I've been reading about surface integrals over vector fields in multivariable calculus, and I've faced a small problem.
The following formula is used to compute the surface integral. Imagine we have a surface $f(x,y,z)=c$ and a vector field $\vec{F}$. Then we have :
$\iint \vec{F}.\hat{n} dS = \iint \vec{F}.\frac{\vec{\nabla}f(x,y,z)}{||\vec{\nabla}f(x,y,z)||}||\vec{\nabla}f(x,y,z)|| dA$
Following this, they have directly computed the integral by plugging $dxdy$ or equivalent, in place of $dA$.
However, this is somewhat wrong, right? We cannot just write $dA=dxdy$, can we?
From my first multivariable calculus class, I remember, $dxdy=\hat{k}.d\vec{S} =(\hat{k}.\hat{n})dS$.
Thus, $dA = \frac{dxdy}{\hat{k}.\vec{n}} = \frac{dS}{||\vec{\nabla}f(x,y,z)||}$.
For different functions and surfaces $\vec{n}.\hat{k} $, would not be $1$ or any number for that matter. It could even be $g(x)$ or some other function. So, we shouldn't directly say $dA=dxdy$, right?
Isn't this more appropriate. Shouldn't the surface integral be :
$\iint \vec{F}.\hat{n} dS = \iint \vec{F}.\frac{\vec{\nabla}f(x,y,z)}{||\vec{\nabla}f(x,y,z)||}||\vec{\nabla}f(x,y,z)|| \frac{dxdy}{\vec{n}.\hat{k}}$
Writing everything in parametric form removes these difficulties and so on, or we could choose an appropriate coordinate system, in which $dA$ is the surface element. But if we want to integrate into cartesian coordinates, shouldn't the transformation from $dS$ to $dxdy$ be like the way I wrote above?
Shouldn't we divide by $\vec{n}.\hat{k}$ while going from $dS$ to $dxdy$ ? Am I correct ?
I am sure this will be clear if you understand the underlying geometric of the formula you have written. Firstly, understand the graphical meaning of a function:
For a single variable function $y=f(x)$, you can think of it as raising up the points a line to a curve. Similarly a multi-variable function of two variables raises up a points on a plane to a surface.
Question: How can we associate the projection of the area on the surface onto plane with the actual area on surface?
Pictures:
From Div, Grad Curl and all that 4th Edition
Intuitively , we can imagine that at each point on the plane, a small patch of area has a different scale factor for it to be related to the area on the surface. We can write : $$dA= P(x,y) dS$$
Question: How do we find $P(x,y)$?
This turns out to be an exercise in geometry.. may remind of you block on ramp from HS physics:
The tilted area on surface is given as $ba=dS$, the area on the flat plane is $b'a=dA$, we know that $b \cos \theta = b'$ and hence:
$$ dS= ba= \frac{b'}{\cos \theta} a = \frac{dA}{\cos \theta}$$
But, how do you get $\cos \theta$? We know that if we are given algebraic curve of the surface $f(x_1,x_2)=C$ then the unit normal is given as $\nabla F$. The angle is just the inclination of the normal with $z$ axis:
$$ dS = \frac{|\nabla f|dA}{\nabla f \cdot \hat{k}}$$
Going back to the original integral:
$$ \int \vec{F} \cdot \hat{n} dS= \int \vec{F} \cdot \frac{ \nabla f}{| \nabla f|} dA \frac{|\nabla f|}{ \nabla f \cdot \hat{k}}$$
Cancel like terms and we find the required equation. Note that all functions in the above expression are in cartesian coordinates.