How $dxdy$ becomes $rdrd\theta$ during integration by substitution with polar coordinates

Polar Coordinates Basic Introduction, Conversion to Rectangular, How to Plot Points, Negative R Valu
Polar Coordinates Basic Introduction, Conversion to Rectangular, How to Plot Points, Negative R Valu

Does anyone know how $dxdy$ becomes $rdrd\theta$ in the example below? I always end up with cosines and sines in the expression no matter how I go about it and I’m not sure how they are disappearing from the expression.

How $dxdy$ becomes $rdrd\theta$ during integration by substitution with polar coordinates

  • $\begingroup$ HINT: “r” is the Jacobian determinant… $\endgroup$ Nov 25, 2019 at 22:31
  • 2$\begingroup$ It is nonsense to write $dx\,dy=(-r\sin(\theta)\,d\theta+\cos(\theta)\,dr)(r\cos(\theta)\,d\theta+\sin(\theta)\,dr)$. $\endgroup$ Nov 25, 2019 at 22:38
  • $\begingroup$ @MarkViola It’s fine if you put in explicit wedge products. $\endgroup$– J.G.Nov 25, 2019 at 22:43
  • $\begingroup$ @J.G. Is wedge products just the generalized name for ijk unit vectors? That had crossed my mind as I was trying to decide whether you should take the derivative of dx and dy with respect to theta or r because it did seem kind of weird to just arbitrarily pick one or the other (or a mix) when calculating dx and dy. $\endgroup$ Nov 25, 2019 at 23:02
  • 1$\begingroup$ @DKNguyen You can apply wedge products on vectors directly, which yields a nice way of describing the “volumes” of higher-dimensional parallelepipeds . You can also have wedge products between “one-forms”, such as $dx$ and $dy$, which send vectors in a vector space to a scalar (which is really in a field). $\endgroup$ Nov 25, 2019 at 23:25

6 Answers

The answers using the Jacobian are (of course) correct. You can get some intuition from this picture:

$\Delta A$ is (approximately) a rectangle with sides $r \Delta \theta$ and $\Delta r$. Its area is proportional to $r$ since it scales as $r$ increases.

Picture from

  • $\begingroup$ So what you’re saying is that the infintesmal area dxdy when converted to polar coordinates ends up being a rectangle with the sides $dr$ and $rsin(dtheta) \approx rd\theta $ with the small angle approximation. That explains it geometrically, but it doesn’t really explain what happens when you want to do a non-geometric substitution (i.e. not a change of coordinates so much as you are just trying to replace either dx or dy or both with something just so you can integrate it). I guess that’s where the jacobian comes in? It’s generalized form of what’s going on? $\endgroup$ Nov 26, 2019 at 17:47
  • $\begingroup$ “Just so you can integrate it” implies figuring out the area element in the new coordinate system, in terms of $dx$ and $dy$. It doesn’t make sense to think of changing those infinitesimals separately. The Jacobian tells you the whole story. (I don’t think you need the $\sin$ and the small angle approximation. The arclength is $rd\theta $.) $\endgroup$ Nov 26, 2019 at 17:57
  • $\begingroup$ Yes, that’s what you are technically doing, but it doesn’t seem like you can rely on geometric intuition if you are going through a table of integrals trying to substitute expressions in to get something that you can work with. $\endgroup$ Nov 26, 2019 at 17:58
  • $\begingroup$ The single valued integrals you look up in tables are often worked out with substitutions. Those are just one dimensional changes of coordinates 0 the Jacobian is just the derivative that the chain rule provides. I don’t know of any tables of multivariable integrals. $\endgroup$ Nov 26, 2019 at 18:03
  • $\begingroup$ Oh yeah. I guess you don’t need the small angle approximation since the arc length is just $$\frac{\theta}{2\pi} \times 2\pi r$$ $\endgroup$ Nov 26, 2019 at 18:40

We are using polar coordinates that is

  • $x=r \cos \theta$
  • $y=r \sin \theta$

and by the Jacobian determinant we have that

$$dxdy=\begin{vmatrix}\cos\theta&-r\sin\theta\\\sin \theta&r\cos\theta\end{vmatrix}drd\theta=rdrd\theta$$

Refer also to the related

  • $\begingroup$ I mean…I see how the determinant is the case, but how is the matrix suddenly just being pulled into this integral in the first place? $\endgroup$ Nov 25, 2019 at 22:35
  • $\begingroup$ @DKNguyen It represens the determinant for the Jacobian that is $$r\cos^2 \theta+r\sin^2 \theta= r(cos^2 \theta+rsin^2 \theta)=r$$ $\endgroup$– userNov 25, 2019 at 22:37
  • 1$\begingroup$ @ DKNguyen It is a standard theorem from vector calculus that when changing coordinates, say from $(x,y)$ to $(u,v)$, one must tack on the determinant of the Jacobian $\textrm{det}(\textrm{D}T)$ in front of the differentials $dudv$, where T is the map that takes $(u,v)$ to $(x,y)$. Of course, this result holds beyond the $\mathbb{R^2}$ case. $\endgroup$ Nov 25, 2019 at 22:42
  • $\begingroup$ @DKNguyen Are you asking for the geometric reason why the change of variables formula works? $\endgroup$ Nov 25, 2019 at 22:50
  • 1$\begingroup$ For the single variable case, (like u-substitution), you are still taking the determinant of Jacobian matrices. However, the Jacobian, in this case, is a 1×1 matrix whose entry is $\frac{dx}{du}$ so the determinant is just $\frac{dx}{du}$. $\endgroup$ Nov 25, 2019 at 22:57

Perhaps seeing the two compared will help.

$\underline{\bf\text{1D Case:}}$

$$ \int_{a}^{b}f(x)dx=\int_{\alpha}^{\beta}f(x(u))\frac{dx}{du}du $$

$\underline{\bf\text{2D Case:}}$

$$ \iint_{D}f(x,y)dxdy=\iint_{D’}f(x(u,v),y(u,v))\Bigl\vert\frac{\partial{(x,y)}}{\partial{(u,v)}}\Bigr\vert{}dudv $$

As for where the actual matrix comes from:

When we transformed our coordinates from Cartesian to polar, we also transformed the differential elements $dx$ and $dy$. By doing this, we are no longer guaranteed that they are perpendicular; therefore, we are no longer guaranteed that $da=dxdy$. To find our true $da$, let’s consider the following equations:

$$ \begin{align} dx&=\frac{\partial{x}}{\partial{r}}dr+\frac{\partial{x}}{\partial{\theta}}d\theta\\ dy&=\frac{\partial{y}}{\partial{r}}dr+\frac{\partial{y}}{\partial{\theta}}d\theta \end{align} $$


$$ \begin{pmatrix} dx\\ dy \end{pmatrix} = \bbox[yellow,5px] { \begin{pmatrix} \partial{x}/\partial{r} & \partial{x}/\partial\theta\\ \partial{y}/\partial{r} & \partial{y}/\partial\theta \end{pmatrix} } \begin{pmatrix} dr\\ d\theta \end{pmatrix}\\ \text{The highlighted matrix ends up being the Jacobian Matrix.} $$

This shows us the form that the basis vectors take under the transformation. Previously, we had $\overrightarrow{dx’}=dx\hat{e}_x= \begin{pmatrix} 1\\ 0 \end{pmatrix} dx\, $ and $\,\overrightarrow{dy’}=dy\hat{e}_y= \begin{pmatrix} 0\\ 1 \end{pmatrix} dy $ , with $da=\Vert{\overrightarrow{dx’}\times\overrightarrow{dy’}}\Vert=dxdy$ . However, now we have $\overrightarrow{dx}= \begin{pmatrix} \partial{x}/\partial{r}\\ \partial{y}/\partial{r} \end{pmatrix} dr $ and $\overrightarrow{dy}= \begin{pmatrix} \partial{x}/\partial\theta\\ \partial{y}/\partial\theta \end{pmatrix} d\theta $ , with $$da=\Vert{\overrightarrow{dx}\times\overrightarrow{dy}}\Vert=\bbox[yellow,5px]{\Bigl(\frac{\partial{x}}{\partial{r}}\frac{\partial{y}}{\partial{\theta}}-\frac{\partial{x}}{\partial{\theta}}\frac{\partial{y}}{\partial{r}}\Bigr)}drd\theta\text{.}\\ \frac{\partial{x}}{\partial{r}}=\cos\theta\,\text{,}\,\frac{\partial{y}}{\partial{\theta}}=r\cos\theta\,\text{,}\,\frac{\partial{x}}{\partial{\theta}}=-r\sin\theta\,\text{,}\,\frac{\partial{y}}{\partial{r}}=\sin\theta\,\text{, so the highlighted portion equals }r\text{.}\\ \therefore{} da=\Vert{\overrightarrow{dx}\times\overrightarrow{dy}}\Vert=rdrd\theta $$ Also notice that this highlighted quantity is equal to the determinant of the previously highlighted matrix. Again, it turns out that this is what we call the Jacobian Matrix, and calculating the Jacobian Determinant uncovers how the coordinate transformation affected the differential area element.

$$ \frac{\partial{(x,y)}}{\partial{(u,v)}}= \begin{vmatrix} \partial{x}/\partial{u} & \partial{x}/\partial{v}\\ \partial{y}/\partial{u} & \partial{y}/\partial{v} \end{vmatrix} =\frac{\partial{x}}{\partial{u}}\frac{\partial{y}}{\partial{v}}-\frac{\partial{x}}{\partial{v}}\frac{\partial{y}}{\partial{u}} $$

(I don’t have enough rep to leave a comment, sorry)

To my knowledge, the Jacobian matrix represents the best approximation (differential) of our Cartesian points represented as polar coordinates. When you find its determinant, you are finding how the Jacobian scales area/volume.

Taylor series mandate $dx^2=0$, from which you can show $dxdy=-dydx$ etc. (Look up the wedge product on differential fotms.) Thus$$\begin {align}dxdy&=(\cos\theta dr-r\sin\theta d\theta)(\sin\theta dr+r\cos\theta d\theta)\\&=r(\cos^2\theta+\sin^2\theta)drd\theta.\end{align}$$In particular, the $drd\theta$ coefficient is a determinant called the Jacobian.

Note, however, the calculation $ds^2=dx^2+dy^2=dr^2+r^2d\theta^2$ takes infinitesimals as commuting, because this time there’s no nilpotency axiom, so we’re not working with the wedge product on differential forms.

  • $\begingroup$ Exterior algebra makes things so much cleaner! $\endgroup$ Nov 25, 2019 at 22:48

Differential element has area

$dA= dx \cdot dy$ in Cartesian system for a rectangle and

$dA= r d\theta \cdot dr$ in Polar coordinates for the trapezium.

Evaluation of Jacobian matrix $((u_x,u_y), (v_x,v_y))$ leaves a multiplier of just $r,$ between $[(u,v)-(x,y)]$ systems.

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