I am trying to gain a better intuition for conversion from Cartesian to polar coordinates during integration. I understand from other readings that the change of variables from Cartesian to polar coordinates requires the transform
$$ dxdy = \frac{dxdy}{drd\theta} drd\theta $$ where $\frac{dxdy}{drd\theta}$ is the determinant of the Jacobian, and is required to scale the area between Cartesian and polar space.
To build further intuition I have been playing around with how increments in $x$, $y$ lead to increments in $r, \theta$. From these changes $dr, d\theta$ I want to check the change in area in polar coordinates. However, I can't quite get the example below to work out.
Start with a vector in Cartesian coordinates at $(1, 1)$ and increment the $x$ component $dx=2$ and the $y$ component $dy=1$, with the resulting vector $(3, 2)$. In polar coordinates, the starting vector is $(\sqrt{2}; \frac{\pi}{4})$
Now I would like to translate this increment $dx$ and $dy$ to polar coordinates.
$$ dr = \frac{\partial r}{\partial x} dx + \frac{\partial r}{\partial y} dy $$
$$ d\theta = \frac{\partial \theta}{\partial x} dx + \frac{\partial \theta}{\partial y} dy $$
$$ r = \sqrt{x^2 + y^2}, \ \theta = \arctan{\frac{y}{x}} $$
$$ \frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} $$
$$ \frac{\partial r}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}} $$
$$ \frac{\partial \theta}{\partial x} = \frac{-y}{x^2 + y^2} $$
$$ \frac{\partial \theta}{\partial y} = \frac{x}{x^2 + y^2} $$
$$ dr = \frac{1}{\sqrt{2}} \cdot 2 + \frac{1}{\sqrt{2}} \cdot 1 = \frac{3}{\sqrt{2}} $$
$$ d\theta = \frac{-1}{2} \cdot 2 + \frac{1}{2} \cdot 1 = \frac{-1}{2} $$
where $x=1$ and $y=1$ used as the initial coordinates. But, the polar coordinates $(\sqrt{2} + \frac{3}{\sqrt{2}}; \frac{\pi}{4} - \frac{1}{2})$ is not $(3, 2)$ in Cartesian coordinates.
Is this because $dx$, $dy$ should not be used in this way and are only to be considered increments meaningful or accurate as their limit goes to zero? Or maybe my conceptualisation is incorrect elsewhere.
