Answer
$$0$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = \sin \left( {x + y} \right) \cr
& {\text{From the image of the rectangle shown below we obtain}} \cr
& {\text{the region }}R \cr
& R = \left\{ {\left. {\left( {x,y} \right)} \right|0 \leqslant x \leqslant \pi ,{\text{ }}x \leqslant y \leqslant \pi } \right\} \cr
& {\text{The area }}A{\text{ of the region is:}} \cr
& A = \left( \pi \right)\left( \pi \right) \cr
& A = {\pi ^2} \cr
& {\text{The average Value of a Function Over a Region }}R{\text{ is}} \cr
& {\text{Average value}} = \frac{1}{A}\iint\limits_R {f\left( {x,y} \right)}dA,{\text{ then}} \cr
& {\text{Average value}} = \frac{1}{{{\pi ^2}}}\int_0^\pi {\int_0^\pi {\sin \left( {x + y} \right)} dydx} \cr
& = \frac{1}{{{\pi ^2}}}\int_0^\pi {\left[ {\int_0^\pi {\sin \left( {x + y} \right)} dy} \right]dx} \cr
& {\text{Integrate with respect to }}y \cr
& = - \frac{1}{{{\pi ^2}}}\int_0^\pi {\left[ {\cos \left( {x + y} \right)} \right]_0^\pi dx} \cr
& = - \frac{1}{{{\pi ^2}}}\int_0^\pi {\left[ {\cos \left( {x + \pi } \right) - \cos \left( {x + 0} \right)} \right]dx} \cr
& = - \frac{1}{{{\pi ^2}}}\int_0^\pi {\left[ {\cos \left( {x + \pi } \right) - \cos x} \right]dx} \cr
& {\text{Integrate}} \cr
& = - \frac{1}{{{\pi ^2}}}\left[ { - \sin \left( {x + \pi } \right) + \sin x} \right]_0^\pi \cr
& = - \frac{1}{{{\pi ^2}}}\left[ { - \sin \left( {\pi + \pi } \right) + \sin \pi } \right] + \frac{1}{{{\pi ^2}}}\left[ { - \sin \left( {0 + \pi } \right) + \sin 0} \right] \cr
& = - \frac{1}{{{\pi ^2}}}\left[ { - \sin \left( {2\pi } \right) + \sin \pi } \right] + \frac{1}{{{\pi ^2}}}\left[ { - \sin \left( \pi \right) + \sin 0} \right] \cr
& = 0 \cr} $$