Is changing variables the same as substitutions? I have asked several on a similar matter. This time the question is tad different.
$$\int_{\mathbb{R}} e^{-x^2} dx$$
We let $x=y \implies dx=dy$
$$\int_{\mathbb{R}} e^{-(x^2 + y^2)} dxdy$$
But then how do we conclude that:
$$x^2 + y^2 = r^2$$
Since $x^2 + y^2 = 2x^2$??
Thanks!
 A: First Problem
While it is true that using the substitution $x\mapsto y$, we get
$$
\int_{\mathbb{R}}e^{-x^2}\,\mathrm{d}x=\int_{\mathbb{R}}e^{-y^2}\,\mathrm{d}y\tag{1}
$$
$(1)$ is not equal to
$$
\int_{\mathbb{R}}\int_{\mathbb{R}}e^{-(x^2+y^2)}\,\mathrm{d}x\,\mathrm{d}y=\int_{\mathbb{R}}e^{-x^2}\,\mathrm{d}x\int_{\mathbb{R}}e^{-y^2}\,\mathrm{d}y\tag{2}
$$
Furthermore, neither $x$ nor $y$ are dependent on the other in either $(1)$ or $(2)$. In $(1)$, we have simply renamed a dummy variable. This is the same as substituting $x\mapsto y$ to go from
$$
f(x)=x^2+2\tag{3}
$$
to
$$
f(y)=y^2+2\tag{4}
$$

Second Problem
To convert the integral
$$
\int_{\mathbb{R}}\int_{\mathbb{R}}e^{-(x^2+y^2)}\,\mathrm{d}x\,\mathrm{d}y\tag{5}
$$
to polar coordinates, we have $r^2=x^2+y^2$ and $\mathrm{d}x\,\mathrm{d}y=r\,\mathrm{d}\theta\,\mathrm{d}r$ where we've used the Jacobian of the change of variables
$$
\begin{align}
x&=r\cos(\theta)\\
y&=r\sin(\theta)
\end{align}\tag{6}
$$
That is,
$$
\begin{align}
\det\frac{\partial(x,y)}{\partial(r,\theta)}
&=\det\begin{bmatrix}\dfrac{\partial x}{\partial r}&\dfrac{\partial y}{\partial r}\\\dfrac{\partial x}{\partial \theta}&\dfrac{\partial y}{\partial \theta}\end{bmatrix}\\
&=\det\begin{bmatrix}\cos(\theta)&\sin(\theta)\\-r\sin(\theta)&r\cos(\theta)\end{bmatrix}\\[12pt]
&=r\tag{7}
\end{align}
$$
Thus, $(5)$ becomes
$$
\begin{align}
\int_0^\infty\int_0^{2\pi}e^{-r^2}\,r\,\mathrm{d}\theta\,\mathrm{d}r
&=2\pi\int_0^\infty e^{-r^2}\,r\,\mathrm{d}r\\
&=\pi\int_0^\infty e^{-s}\,\mathrm{d}s\\
&=\pi\tag{8}
\end{align}
$$
where we have used another substitution: $s\mapsto r^2$.
Using $(1)$, $(2)$, $(5)$, and $(8)$, we get
$$
\left(\int_{\mathbb{R}}e^{-x^2}\,\mathrm{d}x\right)^2=\pi\tag{9}
$$
