Product of partial derivatives of polar coordinates (Demonstration of Normal Density Function)

Question

I am studying the proof of the density function of the normal distribution. In one of the steps, the demonstration goes from

$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(y^2 + x^2)/2}dydx$

to $\int_{0}^{\infty}\int_{0}^{2\pi} e^{-r²/2}rd\theta~dr$

The explanation is that the author changed variables to polar coordinates: $x = r\cos\theta$, $y = r\sin\theta$, and $dydx = rd\theta~dr$

My question is: how did it come to $dydx = rd\theta~dr$ ?

I tried the following:

$\frac{dx}{d\theta} = -r\sin\theta$

$\frac{dy}{dr} = \sin\theta$

$\frac{dx}{d\theta}\frac{dy}{dr} = -r\sin\theta \sin\theta = -r\sin^2\theta$

$dxdy = -r\sin^2\theta d\theta dr$

I do not know how to go from $-r\sin^2\theta d\theta dr$ to $rd\theta dr$.

Someone could help me to see where I am doing it wrongly?

This may help: https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant — Hussain-Alqatari, Aug 08 '20 at 11:55
This question has been asked several times: take a look at Polar Coordinate Transformation - Motivation or Why is this derivation of the transformation from Cartesian to polar coordinates not correct? (these two are just ones which I remember writing, but there are several other answers on some of the other linked questions). Also, what you calculated are partial derivatives $\frac{\partial x}{\partial \theta}=-r\sin\theta$ etc, not total derivatives; but even otherwise it's still wrong method — peek-a-boo, Aug 08 '20 at 12:16

0 Answers0