The region you are integrating over is the part of $y=3-x^2$ that lies above the $x$-axis. If you are going to describe it using polar coordinates, then $\theta$ should only range from $\theta=0$ (to consider the line from $(0,0)$ to $(\sqrt{3},0)$) to $\theta=\pi$ (which gives you the line from $(0,0)$ to $(-\sqrt{3},0)$.
So $\theta$ will only range from $0$ to $\pi$, not from $0$ to $2\pi$. What about $r$?
You want to express $r$ as a function of $\theta$. Your graph is $y=3-x^2$ (you seem to have forgotten the $y$...) So the curve you are trying to express would correspond to:
$$\begin{align*}
y &= 3-x^2\\
r\sin\theta &= 3-r^2\cos^2\theta
\end{align*}$$
So you want to express $r$ as a function of $\theta$, so that you can have that in the "inner" integral (the limits of integration for $r$ may depend on $\theta$, but they should not depend on $r$). So we need to solve for $r$; this gives you a quadratic in $r$:
$$(\cos^2\theta) r^2 + (\sin\theta)r - 3 = 0.$$
Thus, the graph corresponds to
$$r = \frac{-\sin\theta \pm \sqrt{\sin^2\theta +12\cos^2\theta}}{2\cos^2\theta} = \frac{-\sin\theta \pm \sqrt{1 + 11\cos^2\theta}}{2\cos^2\theta}.$$
Since $r$ is positive for the region you want, you would use the $+$ sign. So the region you want would correspond to
$$\begin{align*}
0 &\leq r \leq \frac{-\sin\theta + \sqrt{1+11\cos^2\theta}}{2\cos^2\theta}\\
0 &\leq \theta \leq \pi.
\end{align*}$$
And your integral would be
$$\int_0^{\pi}\int_0^{\frac{-\sin\theta + \sqrt{1+11\cos^2\theta}}{2\cos^2\theta}} 1r\,dr\,d\theta$$
... at which point I would grimace in disgust and switch back to cartesian coordinates, because this is definitely not a nice way to go...
(The issue, of course, is that while circles can be described nicely with polar coordinates, parabolas in general are somewhat nasty.)
Best Answer
$$r^2 \cos (2\theta ) =r^2 \cos^2 \theta -r^2 \sin^2 \theta $$ but $$x=r\cos \theta , y=r\sin\theta $$ so we obtain $$x^2 -y^2 =9.$$