Find the volume of the region contained above $z=1$ and below $x^{2}+y^{2}+z^{2}=4$ Why doesn't this work?

Find the volume of the region contained above $z=1$ and below $x^{2}+y^{2}+z^{2}=4$

going to cylindrical this should be easy. $z=(4-r^{2})^{\frac {1}{2}}$ and $z=1$
Clearly $z=1$ yields  $x^{2}+y^{2}=3 \to r=(3)^{\frac {1}{2}}$
$$\int^{2\pi}_{0}\int^{0}_{(3)^{\frac {1}{2}}} [(4-r^{2})^{\frac {1}{2}} -1] \, r\,dr\,d\theta$$
$$\int^{2\pi}_{0}\left[-\frac {1}{3} (4-r^{2})^{\frac {3}{2}}-\frac{r^{2}}{2}\right]\Bigg|^{0}_{(3)^{\frac {1}{2}}} d\theta$$
$$-\frac{8}{3}-\left[-\frac{1}{3}-\frac{3}{2}\right] \to -\frac{16}{6} +\frac{11}{6} \to -\frac{5}{6} \to -\frac{5\pi}{3}$$ 
I don't get why the bound isn't supposed to be $\int^{0}_{(3)^{\frac {1}{2}}}$ - why is it $\int^{(3)^{\frac {1}{2}}}_{0}?$ Clearly when I start integrating, my radius is root 3 and it reduces to 0?


Here is a cute picture of the question.
