<s>[INST] Find the volume between the surface $x^2+y^2+z=1$ and $ z=x^2+(y-1)^2$ I'm trying to find the volume between the surface $x^2+y^2+z=1$ and $ z=x^2+(y-1)^2$ but nothing works for me.
I made the plot and it looks like this:

How could you start? Any recommendation?
 [/INST] Try checking where the two surfaces intersect. Solve the given equations for $z$ and set them equal:
$$\begin{align*}
1-x^2-y^2&=x^2+(y-1)^2\\
0&=2x^2+2y^2-2y\\
0&=x^2+y^2-y\\
0&=x^2+\left(y-\frac12\right)^2-\frac14\\
x^2+\left(y-\frac12\right)^2&=\frac1{2^2}
\end{align*}$$
which corresponds to the cylinder of radius $\frac12$ with cross sections parallel to the $(x,y)$ plane and centered over the point $\left(0,\frac12,0\right)$.
With this in mind, a change to cylindrical coordinates is the "obvious" move. Take
$$\begin{cases}x=r\cos\theta\\y=\frac12+r\sin\theta\\z=z\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm d\theta\,\mathrm dz$$
The solid (call it $S$) is then given by the set
$$S=\left\{(r,\theta,z)\mid0\le r\le\frac12,0\le\theta\le2\pi,r^2-r\sin\theta+\frac14\le z\le\frac34-r^2-r\sin\theta\right\}$$
where the last inequality follows from
$$\begin{align*}
&x^2+(y-1)^2\le z\le1-x^2-y^2\\
&\implies r^2\cos^2\theta+\left(r\sin\theta-\frac12\right)^2\le z\le1-r^2\cos^2\theta-\left(\frac12+r\sin\theta\right)^2\\
&\implies r^2-r\sin\theta+\frac14\le z\le\frac34-r^2-r\sin\theta
\end{align*}$$
The volume is then
$$\iiint_S\mathrm dx\,\mathrm dy\,\mathrm dz=\int_0^{2\pi}\int_0^{\frac12}\int_{r^2-r\sin\theta+\frac14}^{\frac34-r^2-r\sin\theta}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$$

 $$=\frac\pi{16}$$

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