Why is energy in a wave proportional to amplitude squared?

I'm a mathematics student trying to grasp some basics about wave propagation. A sentence I find very often in introductive physics textbooks is the following:

In a wave, energy is proportional to amplitude squared.

This is something I would like to understand better in the case of mechanical (linear) waves.

The simplest model is the vibrating string of mass density $\mu$ and tension $T$: here an element of string of rest length $dx$ and vertical displacement $y(x, t)$ possesses a kinetic energy $\frac{1}{2}\mu dx \left( \frac{\partial y}{\partial t}\right)^2$ and a potential elastic energy $\frac{1}{2}T dx \left( \frac{\partial y}{\partial x}\right)^2$. So we have the total energy

$$dE=\frac{1}{2}\mu dx \left( \frac{\partial y}{\partial t}\right)^2+\frac{1}{2}T dx \left( \frac{\partial y}{\partial x}\right)^2,$$

which completely explains the sentence. Can we obtain a similar formula for higher-dimensional waves? How should I modify the above formula for a (simplified model of an) elastic membrane, for example? I would expect something like

$$dE=\text{some constant}\left(\frac{\partial z}{\partial t}\right)^2+\text{some other constant}\left\lvert\nabla z\right\rvert^2,$$

where $z=z(x, y, t)$ is the vertical displacement. Am I right?

Answer

You are correct, the equation is generalizable to higher dimensions. The equation you gave is simply the sum of the various forms of energy. In the equation

$$dE = \frac{\mu}{2} dx \left( \frac{\partial y}{\partial t}\right)^2 + \frac{T}{2} dx \left(\frac{\partial y}{\partial x} \right)^2 $$

The first term on the right hand side is the kinetic energy and the second term is the elastic potential energy. As Mike said, the kinetic energy is just $p^2 / 2m$. The elastic potential energy at any location on the string is given by adding up all of the force it took to get that piece there (given by Hooke's law); that is

$$F_s (x) = -T y(x)$$ $$U = \int_0 ^y Ty(x) dy = \frac{T}{2} y(x)^2$$

The total energy is then just the sum of the kinetic and potential energies with appropriate modification using the mass density times a infinitesimal length as the mass.

When you move to higher dimensions, you have to account for that in the kinetic and potential energies. The kinetic energy of a portion of the surface is the square of the momentum of that portion ($mv$) divided by the mass of that portion. The potential energy is the spring constant of the membrane divided by 2, multiplied by the square of the displacement from zero.