Evaluating an integral. I am trying to integrate $$\iint_{D}^\ y^3(x^2+y^2)^{-\frac{3}{2}} dx\, dy$$
on $D$, the region determined by the conditions $1/2\leq y\leq 1$ and $x^2+y^2\leq 1$ 
My book suggested we use $y^2=u$, but I don't know how to evaluate it, and I would really like to know the process.

This may be done in rectangular coordinates.  The integral looks like
$$\int_{1/2}^1 dy \, y^3 \, \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \frac{dx}{(x^2+y^2)^{3/2}} $$
To evaluate the inner integral, use a trig substitution such as $x=y \tan{t}$.  If you do everything correctly, you should get
$$2 \int_{1/2}^1 dy \, y \, \sqrt{1-y^2} = \frac{\sqrt{3}}{4}$$
By the way, this is where that substitution suggested by your book comes in handy.
