# Dx in polar?

Ok, so we know that $x=rcos(\theta)$

So what is dx?

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Furthermore, can I get dS in polar by finding dx and dy in polar and then substituting them into dS for rectangular? Is there an easier way to solve for dS in polar?

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You can figure out the differentials $dx$ and $dy$ from the general formula for a multivariable differential: $$x = x(r, \theta) = r \cos \theta \Leftrightarrow dx = \frac{\partial x}{\partial r}dr+\frac{\partial x}{\partial \theta}d\theta = \cos \theta\,dr - r \sin \theta\,d\theta$$
$$y = y(r, \theta) = r \sin \theta \Leftrightarrow dy = \frac{\partial y}{\partial r}dr+\frac{\partial y}{\partial \theta}d\theta = \sin \theta\,dr + r \cos \theta\,d\theta$$
For the relation between the area elements $dA$, you have to use the Jacobian:
$$dx dy = \frac{\partial(x,y)}{\partial(r,\theta)}dr\,d\theta = (r \cos^2 \theta + r \sin^2 \theta)\,dr\,d\theta = r\,dr\,d\theta$$
This is also easily seen if you sketch out what is happening. If you change $\theta$ by a small amount, you change one side of your area element by $r\,d\theta$. If you change $r$ by a small amount, you change the other side of your area element by $dr$.

arildno