Trying to solve this triple integral: $\iiint (x-1)(y-1) \,dx\,dy\,dz$ Here's the question
$$\iiint (x-1)(y-1) \,dx\,dy\,dz.$$
I am asked to evaluate this integral over the region $$D:=\left \{ (x,y,z) \in\mathbb{R}^3 :x^2+y^2 \leq z \leq 2x+2y+2 \right \}.$$
There are the bounds of integration in set D (the variable $z$ is isolated) and well I tried to find the solution of this integral :
\begin{align*}
&\iint_{Pr_{(y,x)}(D)}\int_{x^2+y^2}^{2x+2y+2}(x-1)(y-1) \,dx\, dy\, dz \\
=&\iint_{Pr_{(y,x)}(D)}\int_{x^2+y^2}^{2x+2y+2}(xy-x-y+1) \,dx\, dy\, dz,
\end{align*}
and integrate only with respect to $z.$ I have that:
\begin{align*}
\int_{x^2+y^2}^{2x+2y+2}(xy-x-y+1) \,dz&=(xy-x-y+1)*(2x+2y+2-(x^2+y^2)) \\
&=3x^2y+3xy^2-2x^3y-2xy^3-3x^2+x^3-2xy-3y^2+y^3+2.
\end{align*}
It looks like this way is too long. The second thing that came to mind when I saw the set $D$ was to apply cylindrical coordinates, but this doesn't make easier the left member of the set $D.$ 
What can I do or what have I done wrong up until now?
Any support for this question would be appreciated.

Hint. Note that the intersection of paraboloid $z=x^2+y^2$ and the plane $z=2x+2y+2$ projected onto $xy$-plane is given by the circle $x^2+y^2=2x+2y+2$ that is
$$(x-1)^2+(y-1)^2=2^2.$$
Hence the given triple integral becomes
$$\iint_{(x-1)^2+(y-1)^2\leq 2^2}\int_{x^2+y^2}^{2x+2y+2}(x-1)(y-1) dx dy dz,$$
and after integrating with respect to $z$ we get
$$\iint_{(x-1)^2+(y-1)^2\leq 2^2}(x-1)(y-1)(4-(x-1)^2-(y-1)^2) dx dy.$$
Now use the polar coordinates centered at $(1,1)$. Can you take it from here?
