You are very close.
The integral in cylindrical coordinates has the right value, but the lower limit for $z$ should be $r$, not $0$.
Notice that in spherical coordinates the angle $\phi$ should be integrated from $0$ to $\pi/4$. This restrict us to the volume inside the cone.
Addendum:
The volume is the region between the cone and the sphere in octant I ($x,y,z\ge 0$).
See the figure below.
In cylindrical coordinates we have
$$\int_{0}^{\frac{\pi}{2}}\int_{0}^{2}\int_{r}^{\sqrt{8-r^2}}\!z\ r\ \mathrm{d}z\ \mathrm{d}r\ \mathrm{d}\theta.$$
In spherical coordinates we have
$$\int_{0}^{\frac{\pi}{4}}\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\sqrt{2}}\!\rho\ \cos\phi\ \rho^2 \sin \phi\ \mathrm{d}\rho\ \mathrm{d}\theta\ \mathrm{d}\phi.$$
Given the set:
$$
\Omega := \left\{ (x,\,y,\,z) \in \mathbb{R}^3 : \sqrt{2\,x^2+2\,y^2} \le z \le y+1 \right\}
$$
it follows that:
$$
I := \iiint\limits_{\Omega} z\,\text{d}x\,\text{d}y\,\text{d}z = \iint\limits_D \frac{(y + 1)^2 - \left(2\,x^2+2\,y^2\right)}{2}\,\text{d}x\,\text{d}y
$$
where:
$$
\small
D := \left\{ (x,\,y) \in \mathbb{R}^2 : \sqrt{2\,x^2+2\,y^2} \le y+1 \right\}
= \left\{ (x,\,y) \in \mathbb{R}^2 : \frac{(x - 0)^2}{(1)^2} + \frac{(y - 1)^2}{\left(\sqrt{2}\right)^2} \le 1 \right\}.
$$
So, opting for the following coordinate transformation:
$$
\Phi :
\begin{cases}
x = 0 + 1\,\rho\,\cos\theta \\
y = 1 + \sqrt{2}\,\rho\,\sin\theta
\end{cases}
\; \; \; \; \; \; \text{with} \;
(\rho,\,\theta) \in [0,\,1] \times [0,\,2\pi]
$$
and Jacobian $J_{\Phi} = 1 \cdot \sqrt{2} \cdot \rho$, all this translates into the following calculation:
$$
\small
I = \int_0^{2\pi} \int_0^1 \frac{\left(2+\sqrt{2}\,\rho\,\sin\theta\right)^2 - 2\,\left(0+1\,\rho\,\cos\theta\right)^2 - 2\,\left(1+\sqrt{2}\,\rho\,\sin\theta\right)^2}{2}\,\sqrt{2}\,\rho\,\text{d}\rho\,\text{d}\theta = \frac{\pi}{\sqrt{2}}\,.
$$
Finally, wanting to verify the result obtained in Wolfram Mathematica:
Ω = ImplicitRegion[Sqrt[2 x^2 + 2 y^2] <= z <= y + 1, {x, y, z}];
Integrate[z, {x, y, z} ∈ Ω] // TraditionalForm
π/√2
Best Answer
The bounds on $z$ correspond to equations:
$$z^2=x^2+y^2\\x^2+y^2+z^2=8$$
The first is a cone, the second is a sphere. They intersect above the $xy$-plane in a circle of radius $2$. The bounds on $x$ and $y$ show that we are considering the region in between the cone and the sphere with $x>0$ and $y>0$. Can you see it?
Let's first just consider the $xy$-plane. As you've said, the given bounds correspond to the quarter circle of $x^2+y^2=4$ that lies in the first quadrant. So whatever region we consider, it will lie completely above this quarter circle. Now $z$ lies in between the cone and the sphere as I mentioned above. You can think of this region as an ice cream cone. However, we're only considering part of the ice cream cone above our quarter circle, so its like a quarter of an ice cream cone.
Hopefully this is helpful, along with Babak's picture.