Answer :

Solution :

V = volume of cube = (1)³

$\overrightarrow{F}=3x\hat{i}+5y\hat{j}+6z\hat{k}$

$\overrightarrow{▽}.\overrightarrow{F}=\frac{\delta }{\delta x}(3x)+\frac{\delta }{\delta y}(5y)+\frac{\delta }{\delta z}(6z)$ = 3 + 5 + 6 = 14

${\int }_{A}F.dA=\int {\int }_V\int \overrightarrow{▽}.\overrightarrow{F}dV$

${\int }_{A}F.dA=\int {\int }_V\int 14dV$ = 14 × ( V )

${\int }_A\overrightarrow{F}.d\overrightarrow{A}$ = 14