Why this integral represents an area in the phase space?

In classical mechanics, and mainly when studying the Hamiltonian formalism, the quantity $$\oint p \ dq$$ is usually referred as an area in the phase space, and can be computed graphically as such. Could someone explain me why this integral represents such an area? Is it simply notation, or does it involve something more?

I know that a general volume in phase space is $$\int dq_1 \dots dq_n dp_1 \dots dp_n$$ and probably this result is related to that area, but I don't see how. Thanks in advance.

• Dimensional analysis alone indicates something important, never mind complex analysis. – J.G. Dec 18 '20 at 23:24

Consider for example the 1-dimensional harmonic oscillator described by the Hamiltonian $$H(q,p)=\frac{1}{2m}p^2+\frac{1}{2}m\omega^2q^2$$.

The general solution is \begin{align} q(t)&=\hat{q}\sin(\omega t + \phi_0) \\ p(t)&=m\omega\hat{q}\cos(\omega t + \phi_0) \end{align}

When you plot $$p(t)$$ versus $$q(t)$$ then you get an elliptical path which is traversed in a clock-wise fashion.

Now for the interpretation of the integral $$\oint p\ dq$$.

During the first half-cycle (the upper half of the ellipsis) $$p$$ and $$dq$$ are both positive, thus contributing a positive narrow rectangle $$p\ dq$$.
And during the second half-cycle (the lower half of the ellipsis) $$p$$ and $$dq$$ are both negative, thus again contributing a positive narrow rectangle $$p\ dq$$.
So $$\oint p\ dq$$ is the total sum of all these rectangles over a full cycle, i.e. just the whole enclosed elliptical area.

These expressions can related via the Stokes' formula. As far as I understand, in general the expression has the following form: $$\oint d q_1 \ldots \oint dq_n \ p_1 \ldots p_n$$ Then apply the Stokes' formula on each pair $$(p_i, q_i)$$: $$\oint p dq = \int dp \ dq$$ In this fashion one gets the expression you wrote.