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Surface area of a sphere using integrals 
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#1
Apr1011, 12:57 PM

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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! 


#2
Apr1011, 01:43 PM

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hello spec00! welcome to pf!



#3
Apr1011, 02:08 PM

P: 3

Thanks for the fast reply, tinytim!
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! 


#4
Apr1011, 02:22 PM

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Surface area of a sphere using integrals
hi spec00!
the error for the volume is the difference between 2πr dh and (roughly) 2π(r + dr/2) dh … a secondorder error of π drdhbut the error for the area is the difference between 2π dh and 2π dh/cosθ … a firstorder error of 2π(1  secθ) dh(in layman's terms, most of the volume is in the middle, and the error is only an edgeeffect, but for the area, it's all edge! ) 


#5
Apr1011, 02:35 PM

P: 3

Wow! Now, that's absolutely clear for me.
Thanks for the huge enlightment..! 


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