Let $A \subseteq \mathbb{R}^2$ a limited set bordered through $x=0, x=1, y=-1+x, y=1-x^2$. Rotate A around the y-axis and define this set with $B$. Calculate the integral $$\int_B y\,\mathrm{d}x\mathrm{d}y\mathrm{d}z$$
In the script the integral was substitute with cylindrical coordinates, that's clearly but I'm struggling to find the limits:
$$\int_B y \,\mathrm{d}x\mathrm{d}y\mathrm{d}z = \int\limits_{0}^{1} \int\limits_{0}^{2\pi} \int\limits_{0}^{\sqrt{1-y}} r\cdot y\, \mathrm{d}r\mathrm{d}\varphi\mathrm{d}y + \int\limits_{-1}^{0} \int\limits_{0}^{2\pi}\int\limits_{0}^{y+1} y\cdot r\,\mathrm{d}r\mathrm{d}\varphi\mathrm{d}y$$
For substitution there used: $\begin{pmatrix} y\\ r\cdot \cos(\varphi) \\ r\cdot \sin(\varphi)\end{pmatrix}$
In every task we did before we substitute with $\begin{pmatrix} x\\ r\cdot \cos(\varphi) \\ r\cdot \sin(\varphi)\end{pmatrix}$. Why there is a different cylindrical coordinates substitution? I think they "turn" the point of view from the x-axis to the y-axis.
Do you have any recommendations how I set the right limits if I substitute with cylindrical coordinates (or polar coordinates)?
First, one comment: the change of variables formula tells you that the differential $dx \, dy \, dz$ will become $r \, dr \, d\varphi \, dy$ (it is $r$, not $r^2$).
To address your question "Why a different cylindrical coordinate substitution?":
Since $y$ is the axis of rotation, consider that the "up" direction, and so the coordinate change is $x = r \cos(\varphi)$ and $z = r\sin(\varphi)$. In other words, do the polar coordinate substitution in the "left" and "right" directions, whatever variables they may be.
Looks like you split the integral in two, for the regions above and below the $x,z$ plane. For the lower one, the bounds should all be constants since the region is a cylinder (and we are in "cylindrical" coordinates). In other words, you should not have $y+1$ in the bounds for the integral. This integral should be:
$$\int_{-1}^0 \int_0^{2\pi} \int_0^1 yr \, dr \, d\varphi \, dy $$
For the upper integral, I would choose a different order of integration. Integrate out $y$ first to get
$$ \int_0^{2\pi} \int_0^1 \int_0^{1-r^2} ry \, dy \, dr \, d\varphi $$ This is natural since the upper boundary of the region is given as the graph of $y=1-r^2$.