Need to find the cylindrical coordinates for the solid that lies under $z=4-x^2-y^2$ and above the region $(x-1)^2+y^2\leq 1$
Then use a triple integral to find the volume.
I don't understand how to convert the coordinates without symmetry about the z axis.
On
If we take usual cylindrical coordinates, then you'll have $r-2\cos \phi \leqslant 0$, which gives on one hand curve for $r(\phi)$ and on another hand restriction $\cos \phi \geqslant 0$, which is $[-\frac{\pi}{2},\frac{\pi}{2}]$ as one possibility. So integral will be $$\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int\limits_{0}^{2\cos \phi}\int\limits_{0}^{4-r^2}\,d\phi dr dz$$
But if we take "shifted" cylindrical coordinates with $x=1+r\cos \phi$, then exactly now for $\phi$ will not be restrictions $$\int\limits_{0}^{2\pi}\int\limits_{0}^{1}\int\limits_{0}^{3-r^2-2r\cos \phi }\,d\phi dr dz$$ Jacobian same for both cases.
Moving the origin to $(1,0,0)$ decreases the $x$-coordinate by $1$. So for the equations to still describe the same surfaces / regions (they should stay where they are and not move along with our origin; if they did, moving the origin would be pointless), we have to substitute $x\mapsto x+1$. This gives us $z=4-(x+1)^2-y^2$ and $(x+1-1)^2+y^2\leq 1$. Tidying up and rewriting to polar form, we have $$ z=4-r^2-2r\cos\theta -1\\ r^2\leq1 $$ This means the integration bounds on $r$ and $\theta$ are both constant (as long as $z$ is the innermost integration variable), and the integral should be rather straight-forward.
As a side note, you are missing a bottom bound on your integration region. As it stands, it is a cylinder with a paraboloid top, stretching infinitely far down. Is it, perhaps, meant to be $z\geq0$?