# Zonal averaging of partial derivatives

Given a quantity $Q(x,y,t)$, the zonal average operator $[Q] = \frac{1}{2\pi}\int_0^{2\pi} Q\:\mathrm{d} \lambda$, and zonal anomaly $Q^\star$ such that $Q = [Q] + Q^\star$, my text book says that zonally averaging

$$\frac{\partial Q}{\partial t} + \frac{\partial}{\partial x}(uQ) + \frac{\partial}{\partial y}(vQ) = S$$ gives $$\frac{\partial [Q]}{\partial t} + [v] \frac{\partial [Q]}{\partial y} + \frac{\partial}{\partial y}[v^\star Q^\star] = [S]$$ I'm unable to see how to arrive at this answer. What is the method or intuition for applying the zonal averaging?

I figured it out. It's clear that the zonally averaged $\partial/\partial x$ is zero. The terms involving $\partial/\partial y$ come from expanding $vQ = ([v] + v^\star)([Q] + Q^\star)$, then noticing that $[v^\star [Q]] = [[v] Q^\star] = 0$. I also had to convince myself that $[\partial Q/\partial t] = \partial [Q] / \partial t$. I'd still be interested in a more rigorous answer, however.