Answer
$0$
Work Step by Step
$\displaystyle \iint_{R}\sin(x-y)dA=\int_{0}^{\pi/2}\int_{0}^{\pi/2}\sin(x-y)dydx=\int_{0}^{\pi/2}[\int_{0}^{\pi/2}\sin(x-y)dy]dx$
... treat $x$ as a constant in the inner integral.$\left[\begin{array}{ll}
t=x-y & \\
dt=-dy & \\
y=0\Rightarrow t=x, & y=\pi/2\Rightarrow t=x-\pi/2
\end{array}\right]$
$=\displaystyle \int_{0}^{\pi/2}[\int_{x}^{x-\pi/2}\sin(t)(-dt)]dx$
$=\displaystyle \int_{0}^{\pi/2}[\cos(x-\frac{\pi}{2})-\cos(x-0)]dx$
$=\displaystyle \int_{0}^{\pi/2}[\cos(x-\frac{\pi}{2})-\cos x]dx$
$=[\displaystyle \sin(x-\frac{\pi}{2})-\sin x]_{0}^{\pi/2}$
$=\displaystyle \sin 0-\sin\frac{\pi}{2}-[\sin(-\frac{\pi}{2})-\sin 0]$
$=0-1-(-1-0)$
$=-1+1$
$=0$
