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We all know that the area element $dA$ in Cartesian coordinates is given by $$dA = dx\ dy.$$ With a bit of geometry, we can also see that the area element $dA$ in polar coordinates is given by $$dA = r\ dr\ d\theta.$$ If this is the case, when why doesn't $dx\ dy = r\ dr\ d\theta$? By implicitly differentiating the equations $x = r \cos \theta$ and $y = r \sin \theta$, we have that $$dx = dr \cos \theta - r \sin \theta\ d\theta$$ and $$dy = dr \sin \theta + r \cos \theta\ d\theta.$$ Thus, \begin{align*} dx\ dy &= (dr \cos \theta - r \sin \theta\ d\theta)(dr \sin \theta + r \cos \theta\ d\theta)\\ &= dr^2 \cos\theta \sin\theta+r\ dr \cos^2 \theta\ d\theta-r\ dr \sin^2 \theta\ d\theta-r^2\sin\theta\cos\theta\ d\theta^2. \end{align*} It is not evident to me that this expression equals $r\ dr\ d\theta$, if it does at all. Why does this discrepancy exist? Is it the case that $dA$ in Cartesian coordinates is simply not the same thing as $dA$ in polar coordinates? If so, why do both correspond to area elements in the plane?

David Zhang
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2 Answers2

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In general, because we want to integrate with sign (for example, we want

$$\int_a^b f(x)dx = - \int_b^a f(x) dx, )$$

so we require that in any coordinate system,

$$dx dy = - dy dx$$

So

$$\begin{align*} dx\ dy &= (dr \cos \theta - r \sin \theta\ d\theta)(dr \sin \theta + r \cos \theta\ d\theta)\\ &= \cos\theta \sin\theta dr\ dr +r\cos^2 \theta dr \ d\theta -r \sin^2 \theta d\theta \ dr -r^2\sin\theta\cos\theta\ d\theta\ d\theta \\ &= rcos^2 \theta dr \ d\theta - r\sin^2 \theta d\theta \ dr \\ &= r\ dr \ d\theta\ . \end{align*}$$

  • I'm not sure I understand your claim that $dx\ dy = -dy\ dx$. Let $C$ be the unit square. Would this imply that $\iint_C dx\ dy = 1$ but $\iint_C dy\ dx = -1$? I have always been told that when possible, the order of integration can be changed without affecting the result. – David Zhang Apr 09 '14 at 16:40
  • @DavidZhang Have you seen differential forms before? Take a look at this: http://en.wikipedia.org/wiki/Differential_form . The wedge product is indeed skew-symmetric so $dx \wedge dy = - dy \wedge dx$. This is why, for example the term $dr^2$ vanished, since one can show that $dr \wedge dr =0$.

    I agree that in your example the order shouldn't matter, so I'm curious to hear John's response!

     – Viktor Vaughn Apr 09 '14 at 18:03
  • @SpamIAm No, I have never seen these differential forms before--only the typical (rather hand-wavy) vector calculus. I have always been taught that $dx$ represents an infinitesimal change in $x$, $dy$ represents an infinitesimal change in $y$, and that $dA = dx\ dy$ is the area of the infinitesimal rectangle formed by a width $dx$ and a height $dy$. It is therefore very counter-intuitive to me that we should define this wedge product to be anticommutative, since the area interpretation is obviously commutative. – David Zhang Apr 10 '14 at 00:15
  • Well, if you like Spencer's approach below, the cross product is similarly skew-symmetric! I believe the negative is just intended to convey orientation. – Viktor Vaughn Apr 10 '14 at 00:46
  • @SpamIAm Unless I am misunderstanding Spencer's notation, I believe his approach uses the absolute value of the cross product, which is indeed commutative. – David Zhang Apr 10 '14 at 18:16
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A more elementary approach:

In Cartesian coordinates:

We have the position vector $\vec{r} = x \vec{e}_x + y \vec{e}_y $. A small area can be defined by the displacement vectors,

$$d\vec{x} = \frac{\partial \vec{r}}{\partial x} dx = dx \vec{e}_x \qquad d\vec{y} = \frac{\partial \vec{r}}{\partial y} dy = dy \vec{e}_y.$$

The area of the parallelogram swept out by these vectors is given by,

$$ dA = \vert d\vec{A} \vert = \vert d\vec{x} \times d\vec{y}\vert = dx dy $$

In polar Coordinates: We have the position vector $\vec{r} = \rho\cos(\theta) \vec{e}_x + \rho\sin(\theta) \vec{e}_y $. A small area can be defined by the displacement vectors,

$$d\vec{\rho} = \frac{\partial \vec{r}}{\partial \rho} d\rho = \left( \cos(\theta) \vec{e}_x + \sin(\theta) \vec{e}_y \right)d\rho, $$

$$ d\vec{\theta} = \frac{\partial \vec{r}}{\partial \theta} d\theta =\left(-\rho\sin(\theta) \vec{e}_x + \rho\cos(\theta) \vec{e}_y\right)d\theta.$$

The area of the parallelogram swept out by these vectors is given by,

$$ dA = \vert d\vec{A} \vert = \vert d\vec{\rho} \times d\vec{\theta}\vert = \rho d\rho d\theta $$

Edit: To answer the question about the general rule for constructing n-dimensional volume elements.

The important mathematical object here isn't cross product, but rather the determinant used to calculate it. Let our coordinates be denoted by $q_j$ where $j=1\dots n$. We will first examine what is going on in the two dimensional case in more detail and then generalize from there.

Our coordinate vector $\vec{r} = x(q_1,q_2) \vec{e}_x + y(q_1,q_2) \vec{e}_y$. The cross product of the two vectors $d\vec{q}_2$ and $d\vec{q}_2$ is given by,

$$d\vec{q}_1 \times d\vec{q}_2 = det \left(\begin{array} _ \vec{e}_x & \vec{e}_y & \vec{e}_z\\ \frac{\partial x}{\partial q_1} & \frac{\partial y}{\partial q_1} & 0 \\ \frac{\partial x}{\partial q_2} & \frac{\partial y}{\partial q_2} & 0 \end{array} \right) dq_1 dq_2 = det \left( \begin{array} \ \frac{\partial x}{\partial q_1} & \frac{\partial y}{\partial q_1} \\ \frac{\partial x}{\partial q_2} & \frac{\partial y}{\partial q_2} \end{array}\right) dq_1 dq_2 \vec{e}_z . $$

The resulting $2\times 2$ matrix at the end of the calculation is called the Jacobian of the transformation. This is usually written as,

$$ \frac{\partial(x,y)}{\partial(q_1,q_2)} \equiv \left( \begin{array} \ \frac{\partial x}{\partial q_1} & \frac{\partial y}{\partial q_1} \\ \frac{\partial x}{\partial q_2} & \frac{\partial y}{\partial q_2} \end{array}\right) . $$

The $n$-dimensional Jacobian is,

$$ \frac{\partial(x_1,x_2,\dots,x_n)}{\partial(q_1,q_2,\dots,q_n)} = \left( \begin{array} \ \frac{\partial x_1}{\partial q_1} & \cdots & \frac{\partial x_n}{\partial q_1} \\ \vdots & \ddots & \vdots \\ \frac{\partial x_1}{\partial q_n} & \cdots & \frac{\partial x_n}{\partial q_n} \end{array}\right) .$$

Notice that this is just made by making a matrix whose rows are $\frac{\partial \vec{r}}{\partial q_j}$.

The $n$-dimensonial volume element is then given by,

$$ dV_n = \left| det \frac{\partial(x_1,x_2,\dots,x_n)}{\partial(q_1,q_2,\dots,q_n)} \right| dq_1 dq_2 \cdots dq_n $$

This formalism using determinants is related to other formalisms you may run into using the levi-cevita tensor or wedge products since all three are closely related.

For more information you can look at,

Spencer
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  • Interesting. So when we say $dx\ dy$, we are implicitly taking the cross product of $d\vec{x}$ and $d\vec{y}$? – David Zhang Apr 09 '14 at 16:48
  • Yup. In ordinary Cartesian coordinates this isn't really necessary since the element of area is a rectangle. The cross product only becomes necessary when dealing with more general curvilinear coordinates since the elements of area become parallelograms (on a really small scale). Volume elements end up becoming the triple scalar product which you may recall gives the volume of a parallelpiped. – Spencer Apr 09 '14 at 16:53
  • Then what is the corresponding operation for general $(n > 3)$-dimensional curvilinear coordinates? I am not aware of an analogue of, say, the cross product or triple scalar product for four- or five-dimensional spaces. – David Zhang Apr 10 '14 at 00:22
  • The wedge product is the answer. Instead of treating, say, a surface by talking about its surface normal vector, the wedge product allows you to directly describe the local tangent plane's orientation with a bivector, which in 4d obviously can't be identified with having any unique, corresponding normal vector. – Muphrid Apr 10 '14 at 05:22