Solving
$$\Gamma\left(\frac{1}{2}\right)=\int_{0}^\infty e^{-x}x^{-\frac{1}{2}}dx$$ $$ \Gamma\left(\frac{1}{2}\right)=2\int_{0}^\infty e^{-x^2} dx$$
$$\begin{align} {\Gamma\left(\frac{1}{2}\right)^2} = 4\left(\int_{0}^\infty e^{-x^2} dx\right) \left(\int_{0}^{\infty} e^{-y^2}dy\right) \\ =4\int_{0}^\infty\left(\int_{0}^\infty e^{-(x^2+y^2)} dx\right)dy \end{align}\\$$
Changing to polar coordinates: $r^2=x^2+y^2$ and $dx\,dy=r\,d\theta\,dr$
$$ \Gamma\left(\frac{1}{2}\right)^2=4\int_0^{\frac{\pi}{2}}\int_0^\infty e^{-r^2}rdrd\theta \\ =4\frac{\pi}{2}\int_0^\infty e^{-r^2}rdr= \pi $$ $$ \Gamma\left(\frac{1}{2}\right)=\sqrt{\pi} $$
The question is, why is the limits of integration $0\to\frac{\pi}{2}$ Instead of $0\to2\pi$?
How is the $\theta$ bounded in such a way?
Let me suggest formal answer to question: having set $\mathbb{R^2}=\{(x,y)\colon x \in \mathbb{R} \land y \in \mathbb{R}\}=\mathbb{R} \times \mathbb{R}$ with Cartesian coordinate system we can consider Polar coordinates in which for each pair $(x,y)$ we would like to put in accordance distance $r$ from reference point and angle $\theta$ from reference direction. For any point besides $(0,0)$ this immediately gives formulas: $$x = r \cos \theta \\ y = r \sin \theta$$ but difficulties only starts here, because we obtain infinite possible solutions $(r,\theta)$ for one $(x,y)$. Accepted solution is to consider separately case $(0,0)$ and for other pairs having restrictions on $(r,\theta)$ taking any half-open $2\pi$ length interval for $\theta$, for example $[0,2\pi)$, and $r > 0$.
De facto, we have one $\mathbb{R^2}_{(x,y)}$ mapped on many its instances on $\mathbb{R^2}_{(r,\theta)}$, but we choose only one to obtain uniqueness.
So, when you consider changing to polar coordinates, from scratch, you consider some restrictions as, for example, $(r,\theta) \in (0, +\infty)\times [0,2\pi)$ .
In your case, to obtain image of set from $\mathbb{R^2}_{(x,y)}$ in $\mathbb{R^2}_{(r,\theta)}$ we formally solve equations from above in existing restriction : having 1-st quadrant $\{(x,y)\colon x>0 \land y>0\}$ on $\mathbb{R^2}_{(x,y)}$ we obtain $$x=r \cos \theta > 0 \\ y=r \sin \theta > 0$$ which gives no restriction on $r$ and $\theta \in \left[ 0, \frac{\pi}{2}\right]$ i.e. $\left\lbrace(r,\theta)\colon r>0 \land 0 \leqslant \theta \leqslant \frac{\pi}{2}\right\rbrace$ on $\mathbb{R^2}_{(r,\theta)}$.