Is the Surface Integral Equivalent to the Traditional Definition?

[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)
?

 

[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.

 

[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!?

 

[qbert]

the Surface integral with ·dr is equivalent to traditional definition, with ·ndS. Correct!?

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