Can someone explain in words, instead of the usual geometry proofs,
how it is that the surface area of a sphere is four times the great
circle?

Hi halejrb,

I have noticed that your question has been ‘locked’  by several
researchers today and no one has offered an answer to your question. 
I suspect the reason is that really there isnīt a reason WHY the
surface area of a sphere is four times the area of its great circle,
it just IS...  or maybe I have just misunderstood your question? 
(please let me know if I have)

You MIGHT find some useful material in these websites below:

The ‘why’ bit of this website should be of interest to you:
"How is the surface area of a sphere calculated, and why?"
http://mathforum.org/library/drmath/view/51754.html


“why the formula for the surface area of a sphere works” **PDF
DOCUMENT**
http://explorer.scrtec.org/explorer/explorer-db/rsrc/817016153-81ED7D4C.2.PDF

As you might already know the discovery for the method of calculating
the surface area of a sphere is generally attributed to Archimedes. 
If you would like to know more about Archimedes and his mathematical
discoveries have a look here:
http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Archimedes.html

I hope this helps you a little.

All the best

THX1138

The easy answer, in words, is: the surface area of a sphere is four
times the great circle simply because of the way we define the terms
"surface area", "sphere, "4 times" and "great circle". It's the same
reason why 1+2=3. It can't possibly be otherwise. In other words, math
is not something that one can apply the law of the excluded middle to
(it would be meaningless to say "either 1+2 equals 3, or does not
equal 3"), nor is it something one needs to confirm through experience
(useless anyway since our lifetimes are finite - I may die after
attempt N, and the results may change on trial N+1). Have fun!

I really wanted to answer this question well since I really feel that
good visualization of concepts is a key to good understanding.  The
second link in the earlier comment did provide one visualization
method that I thought of...  circumference of the sphere (2*pi*r)
multiplied by its height (2*r)...  I abandoned that (even though the
result is correct) because it "unjustly" reaches that answer.  The
fact is that the reasoning does not carry through to other objects..
it just happens to work for a sphere.

I have no idea what math level the original post wanted... I actually
assume as little math as possible.  Regardless, calculus can provide
some insight into how the area/volume of something is built up from
miniscule component pieces.  Now that I've scared away most readers
:)...  The best correlation I can do for showing how the surface areas
relate is by considering two ways of figuring out the VOLUME of a
sphere.  The first way to visualize the volume of a sphere is to
picture a stack of uniformly thick disks (each with a different
radius) stacked on top of each other.  The other way to visualize the
volume of a sphere is like the layers on an onion...  uniformly thick
spherical shells that when all placed inside one another form a
sphere.

The stack of plates uses the area of the great circle, while the
slices of onions use the surface of a sphere...  Building up the
volume from component pieces and breaking down the volume into
component pieces are inverse operations.   The uniform thickness
restriction forces the "build from disks" approach to be a bit more
complicated.  I will call the disk height dh, R the radius of the
sphere, and r the radius of an individual disk.  A disk at height h
has an volume of dh * pi * (R^2 - h^2).... this is just dh * pi *
r^2... again it is the uniform thickness that forces the less natural
form...
This makes the volume of a sphere be the difference of 2 terms: 
dh*pi*R^2  - dh*pi*h^2
This first term becomes the volume of the enclosing cylinder (the area
of the great circle) * (height of 2*R)
The second term becomes the the void between the bounding cylender and
the sphere.... (the area of the great circle) * (2/3 * R)

Decomposing into onion shells effectively divides both terms by
(R/3)... yielding 4*(area of the great sphere)

This may or may not help... but that's why it's just a comment instead
of an answer :)  Good luck on your search for a nice visualization of
the surface area... I'd be interested in the final explanation...

Gee, the comments are so good I don't need to wait for the answer.  I
did notice a lot of "math speak" though in the comments, which fits an
opinion I've had for years:  People good in math can't understand how
people who are bad in math think.  The result is there are very few
good math teachers.  In trying to answer the sphere surface area
question on my own, I got the idea of visualizing a four sided pyramid
composed of equilateral triangles.  If you blow up the pyramid like a
balloon, the triangles will begin to resemble circles and the pyramid
will begin to resemble a sphere.  You can then see that the four
pseudo circles that comprise the surface area are about the same size
as the great circle inside the now inflated sphere.

I'm guessing you want an entirely un-mathematical explanation. Maybe
this will help:

A coin with no width (an infinitely small side) would have a surface
area of twice the area of the circle making up each side.

Now, imagine splitting your sphere up into two hemispheres, by cutting
it down the middle. Each hemisphere is, in effect, the circle (or the
side of the "coin") puffed up so it reaches a curve.
(| |)

The curve itself reaches half of the height of the diameter of the
circle (because it's the radius). At that point the hemisphere's
surface area is twice the area of the original circle. Why is it
twice? Well, obviously there would be a simple expressible
mathematical relationship. This might be voodoo mathematics, but
here's my guess:

You have, in essence:
The circle's area
expanded by two half-circles (because it's two dimensions, not one)
making double the circle's area.

If you put the two semi-circles together, you then have four times the
original area.

Let me rephrase, in case my wording was unclear:
Each hemisphere is like an extended circle.
The amount it is extended by is a half circle in two different
dimensions.
The two hemispheres together make the surface area of the sphere.

Is this more along the lines of what you were looking for?

Good comment, cagliostro-ga. 

I was going to post an answer like your comment, but I also got stuck
at the "voodoo" step.  I am not convinced you can actually get to the
exact answer without invoking some sort of limiting procedure.  I
think I could probably answer the question this way to halejrb’s
satisfaction using a whiteboard, but not in a text editor.

I think starting with a coin is a good way to go, though.  As you have
pointed out, that gets you twice the area of the great circle.  (This
is actually an air-tight lower limit from a strict math point of view
too.)

I have been trying to convince myself that six times the area of the
great circle is an upper limit by arguing that you can look at the
coin in each of the three dimensions.  However, although this is
intuitive, I cannot convince myself it is actually right.   That would
put us between two and six times the area.

If I could convince myself six times was a valid upper limit, I would
then try to figure out how I had over counted.

I would very much like to see an explanation of the “voodoo” if
anybody can come up with it.

8ball-ga

PS.  Great question halejrb-ga.  =)

This still is not an answer, but an upper limit is six times the area
of the great circle.

Imagine you make the smallest possible cylinder that will contain a
sphere.  It seems obvious that this cylinder must have as much or more
surface area than the sphere.  (Actually, I am cheating here.  The
true proof of this is a bit subtle, but I think for purposes of this
question I will leave it out.  It is intuitive.)

Each of the two ends of the cylinder is a circle the same size as the
great circle.  That gives you twice the area of the great circle.  The
link given by thx1138-ga shows how to calculate the area of the round
part of the cylinder.

http://explorer.scrtec.org/explorer/explorer-db/rsrc/817016153-81ED7D4C.2.PDF

I agree with jhouse-ga that this "unjustly" reaches that answer for
the actual surface area of a sphere.  However it does show that the
area of the round part of the cylinder is four times the area of the
great circle.

Thus, we have a lower limit of twice the area of the great circle
(from cagliostro-ga’s comment) and an upper limit of six times the
area.

I’m still not convinced we can find the final answer this way, but I
would be interested if anybody could tighten the range.

How much geometry is not allowed, so to speak? I ask the question
because you 'only' have to invoke 'similar triangles' to be able to
show that the sphere has the same surface area as the curved part of
the enclosing cylinder; the rest is easy (well, 'triangle equals half
rectangle' easy, anyway). (On a more philosophical note, I don't see
that we could possibly prove what is needed using 'words' only - the
words have to refer to things and, the problem being essentially
geometrical, the things are going to be 'geometry', in some sense).

Ok chaps, we are 2/3rds of the way there.  Its easy to prove with
standard geometry that the area of the great circle is pi*r*r.  Its
easy to prove again that the surface area of the enclosing cylinder
(without ends) is 4*pi*r*r.  All we need to prove is that the surface
area of the enclosing cylinder is equal to the surface area of the
sphere.  GO TO IT!

DONE IT!

Create a diagram consisting of a sphere enclosed by a right cylinder
or the same radius (r), 2*r in height.  The cylinder is open at its
ends.  The cylinder touches the sphere along its equator.  Cut across
the great circle at the equator and throw the bottom half away (we
don’t need it).

Cut the great circle into a large number or pizza slices radial from
the centre.  Fold these up at the equator until they form a spiked
crown touching the inside of the cylinder.  It is easy to see that
they cover half of the inside of the cylinder and if we made another
set we could paste them upside down inside the cylinder interlocking
with the first set.

Hence the surface area of the cylinder is TWICE the surface area of
the great circle.

If we can prove that the surface area of the whole cylinder equals the
surface area of the sphere we can then say that the surface area of
the hemisphere is twice the surface area of the great circle.

There is an elegant geometrical proof of this at:

http://mathcentral.uregina.ca/QQ/database/QQ.09.99/wilkie1.html

In other words you need twice as much paint to paint the dome of a
cathedral as you would to paint the floor underneath it!

Thanks to everyone for lots of great comments!

http://www.cs.biu.ac.il/~tsaban/Pdf/mechanical.pdf

check this out for detailed explanation

Consider a point at an azimuth angle phi and polar angle theta.
consider then a differential area around the point. It is RSin(theta)d
phi * Rd theta.
Integrating we get 4 pi R^2 = 4 times area of great circle