Question about a substitution in an integral So studying integrals for a short duration I found this example of an equation. It goes the following way:
$\int_{-r}^r \! \sqrt{r^2-y^2} \, \mathrm{d}y. \stackrel{y=rz}{=} \int_{-1}^1 \! r^2 \sqrt{1-z^2} \, \mathrm{d}z. $
If I get this right $y$ has been substituted with $rz$. Why could this substitution be made and why could $r^2$ be extrated from the root and not just $r$ ? I think I am missing something bigger here, but I just don't get it.
If you need further context for this equation, message me.

By pure textual substitution, then reworking,
$$\int_{y=-r}^r \! \sqrt{r^2-y^2} \, \mathrm{d}y \stackrel{y=rz}{=} \int_{rz=-r}^r \!\sqrt{r^2-(rz)^2} \, \mathrm{d}(rz){=} \int_{z=-1}^1 \!r\sqrt{1-z^2} \, r\mathrm{d}z,$$ hence the claim.
