Surface Area of a Sphere (1 Viewer)

[maths > english]

I tried to work out the surface area of a sphere through calculus and came up with pi^2 * r^2 and I cant see any errors in my working. Can you see where Ive gone wrong? (see attachment)

Thanks.

 

[abdooooo!!!]

ya your workings seems right... same as mine when i did it your way.

so it must be that the surface area cannot be derived using that method.

i know the method that works: just differentiate the volumn of the sphere with repect to r, you should get 4*pi*r^2.

let me have a think of why your method does not work... it seems logical enough... or does anyone else know already???

 

[Affinity]

hmm your method is wrong, the reason is quite subtle though..

for volumes you can take dV = Pi*y^2 dx because the error of this approximation is small compared to the volume of the slice.
but with surface areas, the error is comparable and you need a better approximation by considering the arc length of the sphere. i.e you cannot assume the rings to be horizontal, because sqrt(dy^2 + dx^2 ) is always larger than dx

The area element should be

dS= 2pi*y* [ sqrt( 1 + (dy/dx)^2 ) ] dx

dS = 2pi*sqrt(r^2 - x^2) * sqrt( r^2 / (r^2 - x^2)) dx

dS = 2pi*r dx

S = int (-r, r) 2pi*r dx

S = 4pi*r^2

 

[abdooooo!!!]

cool... thanks affinity... i finally figured it out.

because it is integrating against the wrong delta change since if you stretch the length derived from 2*pi*r into strings horizontally it would not have the same ratio as the area of the circle which need not be stretched. so we have a differing ratio in terms of delta change... if you get what i mean.

sqrt(dy^2 + dx^2 ), is this just the pythag theorem? so that must be the delta change right? but how do you get it down to this: dS= 2pi*y* [ sqrt( 1 + (dy/dx)^2 ) ] dx ?

 

[abdooooo!!!]

ohh... ya... lol... didn't see it. straight forward algebra.

sqrt(dy^2 + dx^2 )/dx = sqrt(dy^2 + dx^2 )/sqrt(dx^2)
= sqrt(1 + dy^2/dx^2)

then substitute that into 2*pi*y*sqrt(dy^2 + dx^2 )/dx = 2pi*y* [ sqrt( 1 + (dy/dx)^2 ) ] dx.

yea... i got it!!! lol

 

[Affinity]

am glad to find ppl who question what they are learning.