Simple proof of integration in polar coordinates?

Question

In every example I saw of integration in polar coordinates the Jacobian determinant is used, not that I have a problem with the Jacobian, but I wondered if there's a simpler way to show this which will also give me some more intuition about the Jacobian.

If I try to simply write the differentials:

\begin{align} x & = r \cos \theta\\ y & = r \sin \theta\\ dx & = dr \cos \theta - r \sin \theta\ d\theta\\ dy & = dr \sin \theta + r \cos \theta\ d\theta\\ \end{align}

In a double integral you integrate $dxdy$, so if I try to plug in the results I'll get something which is not $r d\theta dr$ \begin{align} dxdy & = \left(dr \cos \theta - r \sin \theta\ d\theta \right) \left( dr \sin \theta + r \cos \theta\ d\theta\right)\\ & = dr^2 \cos \theta \sin \theta - r^2 d\theta^2 \cos \theta\ \sin\ \theta + r\ dr\ d\theta\ (\cos^2 \theta\ - \sin^2\theta ) \end{align}

I don't think I can go anywhere from here, I'm not sure if it's just a calculation mistake or the entire logic is bad.

How do I get this right?

Thanx :)

Answer 1

Of course, if you break $\mathbb{R}^2$ into a polar grid

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the small slightly curved rectangles have area $r\,\mathrm{d}\theta\,\mathrm{d}r$.

However, it seems that you are interested in looking at $$ \begin{align} \mathrm{d}y\,\mathrm{d}x &=(\sin(\theta)\,\mathrm{d}r+r\cos(\theta)\,\mathrm{d}\theta)(\cos(\theta)\,\mathrm{d}r-r\sin(\theta)\,\mathrm{d}\theta)\\ &=r\,\mathrm{d}\theta\,\mathrm{d}r \end{align} $$ and why the $\mathrm{d}r^2$ and $\mathrm{d}\theta^2$ terms disappear and the $\mathrm{d}r\,\mathrm{d}\theta$ and $\mathrm{d}\theta\,\mathrm{d}r$ have different signs.

Let's start with $$ \begin{align} \mathrm{d}x&=\cos(\theta)\,\mathrm{d}r-r\sin(\theta)\,\mathrm{d}\theta\\ \mathrm{d}y&=\sin(\theta)\,\mathrm{d}r+r\cos(\theta)\,\mathrm{d}\theta \end{align} $$ rewritten as $$ \begin{bmatrix}\mathrm{d}x\\\mathrm{d}y\end{bmatrix} =\begin{bmatrix}\cos(\theta)\\\sin(\theta)\end{bmatrix}\mathrm{d}r +\begin{bmatrix}-r\sin(\theta)\\r\cos(\theta)\end{bmatrix}\mathrm{d}\theta $$ Therefore, the displacements $\color{green}{\mathrm{d}r}$ and $\color{red}{\mathrm{d}\theta}$ get mapped to $\color{green}{\begin{bmatrix}\cos(\theta)\\\sin(\theta)\end{bmatrix}\mathrm{d}r}$ and $\color{red}{\begin{bmatrix}-r\sin(\theta)\\r\cos(\theta)\end{bmatrix}\mathrm{d}\theta}$ in $\mathbb{R}^2$:

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where the area in gray is given by $\color{green}{\begin{bmatrix}\cos(\theta)\\\sin(\theta)\end{bmatrix}\mathrm{d}r}\times\color{red}{\begin{bmatrix}-r\sin(\theta)\\r\cos(\theta)\end{bmatrix}\mathrm{d}\theta}=r\,\mathrm{d}r\,\mathrm{d}\theta$.

The fact that the cross product is involved is the reason that the $\mathrm{d}r^2$ and $\mathrm{d}\theta^2$ terms disappear and the $\mathrm{d}r\,\mathrm{d}\theta$ and $\mathrm{d}\theta\,\mathrm{d}r$ have different signs. This, and its $n$-dimensional analogs, are why we use wedge products and differential forms when changing variables.

Answer 2

The area element needs to be computed carefully.

This is very informal, but perhaps you should think of the volume element as a pair, rather than just a product, as in: $$\binom{dx}{dy} = \begin{bmatrix} \cos \theta & - r \sin \theta \\ \sin \theta & r \cos \theta \end{bmatrix}\binom{dr}{d\theta}$$ Multiplying a set by a matrix $A$ corresponds to changing the volume by a factor $\det A$. In this case, $\det A = r$, so the volume element computation becomes, informally, $dxdy = r dr d\theta$.

Addendum: The two volume elements are, informally, $[r,r+dr]\times[\theta,\theta+d\theta]$, and $[x,x+dx]\times[y,y+dy]$.

Answer 3

The short answer is that you need to consider the wedge product between the differentials, not a symmetric product as you have written. The reason for this is addressed in Robjohn's excellent answer.

The volume element, $dV$, is formally given by the wedge product of $dx_1, \dots, dx_n$. This means that in $\mathbb{R}^2$ the volume element in Cartesian coordinates should technically be written as $$ dV = dx \wedge dy. $$ Note that the wedge product is antisymmetric, which means that in particular $dx \wedge dy = - dy \wedge dx.$ Taking this into consideration, if you perfrom this wedge product between the $dx$ and $dy$ you calculated, we have the following $$\begin{align*} dV &= dx \wedge dy \\ &= (\cos\theta ~dr - r\sin\theta ~d\theta) \wedge (\sin\theta ~dr + r\cos\theta~ d\theta) \\ &= \cos\theta \sin\theta ~dr\wedge dr + r \cos^2\theta ~dr \wedge d\theta - r\sin^2\theta~ d\theta \wedge dr -r \sin\theta\cos\theta~ d\theta \wedge d\theta\\ &= r(\cos^2\theta + \sin^2\theta) dr \wedge d\theta\\ &= r~ dr \wedge d\theta, \end{align*} $$ where we have also used the fact that $dr \wedge dr = d\theta \wedge d\theta = 0.$ If you calculate the determinant of the Jacobian you'll find $$\det\left(\dfrac{\partial(x,y)}{\partial{(r,\theta)}}\right) = r,$$ which concurs with the above calculation.