Find the volume V of the solid below the hyperboloid $x=7-\sqrt[\phantom{2}]{1x^2{y}^2}$ and above the following region.

$R=\{(r,)$:$2\sqrt[\phantom{2}]{2}r2\sqrt[\phantom{2}]{6},02\}$

Set up the double integral, in polar coordinates, that is used to find the volume.

${\displaystyle \underset{0}{\overset{2}{}}}{\displaystyle \underset{2\sqrt[\phantom{2}]{2}}{\overset{2\sqrt[\phantom{2}]{6}}{}}}((7-\sqrt[\phantom{2}]{1r^2})r)drd$

$($Type exact answers.$)$

$V=()3$

$($Type an exact answer.$)$