# Why this integral represents an area in the phase space?

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.

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.