Let $f\left(x\right)=\sqrt{2x+3}$. Determine the area of the surface generated when the region bounded by $f\left(x\right)$ and the $x$-axis on the interval $[1,2]$ is revolved about the $x$-axis.

$\frac{2\pi }{3}(8\sqrt{2}-3\sqrt{6}){\text{units}}^2$

$\frac{2\pi }{3}(4\sqrt{3}-3\sqrt{2}){\text{units}}^2$

$\frac{4\pi }{3}(8\sqrt{2}-3\sqrt{6}){\text{units}}^2$

$\frac{4\pi }{3}(4\sqrt{3}-3\sqrt{2}){\text{units}}^2$