Volume calculated by different orders of integration How to write  volume of region
$$S=\{(x,y,z): (x-2)^2+y^2<4, z<3-x^2-y^2,z=-1\}$$ in orders x-y-z and y-z-x of integration.
 I tried something but I don't get same result.
First, for x-y-z order I found proection to x-y plane 
so I think integral is: $$\int\limits_{x=0}^1 \int\limits_{y=-\sqrt{4 - (x - 2)^2}}^{\sqrt{4 - (x - 2)^2}} \int\limits_{z=-1}^{3-x^2-y^2} 1\ dz\ dy\ dx + \int\limits_{x=1}^2 \int\limits_{y=-\sqrt{4 - x^2}}^{\sqrt{4 - x^2}} \int\limits_{z=-1}^{3-x^2-y^2} 1\ dz\ dy\ dx $$
For y-z-x orientation,similary I found proection  and integral
$$\int\limits_{y=-2 }^2 dy\int\limits_{-1 }^{3-y^2} dz\int\limits_{x=2- \sqrt{4-y^2}}^{\sqrt{3-y^2-z}} 1\ dx $$


$$\int\limits_{x=0}^2 \int\limits_{y=-\sqrt{4 - (x - 2)^2}}^{\sqrt{4 - (x - 2)^2}} \int\limits_{z=-1}^{3-x^2-y^2} 1\ dx\ dy\ dz = \frac{64}{3} - 4 \pi$$
