Proof of formula for area enclosed by parametric curve Suppose that $\theta \in [0,2\pi]$ and $(x(\theta), y(\theta))$ define a closed parametric curve. There is a formula that says the area enclosed by this curve is equal to $\frac{1}{2} \int\limits_0^{2\pi} (x\frac{dy}{d\theta}-y\frac{dx}{d\theta}) d\theta$. 
I know that this is a simple consequence of Green's Theorem however I would like to know if there is a proof of this statement that does not rely on it.
All the proofs I can find online assume that we can actually parametrise our curve by $x$ as well as $\theta$ but this is not the case for many curves defined parametrically so is there a way of getting around that?
 A: Suppose we express the $2$-vector
$\mathbf r(\theta) = (x(\theta), y(\theta)) \tag 1$
in polar coordinates, thus:
$\mathbf r(\theta) = (x(\theta), y(\theta)) = (r(\theta)\cos \theta, r(\theta)\sin \theta) = r(\theta)(\cos \theta, \sin \theta) =r(\theta)\mathbf n(\theta), \tag 2$
where
$\mathbf n(\theta) = (\cos \theta, \sin \theta) \tag 3$
is a unit vector:
$\mathbf n(\theta) \cdot \mathbf n(\theta) = \cos^2 \theta + \sin^2 \theta = 1, \tag 4$
and
$r(\theta) = \Vert \mathbf r(\theta) \Vert, \tag 5$
as is easily seen from (2) and (4):
$\Vert \mathbf r(\theta) \Vert = \sqrt{r(\theta)\mathbf n(\theta) \cdot r(\theta)\mathbf n(\theta)}$
$= \sqrt{r^2(\theta)(\mathbf n(\theta) \cdot \mathbf n(\theta))} = \sqrt{r^2(\theta)} = r(\theta). \tag 6$
Using the "dot" notation for derivatives with respect to theta, so that
$\dot x(\theta) = \dfrac{dx(\theta)}{d\theta}, \; \dot y(\theta) = \dfrac{dy(\theta)}{d\theta}, \; \dot{\mathbf r}(\theta) = \dfrac{d\mathbf r(\theta)}{d\theta}, \tag 7$
and so forth, we have
$\dot {\mathbf r}(\theta) = \dfrac{d(r(\theta)\mathbf n(\theta))}{d\theta} =  \dot r(\theta)\mathbf n(\theta) + r(\theta)\dot{\mathbf n}(\theta). \tag 8$
We next observe that $x(\theta) \dot y(\theta) - y(\theta) \dot x(\theta)$ is simply the (scalar) $2$-dimansional vector cross product 
$x(\theta) \dot y(\theta) - y(\theta) \dot x(\theta) = (x(\theta), y(\theta)) \times (\dot x(\theta), \dot y(\theta)) = \mathbf r(\theta) \times \dot{\mathbf r}(\theta); \tag 9$
thus, using (2) and (8), we compute
$\mathbf r(\theta) \times \dot{\mathbf r}(\theta) = r(\theta)\mathbf n(\theta) \times (\dot r(\theta)\mathbf n(\theta) + r(\theta)\dot{\mathbf n}(\theta))$
$= r(\theta) \dot r(\theta)\mathbf n(\theta) \times \mathbf n(\theta) + r^2(\theta))\mathbf n(\theta) \times \dot{\mathbf n}(\theta); \tag{10}$
since
$\mathbf n(\theta) \times \mathbf n(\theta) = 0  \tag{11}$
and
$\mathbf n(\theta) \times \dot{\mathbf n}(\theta) =  (\cos \theta, \sin \theta) \times (-\sin \theta, \cos \theta) = \cos^2 \theta + \sin^2 \theta = 1, \tag{12}$
we obtain
$\mathbf r(\theta) \times \dot{\mathbf r}(\theta) = r^2(\theta); \tag{13}$
combining (9) and (13) we arrive at
$\dfrac{1}{2} \displaystyle \int_0^{2\pi} (x(\theta) \dot y(\theta) - y(\theta) \dot x(\theta))d\theta = \dfrac{1}{2} \int_0^{2\pi} (\mathbf r(\theta) \times \dot{\mathbf r}(\theta)d\theta = \dfrac{1}{2} \int_0^{2\pi} r^2(\theta) d\theta; \tag{14}$
the rightmost integral in (14) is in fact the area swept out by the vector $\mathbf r(\theta)$ as $0 \rightarrow \theta \rightarrow 2\pi$; see this for a further explanation.
We can in fact easily show that
$\dfrac{1}{2} \displaystyle \int_0^{2\pi} r^2(\theta)d\theta \tag{15}$
is the requisite area directly from first principles if we accept the fact that
$rdrd\theta = dxdy \tag{16}$
is the area element in polar coordinates, a fact which may easily be checked by performing the coordinate transformation $(x, y) \to (r, \theta)$ using $x = r\cos \theta$ and $y = \sin \theta$; I leave the details to the sufficiently engaged reader.  From (16), the area $A$ swept out by $\mathbf r(\theta)$ as $\theta_1 \rightarrow \theta \rightarrow \theta_2$ will be
$A = \displaystyle \int_{\theta_1}^{\theta_2} \int_0^{r(\theta)} r dr d\theta = \int_{\theta_1}^{\theta_2} (\dfrac{1}{2} r^2(\theta))d\theta = \dfrac{1}{2}\int_{\theta_1}^{\theta_2}r^2(\theta)d\theta; \tag{17}$
if $0 \rightarrow \theta \rightarrow 2\pi$, then (17) yields
$A = \dfrac{1}{2} \displaystyle \int_0^{2\pi} r^2(\theta)d\theta, \tag{18}$
in accord with (14).
