# Infinitesimal area element in polar coordinate

• I
• arpon
In summary, the infinitesimal area element is different in Cartesian and Polar coordinate systems. The mistake in the attempted calculation was in assuming that the infinitesimals can be multiplied together, when in fact they need to be related through the Jacobian determinant or an anti-symmetric tensor product.

#### arpon

We know, that the infinitesimal area element in Cartesian coordinate system is ##dy~dx## and in Polar coordinate system, it is ##r~dr~d\theta##. This inifinitesimal area element is calculated by measuring the area of the region bounded by the lines ##x,~x+dx, ~y,~y+dy## (for polar coordinate ##r, r+dr, \theta, \theta + d\theta##).

Now, I tried to calculate it in a different way. We know,
## x = r \cos {\theta} ; y= r\sin{\theta}; ##
Hence,
## dx = - r \sin{\theta}~ d\theta + \cos{\theta}~ dr##
& ## dy = r \cos{\theta}~ d\theta + \sin{\theta} ~dr##

Now, we have, infinitesimal area element,
## dA = dx ~dy = - r^2 \cos {\theta} \sin {\theta} ~d\theta ^2 + (\cos^2 {\theta}-\sin^2{\theta} ) r ~dr~ d\theta + \sin{\theta}\cos{\theta}~ dr^2##
Where did I make the mistake?

arpon said:
Now, we have, infinitesimal area element,
## dA = dx ~dy = - r^2 \cos {\theta} \sin {\theta} ~d\theta ^2 + (\cos^2 {\theta}-\sin^2{\theta} ) r ~dr~ d\theta + \sin{\theta}\cos{\theta}~ dr^2##
Where did I make the mistake?

This is not how coordinate transformations of area elements work. You cannot just take the infinitesimals and multiply them together. You need to compute the area element spanned by two infinitesimal displacements.

Orodruin said:
This is not how coordinate transformations of area elements work. You cannot just take the infinitesimals and multiply them together. You need to compute the area element spanned by two infinitesimal displacements.
It is not clear to me why it is incorrect to substitute for ##dx## and ##dy##. Could you please explain elaborately?

arpon said:
It is not clear to me why it is incorrect to substitute for ##dx## and ##dy##. Could you please explain elaborately?

This is really about a change of variables, rather than comparing a specific small area in two coordinate systems.

As you have seen, if you have a small rectangle defined by ##dx, dy## and the area of this rectangle is ##dA## then ##dA \ne rdrd\theta##. In fact, it's not even clear what ##dr## and ##d\theta## would be in this case. To see this, draw the rectangle and try to figure out what ##dr, d\theta## would be.

Instead, if you imagine integrating in polar coordinates, you will take the limit of small areas, defined by ##dr, d\theta##. These are not rectangles (although they are close). Each has an area of approximately ##dA_{p} = rd\theta dr ##.

##\int dxdy = \int rdrd\theta##

But, there is not a direct relationship between the area of a specific small rectangular area in Cartesian coordinates and a different specific small (almost) rectangular area in polar coordinates.

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arpon
arpon said:
It is not clear to me why it is incorrect to substitute for ##dx## and ##dy##. Could you please explain elaborately?
Because this is not what an area element means. You need to relate the displacement vectors corresponding to a change in each coordinate to the spanned area. This is done through the Jacobian determinant, not by multiplying infinitesimals together.

PeroK said:
But, there is not a direct relationship between the area of a specific small rectangular area in Cartesian coordinates and a different specific small (almost) rectangular area in polar coordinates.

You can multiply the infinitesimals if you use an anti-symmetric tensor product which defines an area form. For example:
\begin{align}
dx \wedge dy &= [\cos(\theta) dr - r \sin(\theta) d\theta]\wedge [\sin(\theta) dr + r \cos(\theta) d\theta] \nonumber \\
&= \cos(\theta)\sin(\theta) (dr \wedge dr - r^2 d\theta \wedge d\theta) - r \sin^2(\theta) d\theta \wedge dr + r \cos^2(\theta) dr \wedge d\theta \nonumber \\
&= r\, dr \wedge d\theta. \nonumber
\end{align}

Edit: Updated LaTeX display for the anti-symmetric tensor product.

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Samy_A, Buzz Bloom, jim mcnamara and 1 other person

## What is an infinitesimal area element in polar coordinates?

An infinitesimal area element in polar coordinates is a small piece of a two-dimensional surface that is defined using polar coordinates. It is used in mathematics and physics to calculate the area of a surface in a specific location.

## How is an infinitesimal area element calculated in polar coordinates?

In polar coordinates, an infinitesimal area element is calculated by multiplying the radius and the angle of the polar coordinate system. The radius is represented by the letter "r" and the angle is represented by the Greek letter "θ" (theta).

## What is the significance of using polar coordinates in calculating infinitesimal area elements?

Polar coordinates are particularly useful in calculating infinitesimal area elements because they allow us to easily express the position and orientation of a point on a two-dimensional surface. This makes it easier to visualize and manipulate the area element in calculations.

## How is an infinitesimal area element represented graphically in polar coordinates?

In polar coordinates, an infinitesimal area element is represented by a small sector of a circle. The radius of the sector corresponds to the "r" value, and the angle of the sector corresponds to the "θ" value. The area of the sector can then be calculated using the formula A = ½r²θ.

## What are some applications of using infinitesimal area elements in polar coordinates?

Infinitesimal area elements in polar coordinates have various applications in mathematics and physics, such as calculating the area under a curve, finding the volume of a solid of revolution, and determining the force acting on a particle with a given charge in an electric or magnetic field.