Convert the triple integral $to$ spherical coordinates and evaluate $it.$

${\displaystyle \underset{-4}{\overset{4}{}}}{\displaystyle \underset{-\sqrt[\phantom{2}]{16-x^2}}{\overset{\sqrt[\phantom{2}]{16-x^2}}{}}}{\displaystyle \underset{0}{\overset{\sqrt[\phantom{2}]{16-x^2-y^2}}{}}}(x^2+y^2+z^2{)}^{-\frac{1}{2}}dzdydx$

Im confused on what the outer integral $$ would be an why