You have established that $\int_{-\sqrt{3}}^{\sqrt{3}} (3 - x^2) dx = 4 \sqrt{3} $ in the standard Cartesian coordinates, and wished to attempt the same integration using polar coordinates. As Arturo Magidin demonstrated, this is no tidy matter if one makes a direct transformation of the curve.
The difficulty is that curves that do not pass through the origin generally do not have simple expressions in polar coordinates, as one observes even with straight lines $ax + by = c $ , with $a, b, c \ne 0$. Perhaps it will be helpful to find a way to produce a parabola which has one "arm" passing through $( 0, 0)$ .
What we want, after all, is the area of a parabola of the same size, bounded by the x-axis. If we translate $y = 3 - x^2$ to the right by $\sqrt{3}$ units, we obtain $y = 3 - (x - \sqrt{3})^2 = 2 \sqrt{3} \cdot x - x^2$ , which now has intercepts at $x = 0$ and $x = 2 \sqrt{3}$ . The transformation of this expression into polar coordinates produces
$ r \sin \theta = 2 \sqrt{3} \cdot r \cos \theta - r^2 \cos^2 \theta \rightarrow r \cos^2 \theta = 2 \sqrt{3} \ \cos \theta - \sin \theta $
$\rightarrow r(\theta) = \frac{2 \sqrt{3} \ \cos \theta - \sin \theta}{\cos^2 \theta} .$

It is clear that $r(0) = 2 \sqrt{3}$ , so the lower limit of our integration over angle will be $\theta= 0$. We will be able to cover the area under our translated parabola by advancing $\theta$ to the upper limit of integration. Although the radius function appears problematical, we never reach $\theta = \frac{\pi}{2}$, as we have $r(\theta) = 0$ at an angle we will define by $\tan \Theta = 2 \sqrt{3}$ .
We can now set up our "polar area integral" as
$\int_0^{\Theta} \frac{1}{2} \cdot [r({\theta})]^2 \ d\theta \ = \ \frac{1}{2}\int_0^{\Theta} ( \frac{2 \sqrt{3} }{ \cos \theta} - \frac{\sin \theta}{\cos^2 \theta})^2 \ d\theta $
$= \frac{1}{2}\int_0^{\Theta} ( 12 \sec^2 \theta \ - \ 4 \sqrt{3} \cdot \frac{\sin \theta }{ \cos^3 \theta} \ + \ \frac{\sin^2 \theta}{\cos^4 \theta}) \ d\theta $
$= \frac{1}{2} [\ 12 \tan\theta \ - \ \ 4 \sqrt{3} \cdot (\frac{\sec^2 \theta }{ 2}) \ + \ (\frac{\tan^3 \theta}{3}) \ ] |_0^{\Theta} $
$= \ 6 \cdot (\tan\Theta \ - \ 0 ) \ - \ \sqrt{3} \cdot ( \sec^2 \Theta \ - \ \sec^2 0 ) \ + \ \frac{1}{6} \cdot ( \tan^3 \Theta \ - 0 ) $
$= \ 6 \tan\Theta \ - \ \sqrt{3} \cdot ( [ \ \tan^2 \Theta \ + \ 1 \ ] - \ 1 ) \ + \ \frac{1}{6} \cdot \tan^3 \Theta $
$= \ 6 \cdot 2\sqrt{3} \ - \ \sqrt{3} \cdot (\ 2\sqrt{3} \ )^2 \ + \ \frac{1}{6} \cdot ( \ 2\sqrt{3} \ )^3 \ = \ 12 \sqrt{3} \ - \ 12 \sqrt{3} \ + \ 4 \sqrt{3} \ = \ 4 \sqrt{3} \ . $
This method should work for finding the area enclosed by either coordinate axis and a parabola opening toward that axis and with its symmetry axis perpendicular to the coordinate axis. (I just can't imagine why anyone would prefer that to using rectangular coordinates...)