Just curious...

[pi]

Does anyone know why (or is a coincidence) that:
Circle:



and

Sphere:




Haven't done much 'high-powered' calculus in spesh yet, and just thought this was odd.

Thanks in advance!

[onur369]

Ive realised this also, have a look at this, we see the intergrating and differentiation formulas gives us other formulas: http://en.wikipedia.org/wiki/Volume#Cone

[pi]

Ive realised this also, have a look at this, we see the intergrating and differentiation formulas gives us other formulas: http://en.wikipedia.org/wiki/Volume#Cone

Thanks. So its not just random. Will do some reading on this now!

[luken93]

Hahaha I was also thinking about this - thought it couldn't just be random..

[brightsky]

Yes it does have a reason behind it.

For the area:
Think of integrals. You got a function y = sqrt(r^2 - x^2). Find the integral of that from r to -r. Times that by 2 and you get the area formula. If you understand how integration is derived, you will know how the reasoning behind the "first principles" so to speak through taking the limit of a sum.

For the circumference...haven't really thought of it that way, but you can try and use "arc length" to find the formula. But even better, try and understand the "first principles" mentioned above, it will help with deriving the formula here.

For volume:
There's a chapter in spesh relating to solids of revolutions. If you understand that, this should be relatively simple to understand. You have the graph y = sqrt(r^2 - x^2) again. You rotate that around the x-axis once. There's a formula relating to finding the volume, which is in turn derived from cutting the solid into cylinders of infinitesimal height (or 'length' in this case). The derivation is somewhat the same as that of the area. But yeah, to find the volume of the sphere, it should take a few steps using this.

For surface area, again, my knowledge is quite limited...but intuitively, I would probably think that by cutting the above solid (sphere) into cylinders of infinitesimal 'length', the sum of the outside bits (the curved section between the bases) of the cylinders would equal to the surface area. I shall do some pondering, but you can perhaps try work from there to find the surface area.

[Mao]

By definition, circumference

Now, to find the area, we divide the circle like an onion. Each strip is an onion ring of infinitely small width. We break this ring and find that it has a width of dx and a length of , where x is the radius of that ring.

Thus,

The process of SA --> V is similar.

[brightsky]

Hmm...don't know if my logic works for surface area, but here goes:

So we have f(x) = sqrt(r^2 - x^2). Graph that. We grab this graph and rotate it 360 degrees around the x-axis, so that the 'space' it encloses makes a sphere. Now just like for area, we split it into rectangles, this time we split it into cylinders with very very very small 'length'. So you can imagine the sphere, starting from the right hand side, being a lot of cylinders stacked side by side, with the 'radius' of the cylinders going from very small (start of sphere) to big (middle) to very small again. We denote the 'length' of the respective cylinders to be . The 'radius' of the cylinders is denoted as (you can see this by 'flattening the graph again and visualising rectangles again rather than cylinders). Now for surface area we want to find the circular bit of the cylinder that wraps around it (in other words, it's surface area minus the two ends). (Note that to find volume, we directly sum the volumes of each cylinder). So the circular bit of each cylinder is given as: circumference*height = 2*pi*r*h. But in this case r = f(x) and h = . So each one is . We want to sum everything up, but let to get an accurate answer. The sum goes from -r to r. So:





Since f(x) = sqrt(r^2 - x^2),



Solve that and we have our answer.

However, this still requires us to 'prove' the relation between area and circumference...but hope this is of some use.

tl;dr: Above is a way better explanation.

EDIT: Oh yes, similar to the long-winded, sloppy explanation I just gave, instead of finding the sum of the areas of the rectangles to get the area, we can just sum of 's (the tops of the rectangles).