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Surface Area of a Sphere
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Jon G.  
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 More options Aug 23 2008, 3:04 pm
Newsgroups: alt.math, alt.math.recreational, de.sci.mathematik, fj.sci.math, fr.sci.maths, han.sci.math, japan.sci.math, sci.math
From: "Jon G." <jon8...@peoplepc.com>
Date: Sat, 23 Aug 2008 15:04:45 -0400
Local: Sat, Aug 23 2008 3:04 pm
Subject: Surface Area of a Sphere
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
jon8...@peoplepc.com


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hostlocal  
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 More options Aug 23 2008, 4:21 pm
Newsgroups: alt.math, alt.math.recreational, de.sci.mathematik, fj.sci.math, fr.sci.maths, han.sci.math, japan.sci.math, sci.math
Followup-To: alt.math
From: "hostlocal" <rolf...@invalid.invalid>
Date: Sat, 23 Aug 2008 15:21:53 -0500
Local: Sat, Aug 23 2008 4:21 pm
Subject: Re: Surface Area of a Sphere

"Jon G." <jon8...@peoplepc.com> wrote in message

news:YeSdneBl2uY7_C3VnZ2dnUVZ_oPinZ2d@earthlink.com...

> 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
> jon8...@peoplepc.com

you need to show all the steps.

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Jon G.  
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 More options Sep 3 2008, 12:07 pm
Newsgroups: alt.math, alt.math.recreational, de.sci.mathematik, fj.sci.math, fr.sci.maths, han.sci.math, japan.sci.math, sci.math
From: "Jon G." <jon8...@peoplepc.com>
Date: Wed, 3 Sep 2008 12:07:18 -0400
Local: Wed, Sep 3 2008 12:07 pm
Subject: Re: Surface Area of a Sphere

">

> 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.
jon8...@peoplepc.com

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


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