while the problem is a physics problem, the question is a mathematical one.
If a proton is assume to be infinite heavier than an electron, the Hamilton operator for the H-atom is
$\hat{H} = -\frac{\hbar^2}{2m_e}\Delta - \frac{Ze^2}{4\pi \epsilon_0 r} \tag{1}$
whereas $Z$ is the effective nuclear charge and $r$ is the distance between core and electron.
The laplace operator is
\begin{align}\Delta &= \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \tag{2}\\ &= \frac{1}{r}\bigg[\frac{\partial^2}{\partial r^2}\bigg]r + \frac{1}{r^2\sin \theta}\bigg[\frac{\partial}{\partial\theta}\sin\theta\frac{\partial}{\partial\theta}+\frac{1}{\sin\theta}\frac{\partial^2}{\partial \phi^2}\bigg] \tag{3}\end{align}
Using the substitution
$a_0 = \frac{4\pi\epsilon_0 \hbar^2}{m_e e^2}$ and $E_h=\frac{e^2}{4\pi\epsilon_0 a_0}$
we can rewrite (1) to
$\frac{\hat{H}}{E_h} = -\frac{1}{2}\Delta' - \frac{Z}{r'} \tag{4}$
whereas
$x' =\frac{x}{a_0}, y' =\frac{y}{a_0}, z' =\frac{z}{a_0}, r' = \frac{r}{a_0}, r^{'2} = x^{'2} + y^{'2} + z^{'2} $
and
$\Delta' = \frac{\partial^2}{\partial x^{'2}} + \frac{\partial^2}{\partial y^{'2}} + \frac{\partial^2}{\partial z^{'2}}$
We do that to get fancy units which we like. I just write it down for completeness. The $a_0$ is a Bohr (length unit) and the $E_h$ is a Hartree (energy unit).
Now show that the wave equation
$\Psi_{1s}(r',\theta,\phi) = \bigg(\frac{Z^3}{\pi}\bigg)^{1/2} \exp\{-Zr'\}$
a normalized eigen function of (4) is.
Now I just want to check the normalized part. So I check if $\left<\Psi_{1s}|\Psi_{1s}\right>=1$.
Now the solution does compute the above integral like this:
$\left<\Psi_{1s}|\Psi_{1s}\right>=\frac{Z^3}{\pi}\int_0^{2\pi}\int_0^\pi\int_0^\infty e^{-2Zr'}r^{'2}dr'\sin\theta d\theta d\phi$
Which makes sense: We get the $r'$ term and the $\sin\theta$ term from the determinant of the Jacobian of the transformation.
What I did was: $\left<\Psi_{1s}|\Psi_{1s}\right>=\frac{Z^3}{\pi}\int_0^{2\pi}\int_0^\pi\int_0^\infty e^{-2Zr'} d\theta d\phi$
so I did not actually transform it. I thought: The equation is given in spherical coordinates, so we can just compute it "inside the transformed space" and it should still be normalized.
So my question is simple: Why do I have to transform it? Is normalization coordinate dependent?
Normalisation is not coordinate-dependent in the sense that if you transform the functions they will still be normalised. But to do the integral you have to parameterize the space correctly.
You need the Jacobian ($r^2 \sin \theta$) to get the right weighting for each volume element. Think about the integral as a weighted sum of volume elements. If you assign the wrong weights you will get a different sum. Only in Cartesian coordinates is $d^3 r = dxdydz$. In curvilinear coordinate systems, the volume element itself is a function of the coordinates.