Answer
$$\bar f = \frac{4}{{{\pi ^2}}}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( {x,y} \right) = \sin x\sin y;{\text{ }}R = \left\{ {\left( {x,y} \right):0 \leqslant x \leqslant \pi ,{\text{ 0}} \leqslant y \leqslant \pi } \right\} \cr
& {\text{The average value of an integrable function }}f{\text{ over a region }}R \cr
& {\text{is:}} \cr
& \bar f = \frac{1}{{{\text{area of }}R}}\iint\limits_R {f\left( {x,y} \right)}dA \cr
& {\text{The area of the region }}R{\text{ is:}} \cr
& {\text{Area of }}R = \int_0^\pi {\int_0^\pi {dx} dy} \cr
& {\text{Area of }}R = {\pi ^2} \cr
& \cr
& {\text{Therefore}}{\text{, }} \cr
& \bar f = \frac{1}{{{\pi ^2}}}\iint\limits_R {f\left( {x,y} \right)}dA \cr
& \bar f = \frac{1}{{{\pi ^2}}}\int_0^\pi {\int_0^\pi {\sin x\sin y} dxdy} \cr
& {\text{Integrate with respect to }}x \cr
& \bar f = - \frac{1}{{{\pi ^2}}}\int_0^\pi {\left[ {\cos x\sin y} \right]_0^\pi dy} \cr
& \bar f = - \frac{1}{{{\pi ^2}}}\int_0^\pi {\left[ {\cos \pi \sin y - \cos 0\sin y} \right]dy} \cr
& \bar f = - \frac{1}{{{\pi ^2}}}\int_0^\pi {\left[ { - \sin y - \sin y} \right]dy} \cr
& \bar f = \frac{2}{{{\pi ^2}}}\int_0^\pi {\sin ydy} \cr
& {\text{Integrate and evaluate}} \cr
& \bar f = \frac{2}{{{\pi ^2}}}\left[ { - \cos y} \right]_0^\pi \cr
& \bar f = - \frac{2}{{{\pi ^2}}}\left[ {\cos \pi - \cos 0} \right] \cr
& \bar f = - \frac{2}{{{\pi ^2}}}\left[ { - 2} \right] \cr
& \bar f = \frac{4}{{{\pi ^2}}} \cr} $$
