Surface area of a sphere using integrals


by spec00
Tags: integrals, sphere, surface
spec00
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#1
Apr10-11, 12:57 PM
P: 3
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!
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tiny-tim
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#2
Apr10-11, 01:43 PM
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hello spec00! welcome to pf!
Quote Quote by spec00 View Post
I just thougt about a hollow sphere, then we can slice it up to little cylinders
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
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#3
Apr10-11, 02:08 PM
P: 3
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
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#4
Apr10-11, 02:22 PM
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Surface area of a sphere using integrals


hi spec00!
Quote Quote by spec00 View Post
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?
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
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#5
Apr10-11, 02:35 PM
P: 3
Wow! Now, that's absolutely clear for me.

Thanks for the huge enlightment..!


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