# 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?

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See here, Example 3, and also here. –  t.b. May 4 '11 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. –  Lagerbaer May 4 '11 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. –  Giuseppe Negro May 4 '11 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. –  Qiaochu Yuan May 5 '11 at 0:12
$r$ is the "Jacobian" ... when you learn about multi-dimensinoal integration, you should learn how to change variables in that context. –  GEdgar Sep 11 '11 at 13:00

Generally speaking, a double integral always has an area differential, so that you're integrating $\int \int dA$. 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.

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@Jonas: Thanks for editing the picture into my answer. –  Mike Spivey May 5 '11 at 2:44
Is that formula just for aproximation? –  Victor Dec 9 '11 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. –  Mike Spivey Dec 9 '11 at 1:00
i think you ignore the 1/2 (tri(r))^2 (tri(degree)) because it is too small, isn't that call approximate? –  Victor Dec 9 '11 at 1:03
@MikeSpivey Great answer! It really made me get the idea though I'm not familiar with double integration! –  Pedro Tamaroff Feb 20 '12 at 0:41

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 $r \, \delta \theta$.

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you don't need square, you only need rectangle.. –  Aang Jul 10 '12 at 19:18

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

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@Mike Spivey: Thanks! (corrected). –  Américo Tavares May 5 '11 at 8:33
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. –  Fabian May 5 '11 at 9:47
@Fabian: I deliberately omitted it. Both choices (with and without $\land$) have their advantages. Thanks for your reaction. –  Did May 5 '11 at 9:53
Can the "magic rules" be viewed as the cross product of "vectors" $\mathrm{d}r,\mathrm{d}\alpha$ ? –  Américo Tavares May 5 '11 at 17:41