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.
When $z=x^2+y^2=r^2$ and $z=\sqrt{2-x^2-y^2}=\sqrt{2-r^2}$ intersect each other, you will have $$r=1$$ This means that $z=1$ and $\tan(\phi)=1$. So the range of changing for $\phi$ will be $\pi/4\leq\phi\leq\pi$. While typing, I saw @Mhenni did it completely, so I am ending. :-)
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.
A related problem. The title of your problem asks for the use of spherical coordinates. You got the wrong integral limits. Note that, $ \rho $ (I am using $\rho$ instead of $r$) is bounded from above by two different surfaces. The sphere when $ 0 \leq \phi \leq \frac{\pi}{4} $ and the parabolic when $\frac{\pi}{4} \leq \phi \leq \frac{\pi}{2} $. So, we have
$$ V = \left(\int_{0}^{2\pi}\int_{0}^{\pi/4}\int_{0}^{\sqrt{2}}+\int_{0}^{2\pi}\int_{\pi/4}^{\pi/2}\int_{0}^{\frac{\cos(\phi)}{\sin^2(\phi)}}\right)\rho^2\sin(\phi)d\rho d\phi d\theta. $$
I leave it here for you to do the calculations.
that, for $0\leq \rho \leq \frac{\cos(\phi)}{\sin^2(\phi)}$, we used the equation $z=x^2+y^2$ to get . I leave it here for you to do the calculations.
Notes:
1) To get $\rho = \frac{\cos(\phi)}{\sin^2(\phi)}$ use the equation $z=x^2+y^2$ and the spherical coordinates of $x,y,$ and $z$.
2) The integral
$$ \int \frac{\cos^3(\phi)}{\sin^5(\phi)}d\phi = \int \cot^3(\phi) \csc^2(\phi) d\phi=-\frac{1}{4}\cot^4(x). $$
This was a trick problem from the book, use cylindrical. Notice it said, "when suitable". They are not circles.
