+f_2dy+f_3dz)=\int\int_S\vec{f}\cdot d\vec{r}$$

I know, I know... I know that, generally, the definition to Integral Surface is:
$$\int\int_S\vec{f}\cdot \hat{n}\;dS$$
I until like this definition when compared to its respective Line Integral:
$$\int_s\vec{f}\cdot \hat{t}\;ds$$

But, is correct to definite the Surface Integral as:
$$\int\int_S\vec{f}\cdot d\vec{r}$$
being
$$d\vec{r}=(dx,dy,dz)$$
?

2. Dec 17, 2013

### HallsofIvy

No, the surface integral is, as you say, a double integral while the path integral is a single integral. They are NOT the same thing. They can, of course, be connected by, for example, the Stoke's Theorem that says that the integral of the curl of $\vec{F}$ over a surface is the same as the integral of $\vec{F}$ around the boundary of the surface.

3. Dec 17, 2013

### Jhenrique

But I definited the Surface Integral with a double integral (as you can see below or above)
$$\int\int_S\vec{f}\cdot d\vec{r}$$
with the only difference that I used the position vector r, like in Line Integral. But, second the identities above, to define the Surface integral with ·dr is equivalent to traditional definition, with ·ndS. Correct!?

4. Dec 18, 2013

### qbert

what? your definitions make no sense. $dx = dy \wedge dz''$ is literally nonsensical.