# Thread: The surface of a sphere

1. Suppose we have a hollow sphere r = 10 , O =0: its surface is (4pi*r^2) A = 1 256.6, if its thickness =1 that is also its volume, if density = 1 that is also its mass/weight.

Now, suppose we cut 20 horizontal slices (wideness = 1) , we get (2*)10 rings from 0 to +/- 10

How do we calculate the area/wolume/weight/mass of each ring?

At a rough extimate rings have a radius of 2*(10,9.8,9.5,9.2,8.8,8.3,7.6,6.9,5.9,3.9)
even considering that the last two (9,10) have A=50/60 the total is less than 1 100.

in order to get 1256.6 we should have 20 rings of 62.8, or some rings bigger than that.
But the biggest is ring 1 which is a little less than 62.8
can we determine the area/mass of each ring?
does the area of each ring correspond to wolume/weight/mass?

Thanks  2.

3. Originally Posted by whizkid Suppose we have a hollow sphere r = 10 , O =0: its surface is (4pi*r^2) A = 1 256.6, if its thickness =1 that is also its volume, if density = 1 that is also its mass/weight.
No, the volume is 1135.16, The interior surface has a smaller area than the outside surface, so you can't just multiply the exterior surface area by the thickness to get the volume of the shell

Now, suppose we cut 20 horizontal slices (wideness = 1) , we get (2*)10 rings from 0 to +/- 10

How do we calculate the area/wolume/weight/mass of each ring?

At a rough extimate rings have a radius of 2*(10,9.8,9.5,9.2,8.8,8.3,7.6,6.9,5.9,3.9)
even considering that the last two (9,10) have A=50/60 the total is less than 1 100.

in order to get 1256.6 we should have 20 rings of 62.8, or some rings bigger than that.
But the biggest is ring 1 which is a little less than 62.8
can we determine the area/mass of each ring?
does the area of each ring correspond to wolume/weight/mass?

Thanks
You want the full formula? okay, its for all the rings, and for the end caps it's H is the distance from the end of the sphere to the further edge of the ring ( for the first ring after the end cap H=2 = the thickness of the end cap plus the thickness of the ring).

The formula given uses your values of a radius of 10 and thickness of 1 for the spherical shell.  4. Thanks Janus that is a great formula! And volume is of course 4pi/3*(10^3-9^3)  5. Originally Posted by Janus You want the full formula? okay, its Hi Janus, I'm trying to use your formula but I get 10 765.2 + 29pi/3 = 10 795.57 for each hemisphere instead of 628;

integrate pi * (x^2*(10-x/3)-(x-1)^2+(x-2)^2*(9- (x-2)/3)) x from 2 to 10 - Wolfram|Alpha

where did I go wrong?  6. Originally Posted by whizkid  Originally Posted by Janus You want the full formula? okay, its Hi Janus, I'm trying to use your formula but I get 10 765.2 + 29pi/3 = 10 795.57 for each hemisphere instead of 628;

integrate pi * (x^2*(10-x/3)-(x-1)^2+(x-2)^2*(9- (x-2)/3)) x from 2 to 10 - Wolfram|Alpha

where did I go wrong?
I rechecked the formula and did find an error. (when I simplified, I messed up with a sign and ended up canceling out something I shouldn't have.)

So here is the corrected equation: Note the correction in the middle term.

I've double checked this equation by summing up all the rings and end cap in a hemisphere and get the correct answer of 567.58 (NOT 628 as I already stated in my last post.)

But besides that, the mistake you made was integrating the equation to get the volume of a hemisphere. The equation as given directly give the volume for each ring of thickness 1. To get the volume of the hemisphere you just simply add up the volumes of each ring plus the end cap. This is not an integration problem.

To illustrate why, consider the following problem:

You have a stack of 4 disks, 1 unit thick and in decreasing radius (4,3,2,1). what is the total volume of the stack. You basically add up the volumes of each disk, with the equation for each disks volume being (since the thickness is 1. Thus we get volumes of 1 pi, 4 pi, 9 pi and 16 pi for a total volume of 30 pi

However, if we integrate from 1 to 4 we get 21 pi, not even close to correct.

There is a way to get the volume of the hemisphere by integration, but it is not by integrating the formula I gave.

On to something else:

There is something very cool about these rings. I suggest you try the equation with different values of H (from 2 to 10) and compare the results, you are likely to be surprised.  7. Originally Posted by Janus There is something very cool about these rings. I suggest you try the equation with different values of H (from 2 to 10) and compare the results, you are likely to be surprised.
Hi Janus, according to wiki Spherical cap - Wikipedia, the free encyclopedia
all slices have same area.
It is surprising , but it is the only possible solution:
considering each hemisphere, if a slice is smaller than 20pi another should be greater but the one with largest base (a =10, slice 1), has area 20pi.
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