Sphere Surface Area Precision
Date: 04/22/2003 at 19:39:11 From: Alex Beal Subject: Sphere Surface Area Precision How can the formula 4*pi*r^2 be precise? The way that one arrives at this answer seems to have a flaw. Consider a sphere with a pyramid drawn inside it and the point of the pyramid at the center of the sphere and the base of the pyramid on the surface of the sphere. The volume of this pyramid would be r*(1/3)*(area of the base of the pyramid) Now we duplicate this pyramid to fill the entire sphere. This would mean that [r*(1/3)*(area of the base)]+[r*(1/3)*(area of the base)]+[r*(1/3)*(area of the base)]+... should be equal to the volume of the sphere. Now let's use the distributive property. r*(1/3)*(area of the base + area of the base...) This also means r*(1/3)*(surface area of the sphere) This should still equal the volume of the sphere, so let's say that the volume formula for the sphere is equal to it. (4/3)*pi*r^3 = r*(1/3)*(surface area) Now solve for the surface area 4*pi*r^2 = Surface Area of a Sphere But how can the area of the base of the pyramid be found with any precision? You can just do the w*h, but since the base of the pyramid is flat and the surface of a sphere is curved it will be a tiny bit off. You can draw many pyramids inside the sphere, making the answer more and more precise, but the answer will never be exact. Where does my logic break down, or am I right that one can't find the exact surface area of a sphere? I have thought about Zeno's Paradox having some relation to this. An infinite number of points can add up to a finite number in this geometric series (1/2+1/4+1/8+1/16+1/32...). Maybe if you put an infinite number of pyramids in the sphere the bases will add up to the exact surface area.
Date: 04/22/2003 at 20:59:36 From: Doctor Rick Subject: Re: Sphere Surface Area Precision Hi, Alex. The problem you are concerned about arises before you get to the sphere: in the calculation of the value of pi. This was first done by Archimedes, who proved that the ratio of the area of a circle to the square of its radius is between 3 10/71 and 3 1/7. He did this by computing the areas of inscribed and circumscribed 96-sided polygons; the area of the circle must lie between these two areas. In principle we could continue doubling the number of sides and narrowing down the value of pi as far as we wish. By other methods we have in fact gotten billions of digits of pi. Still, we have not found the _exact_ value of pi. But we know that it is a particular number, we just can't express it exactly. The fact that it is a particular number has to do with the theory of limits, which is part of the foundation of calculus. It is indeed related to Zeno's paradoxes. The following Web page discusses both matters in the history of Greek mathematics: Basic Ideas in Greek Mathematics (Michael Fowler) http://www.phys.virginia.edu/classes/109N/lectures/greek_math.htm See also this description of finding the volume of a sphere by a method different from yours: Surface Area and Volume of a Sphere http://mathforum.org/library/drmath/view/55230.html - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
Date: 04/22/2003 at 23:48:22 From: Alex Beal Subject: Sphere Surface Area Precision Are you saying that Archimedes trying to find pi is similar to my problem (can't be found exactly) or are you saying that the formula is imprecise because the value of pi is irrational and we can't give a precise answer in decimal form? If it's the second answer, what I mean is that the surface area formula would be imprecise because when you put the pyramid inside the sphere and find the area of the base of that pyramid, it won't be the same area as the surface of the sphere above it. A 2D-model/cross- section would look like a segment of a circle. The two lines that make up the central angle and extend to the edge of the circle would be the sides of the pyramid and the line that connects those two lines would be the base of the pyramid. The arc of the circle above it would be like the surface of the sphere above the surface of the flat base of the pyramid (the line connecting ends of central angle). The formula 4*pi*r^2 is like measuring the "base of the pyramid" and saying that measure is the same measure as the arc above it, which is false. I'll try to draw a picture: /|\ / | \ / | \ ./ | | \ | | \ | / \ | / \|/ The period is the center of the circle. The slash marks extending out form the central angle and the sides of the pyramid. The line connecting the ends of those slashes is the "base of the pyramid." The curved thing on the right is the arc of the circle. The base of the pyramid is of course not the same measure as the measure of the arc. Another way to visualize it would be like forming a round object out of a lot of pyramids. It would be nearly round but not quite. When you found the surface area of that nearly round object it would be close to the surface area of a corresponding sphere (if you drew a tight fitting sphere around the nearly round object) but it would not be exactly correct. The math is like this (=~ means approximately equal to): r*(1/3)*(area of the base of the pyramid + area of the base of the pyramid...) = r*(1/3)*(surface area of the nearly round object) = (4/3)*pi*r^3 = r*(1/3)*(surface area of the nearly round object) = 4*pi*r^2 = Surface Area of the nearly round object =~ surface area of the sphere The formula 4*pi*r^2 is finding the surface area of the nearly round object and not the sphere around it, so it must be a little bit off. Is there a fault in my logic? Also, even if pi is irrational couldn't you still write an exact answer (assuming that Iím wrong and the formula is precise). For example, if the radius was 2 cm you could write the surface area as: 16*pi cm^2 Thanks.
Date: 04/23/2003 at 08:36:13 From: Doctor Rick Subject: Re: Sphere Surface Area Precision Hi, Alex. >Are you saying that Archimedes trying to find pi is similar to my >problem (can't be found exactly) or are you saying that the formula >is imprecise because the value of pi is irrational and we can't give >a precise answer in decimal form? I'm saying both. Since the volume of a sphere involves pi, it necessarily involves the same issue that Archimedes dealt with; but also the method Archimedes used has enough similarity to the one you describe that I think it's worth your study. This does not mean that either formula is not exact, however! Do I understand correctly that you are referring to the derivation of the volume of the sphere assuming that we have already derived its surface area? This is the approach taken in the Archive page to which I referred you: first derive the surface area, then the ratio of volume to surface area. Do you have no problem with the derivation of surface area? I see the same sort of difficulty in both cases! You're correct that the surface area of a compound of pyramids is not exactly equal to the surface area of the sphere. But the volume of the compound of pyramids is not exactly r/3 times the surface area, either. Note that you used r as the height of each pyramid, but this is not exactly true; r is the slant height of the pyramid, which is slightly greater than the height, as your figure shows. Thus any compound of pyramids with a finite number of faces has a volume that is slightly less than r/3 times the surface area of the sphere. The point is that we imagine continuing the process of adding faces to the polygon without limit. The more faces it has, the closer the ratio of volume to surface area comes to r/3, because the slant height comes closer to the height. To make a solid proof, we need to do the same thing Archimedes did, and calculate another ratio that we can prove to be greater than the ratio for the sphere. Then we need to show that the two ratios both approach r/3, getting as close as you wish as long as you choose a number of faces that is great enough. When we have done this, we have established that the ratio of volume to surface area of a sphere is _exactly_ r/3. Technically, we say that r/3 is the _limit_ of the ratio of volume to surface area as the polygon approaches a sphere. The limit is an exact value, not an approximation. See the following answer in the Dr. Math archives for some further discussion of limits that may help you understand why a limit is exact. The sort of difficulty you are facing is frequently raised in the context of the fact that 0.9999... (repeated without end) equals exactly 1. The Infamous .999... = 1 http://mathforum.org/library/drmath/view/55748.html Here is a more in-depth explanation of limits, in the context of calculus where they are dealt with formally. If it's too much for you, don't worry; I just want you to see that the issue is an important one to which mathematicians do pay careful attention. Understanding the Need for Limits http://mathforum.org/library/drmath/view/53754.html - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
Date: 04/23/2003 at 11:22:57 From: Alex Beal Subject: Sphere Surface Area Precision So you're saying that because the nearly round object's surface area is so incredibly close to the surface area of the corresponding sphere that it might as well be the same using the concept of limits? Thanks.
Date: 04/23/2003 at 15:17:48 From: Doctor Rick Subject: Re: Sphere Surface Area Precision Hi, Alex. Something like that, yes. The focus isn't on any one "nearly round object," but on a *sequence* of polyhedra with increasing numbers of sides. The surface area of each is greater than that of the one before it in the sequence, but the surface area never exceeds that of the sphere. Thus the difference between the surface area of the polyhedron and the surface area of the sphere keeps getting smaller as you go farther in the sequence. The limit concept is like a game: On your move, you get to pick any positive number, as small as you like. On my move, I have to find a polyhedron in the sequence such that the difference between its surface area and that of the sphere is less than the number you picked. If I can always do that, then the limit of the sequence of surface areas is exactly the same as that of the sphere. With care, we could show that this is always possible. In other words, the difference between the areas of a polyhedron in the sequence and the sphere can be made smaller than any number you can name. The only number that is smaller than any positive number is zero. Therefore we say that the surface areas of the polyhedra approach a LIMIT that is EXACTLY the surface area of the sphere. Here's another way to see that the limit of the polyhedron surface areas is exactly 4*pi*r^2. We know that the polyhedron surface area is never greater than 4*pi*r^2. Thus the limit cannot be greater than 4*pi*r^2, either. If the limit is not *exactly* 4*pi*r^2, then it must be *less* than this. But if so, then we could show that at some number of faces, the polyhedron area was greater than 4*pi*r^2. Beyond that point, the polyhedron surface areas would get *farther* from the limit, which contradicts the definition of a limit. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
Date: 04/23/2003 at 17:14:44 From: Alex Beal Subject: Thank you (Sphere Surface Area Precision) Thanks a lot, Dr. Rick! I understand how that formula can be precise now. I first stumbled upon this site when I was looking for a formula to calculate Pi, which I them implemented into a program. Ever since then I've loved this site. It's a great service. - Alex Beal
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum