Answer
a) $0$ and b) $\dfrac{4}{\pi^2}$
Work Step by Step
(a) $ Average =\dfrac{1}{\pi^2} \times \int^{\pi}_{0} \int^{\pi}_{0} \sin(x+y)dy dx $
or, $=\dfrac{1}{\pi^2} \times \int^{\pi}_{0} [-cos(x+y)]^{\pi}_0 \space dx $
or, $= \dfrac{1}{\pi^2} \times 2 \int_0^{\pi} \cos x \space dx $
or, $2 [\sin (x)]_0^{\pi}=0$
(b) $ Average=\dfrac{1}{(\pi^2/2)} \times \int^{\pi}_{0} \int^{\pi/2}_{0} \sin(x+y)dydx $
or, $=\dfrac{1}{(\pi^2/2)} \times \int^{\pi}_{0} [-\cos(x+y)]^{\pi/2}_0dx $
or, $ =\dfrac{1}{(\pi^2/2)} \times [-\sin(x+\dfrac{\pi}{2})+\sin x]^\pi_0$
or, $=\dfrac{2}{\pi}[(-sin \dfrac{3\pi}{2}+\sin \pi)-(-\sin \dfrac{\pi}{2}+\sin (0))]$
or, $=\dfrac{4}{\pi^2}$
