Answer
$$V = 72$$
Work Step by Step
$$\eqalign{
& z = \sqrt y ;\,\,\,\,\,\,\,0 \leqslant x \leqslant 4,\,\,\,\,\,0 \leqslant y \leqslant 9 \cr
& {\text{by the equation just given}}{\text{, the volume is}} \cr
& V = \iint\limits_R {f\left( {x,y} \right)}dxdy \cr
& {\text{replacing the limits for the region }}R{\text{ and }}f\left( {x,y} \right) = z \cr
& V = \int_0^9 {\int_0^4 {\sqrt y dx} dy} \cr
& V = \int_0^9 {\left[ {\int_0^4 {\sqrt y dx} } \right]dy} \cr
& {\text{solve the inner integral by treating }}y{\text{ as a constant}} \cr
& = \int_0^4 {\sqrt y dx} \cr
& {\text{integrate }} \cr
& = \sqrt y \left[ x \right]_0^4 \cr
& {\text{evaluating the limits in the variable }}x \cr
& = \sqrt y \left( {4 - 0} \right) \cr
& {\text{simplifying}} \cr
& = 4\sqrt y \cr
& \cr
& V = \int_0^9 {\left[ {\int_0^4 {\sqrt y dx} } \right]dy} \cr
& V = \int_0^9 {4\sqrt y dy} \cr
& {\text{integrating}} \cr
& V = 4\left( {\frac{{{y^{3/2}}}}{{3/2}}} \right)_0^9 \cr
& V = \frac{8}{3}\left( {{y^{3/2}}} \right)_0^9 \cr
& V = \frac{8}{3}\left( {{9^{3/2}} - {0^{3/2}}} \right) \cr
& V = \frac{8}{3}\left( {27} \right) \cr
& V = 72 \cr} $$
