Surface area of a sphere using integrals

spec00

Hello dear colleagues!

Yesterday i was trying to proof the surface area of a sphere formula, then i got some problems. I know that something is seriously wrong in this concept, but i can't tell what exactly is wrong. Could you guys help me please?

I just thougt about a hollow sphere, then we can slice it up to little cylinders with infinitesimal height (like slicing some onion rings). If we add up these teeny weeny little parts, i thought that we could obtain the area of the sphere.

[tex]x^{2}+y^{2}=r^{2}[/tex]
[tex]\int(2*\pi*x)*dy[/tex]

[tex]2*pi*\int\sqrt{y^{2}-r^{2}}*dy[/tex]

But when i integrate over 0 to R and multiply all by 2, the result is not correct. What did i do wrong?

Thanks!

tiny-tim

hello spec00! welcome to pf!
no, they're not cylinders (with vertical sides), they're little slices of a cone (with sloping sides), which can have a much larger area

spec00

Thanks for the fast reply, tiny-tim!

Well, that's true, but i've seen a proof for the volume of a sphere using slices of it with infinitesimal height. Since the slices are not cylinders, but little cones, that deduction wouldn't get us in the wrong place too?

http://en.wikipedia.org/wiki/Sphere (Proof of the sphere's volume formula)

Thanks again! And sorry about my english, i'm a bit rusty!

tiny-tim

hi spec00!
we get this question quite often! …

the error for the volume is the difference between 2πr dh and (roughly) 2π(r + dr/2) dh …

a second-order error of π drdh

but the error for the area is the difference between 2π dh and 2π dh/cosθ …

a first-order error of 2π(1 - secθ) dh

(in layman's terms, most of the volume is in the middle, and the error is only an edge-effect, but for the area, it's all edge! )

spec00

Wow! Now, that's absolutely clear for me.

Thanks for the huge enlightment..!