Suppose $\dot{x} = f(x)$ is a dynamical system on the state space $X$. My notes define a conservative system as one where there exists a (nontrivial) function $H: X \rightarrow \mathbb{R}$ such that $$\frac{d}{dt}H(x) = 0 = \nabla H(x) \cdot f(x)$$
If we then define $\Sigma_E := H^{-1}(E)$, $E \in \mathbb{R}$ as level sets of $H$, it follows that $\Sigma_E$ is an invariant set due to $H$ being constant along orbits. Immediately after this proof my notes puzzlingly state "the dimension of the problem is reduced by one".
Later on when introducing Hamiltonian systems as a subset of conservative systems the classic example of the pendulum is given:
\begin{align} &\dot{\theta} = p \\ &\dot{p} = -\sin(\theta)\\ \end{align}
It is noted that there is the Hamiltonian $H(p,\theta) = \frac{1}{2}p^2 -\cos(\theta)$ and then shown how any Hamiltonian must be a conserved quantity. Immediately after it is written that "it is useful to notice conserved quantities when they exist as they reduce the dimension of the problem".
Am I missing something obvious? In the case of the pendulum the state space is 2-dimensional, having the Hamiltonian and plotting its level sets gives us the entire phase portrait which is obviously still 2-dimensional. I know a Hamiltonian system has some very nice properties over a generic system, but I do not know one relating to reducing dimensions.
I guess my main question really is "why do we care about conserved quantities?".
Once you start at a given point, you are confined to the level set of $H$ passing through that point, which (typically) is an $(n-1)$-dimensional hypersurface in the $n$-dimensional state space $X$. If you introduce an $(n-1)$-dimensional coordinate system on that hypersurface, you can write the ODEs for the motion on the surface in terms of just those $n-1$ variables.
If you know several constants of motion, you can reduce the order further, and if you know sufficiently many, you can (in principle) integrate the system of ODEs exactly.