<s>[INST] 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?
 [/INST] $$\displaystyle{\int _0^{2\pi }\int _0^{\sqrt{3}}\int _1^{\sqrt{4-r^2}}rdzdrd\theta} $$
$$\displaystyle{2\pi \int_0^{\sqrt{3}} \left(r\sqrt{4-r^2}-r\right) \, dr} $$
The first part you solve it by changing variables: 
$$u=4-r^2   \qquad du=-2rdr    \qquad    \frac{-du}{2}=rdr$$
After solving that, the volume is $\frac{5\pi}{3}$
There are 2 ways to solve the problem using cylindrical coordenates
$$0 < \theta <2\pi$$
$$0 < r < \sqrt{3} $$
$$ 1 < z < \sqrt{4-r^2} $$
or
$$0 < \theta <2\pi$$
$$1 < z < 2 $$
$$ 0 < r < \sqrt{4-z^2} $$
r will always go from the inside to the outside, z from de first plane it intersects to the other, and $\theta$ will spin totally in this case
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