Your setup is right. Here is the method you could have done to compute the volume.
Assume the density is $f(x,y,z) = 1$, so
$$V = \iiint_D \,dx\,dy\,dz$$
We are given that the solid is bounded by $z = x^2 + y^2 + 1$ and $z = 2 - x^2 - y^2$. As I commented under your question, you need to use cylindrical coordinates to evaluate the volume integral. Using the substitutions $x = r\cos(\theta)$, $y = r\sin(\theta)$ and $z = z$, we have $z = r^2 + 1$ and $z = 2 - r^2$. With some knowledge in graphs and functions, we see that the bounds are
$$\begin{aligned}
r^2 + 1 \leq z \leq 2 - r^2\\
0 \leq \theta \leq 2\pi\\
0 \leq r \leq \dfrac{1}{\sqrt{2}}
\end{aligned}$$
where $r = \frac{1}{\sqrt{2}}$ is found by solving for $r$ when $r^2 + 1 = 2 - r^2$. So for the volume triple integral, we have
$$\begin{aligned}
V &= \iiint_D \,dx\,dy\,dz\\
&= \iiint_D r\,dr\,d\theta\,dz\\
&= \int_{0}^{2\pi}\int_{0}^{\frac{1}{\sqrt{2}}}\int_{r^2 + 1}^{2 - r^2}r\,dz\,dr\,d\theta\\
&= \int_{0}^{2\pi}\int_{0}^{\frac{1}{\sqrt{2}}} r(2 - r^2 - r^2 - 1)\,dr\,d\theta\\
&= \int_{0}^{2\pi}\int_{0}^{\frac{1}{\sqrt{2}}} r(1 - 2r^2)\,dr\,d\theta\\
&= \int_{0}^{2\pi}\int_{0}^{\frac{1}{\sqrt{2}}} (r - 2r^3)\,dr\,d\theta\\
&= \int_{0}^{2\pi}\,d\theta\left.\left(\dfrac{1}{2}r^2 - \dfrac{1}{2}r^4 \right)\right\vert_{r = 0}^{r = \frac{1}{\sqrt{2}}}\\
&= 2\pi \left.\left(\dfrac{1}{2}r^2 - \dfrac{1}{2}r^4 \right)\right\vert_{r = 0}^{r = \frac{1}{\sqrt{2}}}\\
&= 2\pi \cdot \dfrac{1}{8}\\
&= \dfrac{\pi}{4}
\end{aligned}$$
Let
$(x,y,z) \in \mathbb{R}^{3}$ and let $S$ be the region enclosed by the surfaces $z = x^{2}+y^{2}$ and $z=4$. Then $(x,y,z) \in S$ if and only if $|x| \leq 4, |y| \leq \sqrt{4-x^{2}}, 0 \leq z \leq 4$. Thus
$$4\int_{0}^{4}\int_{0}^{\sqrt{4-x^{2}}} x^{2} + y^{2} dy dx = 4\int_{0}^{4} x^{2}\sqrt{4-x^{2}} + \frac{(4-x^{2})^{3/2}}{3} dx$$
is the content of $S$.
Let $T$ be the region enclosed by $z=4$ and $z = 4x^{2} + 4y^{2}$ and do the above for $T$. Then the absolute value of the difference of the resulting two contents is the desired content.
Best Answer
If you express your region in cylindrical coordinates, then $x^2+y^2=1$ becomes $\rho=1$ and $z=y^2$ becomes $z=\rho^2\sin^2(\theta)$. So, you can compute your volume as follows:\begin{align}\int_0^{2\pi}\int_0^1\int_0^{\rho^2\sin^2(\theta)}\rho\,\mathrm dz\,\mathrm d\rho\,\mathrm d\theta&=\int_0^{2\pi}\int_0^1\rho^3\sin^2(\theta)\,\mathrm d\rho\,\mathrm d\theta\\&=\left(\int_0^{2\pi}\sin^2(\theta)\,\mathrm d\theta\right)\left(\int_0^1\rho^3\,\mathrm d\rho\right)\\&=\frac\pi4.\end{align}