Let $r = \theta$, if we want to compute the area inside the spiral up to the negative $y$-axis we can set up the integral: $$\int\limits_0^{\frac{3\pi}{2}} \!\!\! \int\limits_0^\theta r\ dr\ d\theta$$ and we could solve that.
Is there any way I can represent this integral in Cartesian coordinates? Maybe I'll have to break it into a sum of multiple integrals? I'm pretty sure there's only an implicit representation of the spiral of Archimedes in Cartesian so this will certainly make this more difficult. We could start with $r = \theta$ and then maybe take tangent of both sides. Then we have $\frac{y}{x} = \tan{\sqrt{x^2 + y^2}}$ and you can't solve for $x$ or $y$ explicitly in terms of the other variable.
Any thoughts would be much appreciated!
You can rotate the spiral clockwise by ${\pi}/2$ by defining $r = \theta + \pi/2$ and the area can be written in cartesian coordinates as
\begin{equation}I = \int_{}^{}\int_{}^{}{{\chi}}_{A} \left(x , y\right) d x d y\end{equation}
where $A$ is the set
\begin{equation}A = \left\{\left(x , y\right) \in {\mathbb{R}}^{2} , \sqrt{{x}^{2}+{y}^{2}} \leqslant \frac{{\pi}}{2}+2 \arctan \left(\frac{y}{\sqrt{{x}^{2}+{y}^{2}}+x}\right)\right\}\end{equation}
Remark Coming back to the unrotated spiral by replacing $x$ by $y$ and $y$ by $-x$, you could as well integrate over the domain
\begin{equation}B = \left\{\left(x , y\right) \in {\mathbb{R}}^{2} , \sqrt{{x}^{2}+{y}^{2}} \leqslant \frac{{\pi}}{2}-2 \arctan \left(\frac{x}{\sqrt{{x}^{2}+{y}^{2}}+y}\right)\right\}\end{equation}