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 Subject: Surface area of sphere Category: Science > Math Asked by: halejrb-ga List Price: \$4.00 Posted: 20 Jun 2002 10:35 PDT Expires: 27 Jun 2002 10:35 PDT Question ID: 29824
 ```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?```
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 Subject: Re: Surface area of sphere From: thx1138-ga on 20 Jun 2002 15:24 PDT
 ```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```
 Subject: Re: Surface area of sphere From: zeno-ga on 20 Jun 2002 15:49 PDT
 ```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!```
 Subject: Re: Surface area of sphere From: jhouse-ga on 20 Jun 2002 17:10 PDT
 ```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...```
 Subject: Re: Surface area of sphere From: halejrb-ga on 20 Jun 2002 18:26 PDT
 ```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.```
 Subject: Re: Surface area of sphere From: cagliostro-ga on 21 Jun 2002 02:42 PDT
 ```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?```
 Subject: Re: Surface area of sphere From: 8ball-ga on 21 Jun 2002 17:26 PDT
 ```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. =)```
 Subject: Re: Surface area of sphere From: 8ball-ga on 23 Jun 2002 15:46 PDT
 ```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.```
 Subject: Re: Surface area of sphere From: moltonian-ga on 24 Jun 2002 05:13 PDT
 ```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).```
 Subject: Re: Surface area of sphere From: black_shadow-ga on 02 Jul 2002 05:01 PDT
 ```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!```
 Subject: Re: Surface area of sphere From: black_shadow-ga on 02 Jul 2002 07:23 PDT
 ```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!```
 Subject: Re: Surface area of sphere From: halejrb-ga on 02 Jul 2002 08:15 PDT
 `Thanks to everyone for lots of great comments!`
 Subject: Re: Surface area of sphere From: tne-ga on 26 Jul 2002 17:01 PDT
 ```http://www.cs.biu.ac.il/~tsaban/Pdf/mechanical.pdf check this out for detailed explanation```
 Subject: Re: Surface area of sphere From: vizguy-ga on 06 May 2004 13:35 PDT
 ```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```