Findind Area element in Cylindrical Coordinate System

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SUMMARY

The discussion focuses on calculating area elements in cylindrical and spherical coordinate systems without relying on graphical methods. It establishes that area elements in Cartesian coordinates can be expressed as Adxdy + Bdxdz + Cdydz, where A, B, and C are functions of x, y, and z. The transformation to cylindrical coordinates is detailed, with specific equations for x, y, and z, and the corresponding differential changes. The conversation highlights the need for clarity on the two-dimensional object in question to derive the appropriate area element.

PREREQUISITES
  • Cylindrical coordinate system fundamentals
  • Understanding of differential geometry
  • Knowledge of Cartesian to cylindrical coordinate transformations
  • Familiarity with the wedge product of differentials
NEXT STEPS
  • Study the derivation of area elements in spherical coordinates
  • Learn about the applications of the wedge product in differential forms
  • Explore the implications of changing variables in multivariable calculus
  • Investigate the geometric interpretations of area elements in various coordinate systems
USEFUL FOR

Mathematicians, physicists, and engineering students who are working with multivariable calculus and need to understand area elements in cylindrical and spherical coordinate systems.

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Hi
I would like to know is there any way except using graph to find area element in cylindrical ( or Spherical) coordinate system?
Thanks.
 
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Cylindrical and spherical coordinate systems are three dimensional so you would have to say what two dimensional object you want to find the area of before an area element can be given. However, we can say this- any area element is Cartesian coordinates can be written as Adxdy+ Bdxdz+ Cdydz for some A, B, C, which may be functions of x, y, and z, depending on the surface.

In cylindrical coordinates, we have x= r cos(\theta), y= r sin(\theta), and z= z so that dx= cos(\theta)dr- r sin(\theta)d\theta, dy= sin(\theta)dr+ rcos(\theta)d\theta, dz= dz. From that, we can compute, remembering that the "wedge product" of differentials is skew-commutative,
dxdy= r cos^2(\theta)drd\theta- r sin^2(\theta)d\theta dr= r cos^2(\theta)drd\theta+ r sin^2(\theta)drd\theta= r dr d\theta
dxdz= cos(\theta)drdz- r sin(\theta)d\theta dz
dydz= sin(\theta)drdz+ rcos(\theta)d\theta dz
and, of course, changing the variables in A(x,y,z), B(x,y,z), C(x,y,z) to r, \theta, and z.
 
Thanks I know rdrdθ is a valid statement but what about other area element, you got something that both depends on cosθ and sin but we know area element in the θ direction is just drdz. how can I calculate those?
Thank you.
 

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