# Simple proof of integration in polar coordinates?

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 :)

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In the math mode use \sin instead of sin (similarly \cos instead of cos, \log instead of log, \tan instead of tan, etc) –  user17762 May 16 '12 at 18:24
There is an approximation argument for the relevant Riemann sum in a two dimensional figure that probably came before the Jacobian calculation in history. This should be in many vector calculus books, but will surely be in any physics books that are just at the point of needing 2-D and 3-D integrals. Informally, to get the average value of a function, you sum up a representative value on each chosen subregion times the area of the subregion divided by the (fixed) area of the whole region. To get the integral, you just multiply function average value times area of the region. –  Will Jagy May 16 '12 at 18:32
There's a subtlety in writing the volume element as $dx dy$ in that there's actually a wedge product taking place between the differentials, i.e. it should appear as $dx \wedge dy$. I think the wedge symbol is dropped for convenience of writing the integral.. –  tentaclenorm May 16 '12 at 18:58
Also, see volume element for a quick derivation of the volume element in $\mathbb{R}^3$ under a coordinate transformation, and the appearance of the Jacobian. –  tentaclenorm May 16 '12 at 19:00
The overlap between the answers below and the answers to this question is impressive. –  Did May 17 '12 at 18:04

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

$\hspace{3.5cm}$

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 $\mathbb{R}^2$:

$\hspace{3cm}$

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.

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Very nice graphics. (Thanks for the ping @tentaclenorm.) –  copper.hat May 17 '12 at 17:29

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.

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I think if one understands the wedge product, then the question would not arise? –  copper.hat May 16 '12 at 19:18
@copper.hat Yes, I just edited my answer to include the explicit calculation. Do you think I should include more detail? The OP's logic is good, but they need to use the wedge product for it to work out. –  tentaclenorm May 16 '12 at 19:28
Well, I may be struck dead for writing this, but the wedge product to me is one of those constructions that works nicely but provides no intuition. I am not sure that the determinant as volume measurement is an improvement, but it is familiar. –  copper.hat May 16 '12 at 20:00
@copper.hat I see what you mean. The wedge product isn't introduced to make things work out nicely, it's the reason that these calculations do work out nicely. I think this is one of those situations where intuition gets hindered by the lack of full disclosure when introducing these topics. Any course in multivariable/vector calculus should really be a course in differential geometry, but I think that's rarely the case. –  tentaclenorm May 16 '12 at 20:52
@copper.hat just wanted to ping you to also checkout Robjohn's answer. :) –  tentaclenorm May 17 '12 at 17:17

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 $[x,x+dx]\times[y,y+dy]$.

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We know that $$x=r\cos\theta,\quad y= r\sin\theta$$ Differentiate b.s respectively, we get $$dx= -r\sin\theta\, d\theta+ \cos\theta\, dr,\quad dy= r\cos\theta\, d\theta + \sin\theta\, dr$$
Then \begin{align}dx\cdot dy &= (-r\sin\theta\, d\theta+ \cos\theta\, dr).(r\cos\theta\, d\theta + \sin\theta\, dr)\\ &= - r\cdot r \sin\theta\cdot\cos\theta\, d\theta\cdot d\theta + r\cos\theta\cdot\cos\theta\, dr\cdot d\theta - r \sin\theta\sin\theta\, dr\,d\theta +\sin\theta\cdot\cos\theta\,dr\,dr\end{align}