Answer
$$\frac{{224}}{9}$$
Work Step by Step
$$\eqalign{
& \int_1^4 {\int_0^4 {\sqrt {uv} } } dudv \cr
& = \int_1^4 {\left[ {\int_0^4 {\sqrt {uv} } du} \right]} dv \cr
& {\text{solve the inner integral}}{\text{, treat }}v{\text{ as a constant}} \cr
& = \int_0^4 {\sqrt u \sqrt v } du \cr
& = \sqrt v \int_0^4 {\sqrt u } du \cr
& {\text{integrating}} \cr
& = \sqrt v \left( {\frac{{{u^{3/2}}}}{{3/2}}} \right)_0^4 \cr
& = \sqrt v \left( {\frac{{{u^{3/2}}}}{{3/2}}} \right)_0^4 \cr
& = \frac{2}{3}\sqrt v \left( {{u^{3/2}}} \right)_0^4 \cr
& {\text{evaluating the limits in the variable }}u \cr
& = \frac{2}{3}\sqrt v \left( {{4^{3/2}} - {0^{3/2}}} \right) \cr
& {\text{simplifying}} \cr
& = \frac{{16}}{3}\sqrt v \cr
& \cr
& = \int_1^4 {\left[ {\int_0^4 {\sqrt {uv} } du} \right]} dv = \int_1^4 {\frac{{16}}{3}\sqrt v } dv \cr
& = \frac{{16}}{3}\int_1^4 {{v^{1/2}}} dv \cr
& {\text{integrating}} \cr
& = \frac{{16}}{3}\left[ {\frac{{{v^{3/2}}}}{{3/2}}} \right]_1^4 \cr
& = \frac{{32}}{9}\left[ {{v^{3/2}}} \right]_1^4 \cr
& {\text{evaluate}} \cr
& = \frac{{32}}{9}\left[ {{4^{3/2}} - {1^{3/2}}} \right] \cr
& = \frac{{32}}{9}\left( 7 \right) \cr
& = \frac{{224}}{9} \cr} $$
