This is a follow-on to Dr. Shifrin's comment about depicting the derivative of a polar coordinate transformation. He said "The row vectors you don't see so much geometrically." I agree with him, when it comes to the depiction in $x\times{y}$-space. And I'm not suggesting he should have taken the tangent I took during his lecture.
https://youtu.be/dxt3eZ2zLq4?t=415
But I did think: the rows are one-forms. They are the gradients of the $x\times{y}$ coordinates with respect to the $r\times{\phi}$ coordinates. The place to depict them is in $r\times{\phi}$-space. A differential-forms purist would not depict them as arrows, but rather as level surfaces. Nonetheless, I used arrows. The representations are not to scale. Nor do they represent the same point of evaluation. I am confident the component representations are correct.
My question is at the bottom
The next part is just context, and my demonstration that I have some clue as to what this all means.
Beginning with the transformation $$\vec{r}^{\prime}=\begin{bmatrix}\cos\phi & -r\sin\phi\\ \sin\phi & r\cos\phi \end{bmatrix}=\begin{bmatrix}\partial_{r}\vec{r} & \partial_{\phi}\vec{r}\end{bmatrix}=\begin{bmatrix}d_{\mathbf{u}}x\\ d_{\mathbf{u}}y. \end{bmatrix}$$
It seems correct to say $$\left|\vec{r}^{\prime}\right|=\partial_{r}\vec{r}\wedge\partial_{\phi}\vec{r}=d_{\mathbf{u}}x\wedge d_{\mathbf{u}}y.$$
So the determinant is a differential two-form. Were the depictions drawn to scale, the vector parallelograms in both diagrams would contain the same number of squares to the first order. That relation becomes equality in the limit. So my pictures say the same thing as
$$\begin{aligned} dx&=\partial_{r}xdr+\partial_{\phi}xd\phi\\ dy&=\partial_{r}ydr+\partial_{\phi}yd\phi\\ dx\wedge dy&=\left(\partial_{r}x\partial_{\phi}y-\partial_{r}y\partial_{\phi}x\right)dr\wedge d\phi . \end{aligned}$$
If there were a one-form $\omega$ such that $$\begin{aligned} \omega&=w_{x}dx+w_{y}dy\\ d\omega&=\left(\partial_{x}w_{y}-\partial_{y}w_{x}\right)dx\wedge dy=dx\wedge dy , \end{aligned}$$
then we could find the area of the depicted region $\mathcal{R}$ by integrating over the boundary $$\int_{\partial\mathcal{R}}\omega=\int_{\mathcal{R}}d\omega=a\left(\mathcal{R}\right).$$
So lets try this: $$\begin{aligned} \omega&=\frac{-ydx+xdy}{2}\\ d\omega&=\left(\partial_{x}x-\partial_{y}\left(-y\right)\right)dx\wedge dy=dx\wedge dy\\ dx&=\partial_{r}xdr+\partial_{\phi}xd\phi=\cos\phi dr-r\sin\phi d\phi\\ dy&=\partial_{r}ydr+\partial_{\phi}yd\phi=\sin\phi dr+r\cos\phi d\phi\\ \omega&=\frac{-ydx+xdy}{2}\\ &=\frac{-r\sin\phi\left(\cos\phi dr-r\sin\phi d\phi\right)+r\cos\phi\left(\sin\phi dr+r\cos\phi d\phi\right)}{2}\\ &=\frac{r^{2}}{2}d\phi .\end{aligned}$$
Eyeballing the integral gives
$$\frac{\left(r_{o}+\Delta r\right)^{2}-r_{o}^{2}}{2}\Delta\phi=\left(r_{o}+\frac{\Delta r}{2}\right)\Delta r\Delta\phi .$$
Great! I spent a day doing what I could have done in less than 5 minutes!
Is there a way to see this using Green's theorem and pictures?
In books on general relativity I have read that the number of grid cells enclosed in the depicted area can be determined by keeping track of the number of coordinate curves encountered during the traversal of the boundary. But I have never understood how to do that. Can someone please explain this?
