Answer
$\dfrac{2}{9}$
Work Step by Step
We write the region in the polar co-ordinates as: $0 \leq r \leq 1$ and $0 \leq \theta \leq \pi$
$\iint_{D} f(x,y) \ dA=\int_0^{\pi} \int_0^{1} (r^7 \sin \theta) (r) \ dr \ d\theta$
Now, we have:
$\int_0^{\pi} \int_0^{1} (r^7 \sin \theta) (r) \ dr \ d\theta =\int_0^{\pi} [\dfrac{r^9}{9}]_0^{1} \sin \theta d\theta$
or, $=\int_0^{\pi} \dfrac{\sin \theta}{9} d\theta$
or. $=(1/9) \times[-\cos \theta]_0^{\pi}$
or. $=\dfrac{1}{9}[-(-1)-(-1)]$
or. $=\dfrac{2}{9}$
