Flux Integral: How to find ds for line integrals in general

AI Thread Summary
To find ds for line integrals, it is essential to understand that ds represents an infinitesimal area element in the chosen coordinate system. In Cartesian coordinates, ds is derived from the area of a rectangular element, expressed as dx by dz. For polar coordinates, the area element is represented as rdrdθ, accounting for the angular relationship between the coordinates. The general procedure for determining ds involves considering the product of differentials in the coordinate system being used, and applying a Jacobian if necessary. Understanding these principles is crucial for accurately calculating flux integrals in various coordinate systems.
Marcin H
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Homework Statement


Screen Shot 2017-10-15 at 1.49.35 PM.png


Homework Equations


flux = int(b (dot) ds)

The Attempt at a Solution


I just wanted clarification on finding ds. I understand why ds is in the positive yhat direction (just do rhr) but I don't understand where the dxdz come from. How do we find ds in general?
 
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Marcin H said:
but I don't understand where the dxdz come from.
It s just the area of a rectangular element, dx by dz, in the xz plane. In polar it would have been rdrdθ, etc.
 
haruspex said:
It s just the area of a rectangular element, dx by dz, in the xz plane. In polar it would have been rdrdθ, etc.
So what is the general procedure for finding ds? I feel like I struggle determining ds for problems.
 
Marcin H said:
So what is the general procedure for finding ds? I feel like I struggle determining ds for problems.
It depends on your coordinate system. With general coordinates, ξ, η say, you consider the product dξdη. In general, the area enclosed by the points (ξ,η), (ξ+dξ,η), (ξ+dξ,η+dη), (ξ,η+dη) might have area equal to dξdη. To make the right area you may need to multiply by a Jacobian.

In the case of Cartesian coordinates dxdy does have the right area.
For plane polar coordinates you need a factor r. This works because the increments dr and rdθ are at right angles, so form a near-enough rectangle.
For the surface of a sphere in spherical polar a different Jacobian is needed.
 
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