double integral (D) of y^3(x^2+y^2)^(-3/2) DXDY where (D) is the region determined by the ocnditions .5 less than or equal to y less than or equal to 1 and x^2+y^2 less than or equal to 1.
Have you drawn the region of integration? It's the interior of the unit circle that lies above the line y = 1/2.
One possible set up is $\displaystyle \int_{x = -\sqrt{3}/2}^{x = \sqrt{3}/2} \int_{y=1/2}^{y=+\sqrt{1 - x^2}} y^3 (x^2 + y^2)^{-3/2} \, dy \, dx$.
Another is $\displaystyle \int_{x = - \sqrt{1 - y^2}}^{x=+\sqrt{1 - y^2}} \int_{y = 1/2}^{y=1} y^3 (x^2 + y^2)^{-3/2} \, dx \, dy$.
But you should consider switching to polar coordinates. In which case you'll need to express the line y = 1/2 in polar coordinates. Note that $\displaystyle \frac{\pi}{6} \leq \theta \leq \frac{5 \pi}{6}$.
Have you drawn a sketch of the region?
Switching to polar coordinates: $\displaystyle I = \int_{\theta = \pi/6}^{5 \pi/6} \int_{r = 1/(2 \sin \theta)}^{r=1} r \sin^3 \theta \, dr \, d\theta $.
If you're still stuck now, you'll need to say what part(s) of this expression you're struggling to get.