<s>[INST] Find the volume of the solid bounded by $z=x^2+y^2+1$ and $z=2-x^2-y^2$. Question: Find the volume of the solid bounded by $z=x^2+y^2+1$ and $z=2-x^2-y^2$.
Setting the 2 equations equal w.r.t. $z$, $x^2+y^2+1=2-x^2-y^2 \rightarrow x=\pm\sqrt{\frac 12-y^2}$
Therefore the boundary of $y=\pm\frac {1}{\sqrt2}$.
So to find the volume of the solid, take the integration by subtracting the volume above and below the boundaries.
$\displaystyle V=\int_{-\frac {1}{\sqrt2}}^{+\frac {1}{\sqrt2}}\int_{-\sqrt{\frac 12-y^2}}^{+\sqrt{\frac 12-y^2}}(2-x^2-y^2)dxdy-\int_{-\frac {1}{\sqrt2}}^{+\frac {1}{\sqrt2}}\int_{-\sqrt{\frac 12-y^2}}^{+\sqrt{\frac 12-y^2}}(x^2+y^2+1)dxdy$
This is what I did. Without solving the equation, can someone tell me if it is correct?
Thank you!
 [/INST] 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
$$\begin{aligned}
V &= \iiint_D \,dx\,dy\,dz\\
&= \iiint_D r\,dr\,d\theta\,dz\\
&= \int_{0}^{2\pi}\int_{0}^{\frac{1}{\sqrt{2}}}\int_{r^2 + 1}^{2 - r^2}r\,dz\,dr\,d\theta\\
&= \int_{0}^{2\pi}\int_{0}^{\frac{1}{\sqrt{2}}} r(2 - r^2 - r^2 - 1)\,dr\,d\theta\\
&= \int_{0}^{2\pi}\int_{0}^{\frac{1}{\sqrt{2}}} r(1 - 2r^2)\,dr\,d\theta\\
&= \int_{0}^{2\pi}\int_{0}^{\frac{1}{\sqrt{2}}} (r - 2r^3)\,dr\,d\theta\\
&= \int_{0}^{2\pi}\,d\theta\left.\left(\dfrac{1}{2}r^2 - \dfrac{1}{2}r^4 \right)\right\vert_{r = 0}^{r = \frac{1}{\sqrt{2}}}\\
&= 2\pi \left.\left(\dfrac{1}{2}r^2 - \dfrac{1}{2}r^4 \right)\right\vert_{r = 0}^{r = \frac{1}{\sqrt{2}}}\\
&= 2\pi \cdot \dfrac{1}{8}\\
&= \dfrac{\pi}{4}
\end{aligned}$$
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