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I have a couple of questions about area elements and volume elements and why they are the form they are when we transform between different coordinate systems. Say we have some curve parametrized in terms of $ x$ and $ y$. So the corresponding area element if I were to evaluate a surface integral, say would be $ dA = dx\,dy$. Now transform the coordinates to polars, the corresponding dA would be $ r\,dr\,d\theta$.

But in cylindrical coordinates, the corresponding volume element is $ r \,dr\,d\theta$ .How can $ r\,dr\,d\theta $ be both an area element and a volume element?

Perhaps a more specific question, but say I was computing a surface integral in x and y. Then as above when I transform to polars, $ dA = r\,dr\,d\theta$. Now say I start the surface integral in polars. Then the area element would just be $ dr\,d\theta$. Why is this the case? (Also $dr\,d\theta$ does not have the right dimensions to be an area?)

As you can tell, I am rather confused at the moment. Any help is greatly appreciated.

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1 Answer 1

The volume element in cylindrical coordinates is $r \; dr \; d\theta \; dz$. That extra $dz$ is what's necessary to give the correct dimensions.

In polar coordinates, a surface element is always $r \; dr \; d\theta$. You cannot choose to just use $dr \; d\theta$. As you observed, this is not an area (in terms of correct dimensions), but one could also deduce the correct area element geometrically.

In short, one can set up integrals however one likes, but to be consistent with the geometric interpretation of the given coordinates, each coordinate system has only one area element. You do not have the freedom to choose it, only the freedom to choose your coordinates.

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Thanks. There have been times, however, when I have used just $dr\,d\theta$ for the area element and got the correct answer. E.g if I start the question off in polars and remain in polars throughout. Why would this be the case? –  CAF Nov 23 '12 at 7:40
    
Poorly designed problems where both integrals yield the same result. –  Muphrid Nov 24 '12 at 4:41
    
I'll check it the correct way. Why is it, also, that if we have a parametrization $r(z,\theta)$ we take the area element as $ dz\,d\theta$? –  CAF Nov 24 '12 at 9:36
    
I checked it the other way. In one typical Q, I had $\int_{0}^{2\pi} \int_{0}^{4} r\,dr\,d\theta.$, which when evaluated gives $16\pi$, the correct answer. Replacing $dr\,d\theta$ by $r\,dr\,d\theta$ gives $128\pi/3$, incorrect. What is going on? –  CAF Nov 24 '12 at 17:09
    
The former is an area integral; the latter is not. If it's already in the form of $r \, dr \, d\theta$, why would you think it needs to be changed? –  Muphrid Nov 24 '12 at 18:29

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