Line charge density expressed via Dirac delta function

[Heirot]

Homework Statement

Let's say we have a wire of finite length L with total charge Q evenly spread along the wire so that lambda=Q/L, linear charge density, is constant. The wire is shaped in x-y plane in some well behaved curve y = f(x). Find the surface charge density sigma(x,y).

Homework Equations

Q = Int(lambda dl) = Int ( Sqrt(1+(dy/dx)^2) dx) for line charge density
Q = Int(sigma(x,y) dx dy) for surface charge density

The Attempt at a Solution

The most logic thing to do would be to write sigma(x,y) = lambda * delta(y-f(x)), where delta(x) is Dirac's delta function. Unfortunately, this doesn't give the right Q, because integration of the delta function over y only give 1 and not the required Sqrt(1+(dy/dx)^2).

Where's the error in reasoning?

 

[George Jones]

But

$$\begin{equation*}<br /> \begin{split}<br /> Q &= \lambda L \\<br /> &= \lambda \int \sqrt{1 + \left( \frac{dy}{dx}\right)^2} dx.<br /> \end{split}<br /> \end{equation*}$$

Heuristically, the amount of charge on the wire between $x$ and $x + dx$ is

$$dQ = \lambda dL = \lambda \sqrt{1 + \left( \frac{dy}{dx}\right)^2} dx.$$

 

[Heirot]

I agree. So, dQ = sigma(x,y) dx dy. What's sigma(x,y)?