Proof of Area Invariance of Closed Curve

  • Context: Graduate 
  • Thread starter Thread starter anantchowdhary
  • Start date Start date
  • Tags Tags
    Area Invariance
Click For Summary
SUMMARY

The discussion centers on the proof of area invariance for closed curves, specifically addressing the independence of area calculations based on the orientation of spliced rectangles. The participants clarify that the area of a closed curve can be defined as the limit of the sum of areas of non-overlapping rectangles as their number approaches infinity. However, the proof presented is critiqued for lacking rigor, as it does not account for the necessity of integration and proper definitions in measure theory. The consensus emphasizes the importance of precise definitions and the role of limits in establishing area invariance.

PREREQUISITES
  • Understanding of integral calculus and limits
  • Familiarity with the concept of non-overlapping regions
  • Knowledge of measure theory and its definitions
  • Basic principles of geometry, specifically area calculation
NEXT STEPS
  • Study the fundamentals of measure theory to understand area definitions
  • Learn about the properties of integration and its application in area calculations
  • Explore the concept of limits in calculus, particularly in relation to infinite sums
  • Investigate the relationship between geometry and calculus in defining areas of complex shapes
USEFUL FOR

Mathematicians, students of calculus, and anyone interested in the rigorous foundations of geometry and area calculations.

anantchowdhary
Messages
372
Reaction score
0
Hello!

Quite some time ago I'd asked for help with a proof that proves that area of a closed curve is invariant i.e : its independent of the way it is spliced into.

Say we splice a closed curve into one set of rectangles with parallel sides and we then splice an identical curve with rectangles with some different orientation, I basically sought to prove that area calculated by summing up areas in both cases would be equal.


Here i present a proof

I'd be grateful to members who could comment on the proof and check it for validity.
Thanks :D
 

Attachments

Mathematics news on Phys.org
Please Help! 127 Views and not a single reply! :S

Kindly Help!
 
I have no idea what you mean by "spliced into". "Sliced into"? Do you mean "divided into non-overlapping subregions"? Also, how are you defining the "area" of a plane region?
 
Thanks a Lot for the reply! by splice i mean ''slice'' into non overlapping regions.

Firstly defining the area of a rectangle as its length*breadth, and then for any general closed curve defining its area as the sum of areas of 'n',non overlapping rectangles that it can be divided into.Where n--> infinity
 
Your proof fails because you can never get all the rectangles to be true rectangles so the formula lnwn + ln+1wn+1 + ... = A is not true. This is why you need a limit as w -> 0, and then you simply have integration which needs no proof.
 
@ mu naught : yes that is why i said 'n' rectangles where n-->infinity.In integral calculus nowhere do we prove that the area is invariant (i mean irrespective of orientation of coordinate axes in the case of integration)
 
It's tricky, but you need to be more precise than saying your definition of the area of a region A is the sum of the areas of n non-overlapping rectangles in A as n tends to infinity. What if I give you 10 rectangles that roughly provide the shape of the outer boundary of A and and then keep dividing the inner rectangles? The number of rectangles approaches infinity, but the area never gets closer to the intuitive area.
 
Even if you do get this to work, all you will have proven is that the sum of the areas of some rectangles tends to the same limit as that of some differently oriented rectangles, as their number goes to infinity. If you're interested in defining area correctly and proving its properties, you need to learn about measure.
 
@werg22 : Yes! This is exactly what i intended to prove,for which i have never seen a proof.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 33 ·
2
Replies
33
Views
8K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K