How Is the Surface Area of Spherical Cap Slices Calculated?

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Discussion Overview

The discussion revolves around the calculation of the surface area of spherical cap slices, particularly in the context of dividing a hemisphere into horizontal slices. Participants explore the implications of varying slice thickness and the use of calculus in determining surface areas.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant references a formula for the curved surface area of a spherical cap and questions whether all slices of a hemisphere cut into ten equal horizontal sections would have the same surface area.
  • Another participant suggests that it is possible to achieve equal surface areas by varying the spacing of the slices and recommends using calculus for precise calculations.
  • A participant expresses uncertainty about whether equal thickness slices would yield equal surface areas and requests help in framing the relevant equations.
  • Further contributions confirm that the areas of the slices are unlikely to be the same if the slices are of equal thickness, and they discuss the method of using thin disks to derive the surface area in terms of height and radius.
  • One participant reflects on their misunderstanding of the initial formula and attempts to reconcile it with their calculations, leading to confusion about where they went wrong.
  • Several participants express surprise at the results of their calculations and encourage the use of calculus to verify their findings.
  • Links to external resources are shared for further exploration of the topic.

Areas of Agreement / Disagreement

Participants generally agree that equal thickness slices do not yield equal surface areas, but there is no consensus on the specific calculations or methods to derive the surface area for each slice. The discussion remains unresolved regarding the correct approach to the problem.

Contextual Notes

Participants mention the need for calculus to accurately determine the surface area of the slices, indicating that the discussion may depend on mathematical assumptions and definitions that are not fully resolved.

bobie
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from wiki:
If the radius of the base of the cap is a, and the height of the cap is h, then the curved surface area of the spherical cap is
A = 2 \pi r h.

Suppose we have a hemisphere of radius 10 r10 (a) and cut it in ten horizontal slices (1 is on the top), does that mean that all slices have the same surface ?

even slice 1 has surface 62.8 (2\pi *10*1)? and its a (r1) =4.36?
so, the area of slice 4 (like all others) is
2pi*10*4-2pi*10*3 = 2pi*10= 62.8

is this correct?
If it is not, what is the formula to find the area and a (r1) of slice 1?

Thanks
 
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It is possible to cut the horizontal slices so that each has the same surface area by varying the spacing.
To work out the surface area of each slice - use calculus.
 
Simon Bridge said:
It is possible to cut the horizontal slices so that each has the same surface area by varying the spacing.
To work out the surface area of each slice - use calculus.
Hi Simon, thanks. I am just starting to learn calculus.
If I understood what you said, if we cut 10 equal slices of 1 cm , they will not have the same surface?
Could you show me how to frame the equation(s)?
Thanks
 
That's right - I would be surprised if the areas came out the same.

If we say that the floor is the x-y plane and up is the +z axis, then you start by dividing the whole hemisphere (radius R) into very thin disks - thickness "dz". Then you want to work out the equation for the surface area "dS" of the disk between z and z+dz in terms of z and R.
 
Simon Bridge said:
That's right - I would be surprised if the areas came out the same.
Then I misinterpreted wiki?

If the radius of the base of the cap is a, and the height of the cap is h, then the curved surface area of the spherical cap is
A = 2 \pi r h.
Because if we find the area of the slice on the 'floor' S 9 subtracting the cap with h = 9 (2pi*10*9) = 565.48 from the hemisphere 628,3 we get 62.8
and the same happens all the way to the top to S1
Wher did I go wrong?
 
Hah - I just tried it out and I am surprised ;) - try of for 2 slices.
I still think your best proof involves doing the calculus.
 
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Simon Bridge said:
Hah - I just tried it out and I am surprised .

I was, too, that's why I checked here, it seems amazing, right!
If you are intrigued, check by calculus, and let me know!
 
Yes. This is one of the reasons I like to answer questions here - sometimes someone surprises me.
This is the sort of thing that is obvious in retrospect.
 

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