How Is the Surface Area of Spherical Cap Slices Calculated?

[bobie]

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

 

[Simon Bridge]

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.

 

[bobie]

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

 

[Simon Bridge]

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.

 

[bobie]

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?

 

[Simon Bridge]

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.

 

[bobie]

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!

 

[pbuk]

http://mathworld.wolfram.com/Zone.html

 

[Simon Bridge]

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.

 

[bobie]

http://mathworld.wolfram.com/Zone.html

Thanks, MrAnchovy