This solution may seem trivial, since it's already been done, but for those
interested, this page shows the calculus for deriving the surface area of a
sphere.

http://mypeoplepc.com/members/jon8338/math/id17.html


Jon Giffen
jon8338***peoplepc.com

"Jon G." <jon8338***peoplepc.com> wrote in message
news:YeSdneBl2uY7_C3VnZ2dnUVZ_oPinZ2d***earthlink.co m...
> This solution may seem trivial, since it's already been done, but for
> those interested, this page shows the calculus for deriving the surface
> area of a sphere.
>
> http://mypeoplepc.com/members/jon8338/math/id17.html
>
>
> Jon Giffen
> jon8338***peoplepc.com
>

you need to show all the steps.

">
> you need to show all the steps.
>

Suppose a sphere with center located at the origin is sliced by a plane
parallel with the xz axis. This forms a circle. The xy plane slices the
circle at a point in Quadrant I. The angle between this point, the origin
and the x axis is angle w.

Angle w = (Arclength A)/(radius r)

w = A/r

wr = A differentiating,

rdw + wdr = dA The radius doesn't change, so dr=0

rdw = dA

C is the circumference of the parallels at each increment of dA. Each
parallel has a radius of r cos w. A band of surface area dS of
Circumference C is,

dS=CdA

dS=2pi r cos w dA

dS=2pin r^2 cos w dw

Integrating from w= -pi/2 to pi/2,

S = 4pi r^2

http://mypeoplepc.com/members/jon8338/math/id17.html

has a diagram to facilitate this spoon-feeding.

--
Jon G.
jon8338***peoplepc.com

Where is she?
http://www.charleyproject.org/cases/...n_cynthia.html