If $y=r\cos\theta$ and $z=r\sin\theta$, why $dydz=rdrd\theta$? Let random variable $X$ has a normal distribution with mean $\mu$ and variance $\sigma^2$. Then PDF of $X$ is given by:
$$f(x)=\frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2\right).$$
In the book “Probability and Statistics”, 2nd edition, by Morris H. DeGroot, the proof that $\int_{-\infty}^{\infty}f(x)dx=1$ follows like this:

Let $I=\int_{-\infty}^{\infty}\exp(-\frac{1}{2}y^2)dy,$ then
\begin{align}I^2&=\int_{-\infty}^{\infty}\exp\left(-\frac{1}{2}y^2\right)dy
> \int_{-\infty}^{\infty}\exp\left(-\frac{1}{2}z^2\right)dz\\&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp\left(-\frac{1}{2}\left(y^2+z^2\right)\right)dydz.\end{align}
Let $y=r\cos\theta$ and $z=r\sin\theta$, then, since $y^2+z^2=r^2,$
$$I^2=\int_{0}^{2\pi}\int_{0}^{\infty}\exp\left(-\frac{1}{2}r^2\right)rdrd\theta=2\pi.$$

I don't understand, why $dydz=rdrd\theta$?
Since $y=r\cos\theta$, we can find that $\frac{dy}{dr}=\cos\theta$ and $\frac{dy}{d\theta}=-r\sin\theta$. Also, $\frac{dz}{dr}=\sin\theta$ and $\frac{dz}{d\theta}=r\cos\theta$.
Then we can have the following combinations: $dydz=r\cos^2\theta drd\theta$, $dydz=r\cos^2\theta=-r\sin^2\theta drd\theta$.
 A: $\space\space\space$Rather than thinking about the Jacobian or partial derivatives, try and reason purely geometrically. In Cartesian coordinates, you are dividing $\mathbb{R}^2$ into squares aligned with the $x$ and $y$ axes. Therefore, your differential area element is simply an infinitesimal change in $x$ times an infinitesimal change in $y$ which gives you the familiar $dxdy$.
$\space\space\space$Now, in polar coordinates the infinitesimal area element must be defined in terms of your new coordinates $r$ and $\theta$. An infinitesimal change in $r$ will simply be $dr$. However, the effect that an infinitesimal change in $\theta$ will have on your area element will depend on how far away you are radially from the origin. The formula for arclength along a circular path is $L=r\cdot \theta $ which means that $dL=rd\theta$.
$\space\space\space$So, our infinitesimal change in $\theta$ must be scaled by a factor of $r$. Therefore, in polar coordinates, the infinitesimal area element is $rdrd\theta$.
