<s>[INST] Using Spherical coordinates find the volume: Inside the surfaces $z=x^2+y^2$ and $z=\sqrt{2-x^2-y^2}$
I integrated over the ranges:
$0 \leq \theta \leq 2\pi$
$ 0 \leq \phi \leq \frac{\pi}{2}$
$0 \leq r \leq \sqrt{2}$
I get $\frac{\pi}{2}(4\sqrt{2} -4).$
There answer is the same except a $-\frac{7}{2}$ instead of the 4 at the end. Obviously I'm missing a 1\2 but I seems, I can not find it.
 [/INST] This problem is actually better suited for cylindrical coordinates:
$$\begin{align}\int_{0}^{2\pi}\int_0^1\int_{r^2}^{\sqrt{2-r^2}}rdzdrd\theta&=2\pi\int_0^1(r\sqrt{2-r^2}-r^3)dr\\&=2\pi(-\frac{1}{3}(2-r^2)^{3/2}-\frac{r^4}{4})\mid_{0}^1\\&=2\pi(-\frac{1}{3}-\frac{1}{4}+\frac{2\sqrt{2}}{3})\\&=\frac{\pi}{3}(4\sqrt2-\frac{7}{2})\end{align}$$
The problem with spherical coordinates here is that the radius is a (piecewise) function of the azimuthal angle, which makes the integration a bit more difficult.
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