# Explain $\iint \mathrm dx\,\mathrm dy = \iint r \,\mathrm \,d\alpha\,\mathrm dr$

It is changing the coordinate from one coordinate to another. There is an angle and radius on the right side. What is it? And why?

I got:

$2\,\mathrm dy\,\mathrm dx = r(\cos^2\alpha-\sin^2\alpha)\,\mathrm d\alpha \,\mathrm dr$,

where $x = r \cos(\alpha)$ and $y = r \sin(\alpha)$.

but cannot understand and get the right side. The problem emerged when trying to integrate $\displaystyle \int_0^\infty e^{\frac{-x^2}{n^2}}\,\mathrm dx$ where I tried to change the problem knowing $r^2=x^2+y^2$ but stuck to this part. What is the change in the title called and why is it so?

• See here, Example 3, and also here.
– t.b.
May 4, 2011 at 21:54
• If you think of $x$ and $y$ being cartesian coordinates for the plane, then what this does is a change to polar coordinates. May 4, 2011 at 22:12
• A physicist would note that this is dimensionally correct. Typically $dx, dy,r$ and $dr$ have dimensions of lenght, while $d \alpha$ is dimensionless (it is an angle). So $dxdy=r drd\alpha$ is dimensionally correct: both sides of the equation are areas. On the contrary $dx dy=drd\alpha$ is not dimensionally correct: you have an area equal a lenght. This is useful to quickly spot computational errors. May 4, 2011 at 23:07
• You don't have to be a physicist to talk about dimensional analysis. $\mathbb{R}^2$ is acted on by scaling and this extends to an action on differential forms, etc. And of course if two things are equal then the corresponding scaling action needs to agree. May 5, 2011 at 0:12
• $r$ is the "Jacobian" ... when you learn about multi-dimensinoal integration, you should learn how to change variables in that context. Sep 11, 2011 at 13:00

Generally speaking, a double integral always has an area differential, so that you're integrating $$\iint dA$$. Another way to view the question, then, is why $$dA = dx dy$$ in Cartesian coordinates and $$dA = r dr d\theta$$ in polar coordinates.

An area element in Cartesian is a rectangle, as Qiaochu describes in his answer. The area of the rectangle is the small change in $$x$$ times the small change in $$y$$, or $$\Delta x \Delta y$$.

An area element in polar, however, is a piece of a circle sector. There's a nice picture below taken from here. (The area element is the shaded part.) If the angle is measured in radians, we know that the area of a sector of angle $$\theta$$ of a circle of radius $$r$$ is $$\frac{1}{2}r^2 \theta$$. So the area of the shaded piece in the picture is $$\Delta A = \frac{1}{2}(r + \Delta r)^2 \Delta \theta - \frac{1}{2}r^2 \Delta \theta = r \, \Delta r \, \Delta \theta + \frac{1}{2}(\Delta r)^2 \, \Delta \theta.$$ The quadratic factor of $$\Delta r$$ makes the second term negligible compared to the first term for small enough $$\Delta r$$ and $$\Delta \theta$$. Thus in the limit we get $$dA = r \, dr \, d\theta$$.

As others have said, you can also use the multivariate change of variables formula involving the Jacobian of the transformation directly. I like the geometric argument when first introducing the polar element in a calculus course, though.

• @Jonas: Thanks for editing the picture into my answer. May 5, 2011 at 2:44
• Is that formula just for aproximation? Dec 9, 2011 at 0:55
• @Victor: I'm not sure I understand your question. The expression for $\Delta A$ becomes $dA = r dr d\theta$ in the limit, as explained in the sentence following the formula. Dec 9, 2011 at 1:00
• @MikeSpivey Great answer! It really made me get the idea though I'm not familiar with double integration!
– Pedro
Feb 20, 2012 at 0:41
• Hm i guess to ease the computation as in the limit it will not matter whether those lines are straight or arc-like
– Naz
May 17, 2016 at 19:40

It's a special case of the multivariate change of variables formula. Intuitively you can think of it as follows: starting from a point $(x, y)$, you make an infinitesimal change in $x$ and then an infinitesimal change in $y$ to get to $(x + \delta x, y + \delta y)$. Those changes trace out a little square whose area is $\delta x \delta y$, and you're summing over a bunch of these little squares.

So what happens when you do the same thing in polar coordinates? Starting from $(r, \theta)$ you move to $(r + \delta r, \theta + \delta \theta)$. Now moving $\delta r$ is just like moving $\delta x$ in an appropriately rotated set of axes. But when you move $\delta \theta$, the actual distance you move by is multiplied by $r$ (draw a diagram to see this), and it's in a direction orthogonal to the direction you move when changing $r$. You're actually moving by $r \, \delta \theta$. The result is not quite a square, but for small enough values of $\delta r, \delta \theta$ it approaches a square with side lengths $\delta r$ and $r \, \delta \theta$.

• you don't need square, you only need rectangle..
– Aang
Jul 10, 2012 at 19:18

Let us approach this problem like some physicists would, that is, let us consider the differential $\mathrm{d}$ as an operator which obeys the rules of derivation, only with a sign.

In the present case, $x=r\cos\alpha$ and $y=r\sin\alpha$ hence $$\mathrm{d}x=\cos\alpha\mathrm{d}r-r\sin\alpha\mathrm{d}\alpha, \quad \mathrm{d}y=\sin\alpha\mathrm{d}r+r\cos\alpha\mathrm{d}\alpha.$$ Now, there are some magic rules which allow to multiply $\mathrm{d}r$ and $\mathrm{d}\alpha$ elements. These are $$\mathrm{d}r\mathrm{d}r=0,\quad \mathrm{d}\alpha\mathrm{d}r=-\mathrm{d}r\mathrm{d}\alpha,\quad \mathrm{d}\alpha\mathrm{d}\alpha=0.$$ Hence, $$\mathrm{d}x\mathrm{d}y=(\cos\alpha\mathrm{d}r-r\sin\alpha\mathrm{d}\alpha)(\sin\alpha\mathrm{d}r+r\cos\alpha\mathrm{d}\alpha).$$ The $\mathrm{d}r\mathrm{d}r$ and $\mathrm{d}\alpha\mathrm{d}\alpha$ terms disappear and the terms which interest us are the $\mathrm{d}r\mathrm{d}\alpha$ and $\mathrm{d}\alpha\mathrm{d}r$ ones. One gets $$\mathrm{d}x\mathrm{d}y=(\cos\alpha\cdot r\cos\alpha-r\sin\alpha\cdot (-1)\sin\alpha)\mathrm{d}r\mathrm{d}\alpha,$$ that is, $$\color{green}{\mathrm{d}x\mathrm{d}y=r\mathrm{d}r\mathrm{d}\alpha},$$ a formula which yields $$\color{red}{\iint f(x,y)\mathrm{d}x\mathrm{d}y=\iint f(r\cos\alpha,r\sin\alpha)r\mathrm{d}r\mathrm{d}\alpha}.$$ One can also transform integrals the other way round, all there is to do is to compute a formula for $\mathrm{d}r\mathrm{d}\alpha$ as a multiple of $\mathrm{d}x\mathrm{d}y$. Our formula for $\mathrm{d}x\mathrm{d}y$ in terms of $\mathrm{d}r\mathrm{d}\alpha$ yields $$\mathrm{d}r\mathrm{d}\alpha=\frac1r\mathrm{d}x\mathrm{d}y=\frac1{\sqrt{x^2+y^2}}\mathrm{d}x\mathrm{d}y,$$ hence $$\color{blue}{\iint f(r,\alpha)\mathrm{d}r\mathrm{d}\alpha=\iint f(x,y)\frac1{\sqrt{x^2+y^2}}\mathrm{d}x\mathrm{d}y}.$$ Once again, this only describes the computational side of the story, but this recipe is supported by a well established theory of differential forms which we omitted.

• it would be maybe helpful to use the symbol $\wedge$ (or $\cdot$, or $\otimes$, or ...) for the multiplication of two differentials just to make clear that the "usual" rules for product do not apply. May 5, 2011 at 9:47
• @Fabian: I deliberately omitted it. Both choices (with and without $\land$) have their advantages. Thanks for your reaction.
– Did
May 5, 2011 at 9:53
• Can the "magic rules" be viewed as the cross product of "vectors" $\mathrm{d}r,\mathrm{d}\alpha$ ? May 5, 2011 at 17:41
• @Américo Rather as a differential form (a 2-form, in this case). This is explained rigorously in many lecture notes, a congenial one might be math.berkeley.edu/~wodzicki/H185.S11/podrecznik/2forms.pdf
– Did
May 5, 2011 at 18:09
• Thanks! +1 for your approach. May 5, 2011 at 18:33

UPDATE: The series have been replaced by limits of sums.

Geometric interpretation. By definition of a double integral of a continuous function over a bounded closed region $R$ of the $xy$-plane, we have

$$\iint_R f(x,y)\;\mathrm{d}x\;\mathrm{d}y=\lim_{n\to\infty}\; \sum_{i=1}^{n }f(x_{i},y_{i})\Delta A_{i},$$

where $\Delta A_{i}$ is the area of a generic rectangular cell and $n$ the number of cells.

If $f(x,y)=1$, we get the area of $R$

$$\iint_R \mathrm{d}x\;\mathrm{d}y=\lim_{n\to\infty}\; \sum_{i=1}^{n }\Delta A_{i}.$$

If we decompose $R$ into cells with a shape of sectors of a circle defined by two radii whose difference is $\Delta r_{i}$ for the generic $i^{th}$ cell and two rays making an angle $\Delta \theta _{i}$ with each other, the area of the cell, using the formula of a circle sector, is

$$\frac{1}{2}\left[ \left( r_{i}+\frac{1}{2}\Delta r_{i}\right) ^{2}-\left( r_{i}-\frac{1}{2}\Delta r_{i}\right) ^{2}\right] \Delta \theta _{i}=r_{i}\Delta r_{i}\Delta \theta _{i}\text{,}$$

where $r_{i}$ is the radius of the middle point of the cell. The same area $R$ can be expressed as the limit $\lim_{n\to\infty}\;\sum_{i=1}^{n }r_{i}\Delta r_{i}\Delta \theta _{i}$, which by definition of a double integral is equal to $$\iint_R r\;\mathrm{d}r\;\mathrm{d}\theta.$$ Figure: Generic $i^{th}$-cell in polar co-ordinates with the shape of a circle sector

This transformation is defined rigorously by the absolute value of the Jacobian of the transformation $\left\vert \frac{\partial (x,y)}{\partial (r,\theta )}\right\vert =r$ from the Cartesian to the polar system of co-ordinates ($x=r\cos \theta ,y=r\sin \theta$):

$$\iint_R \mathrm{d}x\;\mathrm{d}y=\iint_R \left\vert \frac{\partial (x,y)}{\partial (r,\theta )}\right\vert \;\mathrm{d}r\;\mathrm{d}\theta = \iint_R r\;\mathrm{d}r\;\mathrm{d}\theta.$$

Added: Evaluation of the Jacobian determinant: $$\begin{eqnarray*} \frac{\partial \left( x,y\right) }{\partial \left( r,\theta \right) } &=&\det \begin{pmatrix} \partial x/\partial r & \partial x/\partial \theta \\ \partial y/\partial r & \partial y/\partial \theta \end{pmatrix} \\ &=&\det \begin{pmatrix} \cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{pmatrix} \\ &=&r\cos ^{2}\theta +r\sin ^{2}\theta \\ &=&r. \end{eqnarray*}$$

• @Mike Spivey: Thanks! (corrected). May 5, 2011 at 8:33