I was trying to figure out how the surface area of a sphere was derived. I was wondering if anyone can point me towards a book or website or any kind of help to help me see the derivation

A nonrigorous but thorough discussion of surface area can in fact be found in Klines book. He gives the formula Surface area = integral :a...b: 2 pi y sqrt( 1 + y'^2) dx where y here is the curve that is a function of x. Given the equation of a circle centered upon the origin, y^2 + x^2 = r^2, its a straightforward integration to find that surface area is 4 pi r^2. Note that it's the derivative of y that's inside the square root.

The equation of the volume of a sphere is 1/3*PI*R^3=V The derivative is PI*R^2=A or the area of a circle The dervative of this is 2*PI*R=C or the circumference of a circle Kind of defining a circle in 3d, 2d and1d I think in caculus there was something called solids of revelution, or something like that. you would revolve an equation about an axis, or part of and equation. You had to define the axis and the limits of the equation. like X^2+Y^2=1 substituting numbers between 0 and one we create a circumference of the unite circle of 1. and subsiquently derive the sine or cosine by imputing any decimal between ) and one. the basis of trigonometry.

Really, this needs a diagram. You can set up a surface integral using spherical co-ordinates r, theta and phi. An element of surface area in this system is dA = R^2 sin(theta) dr d(theta) d(phi) Then, integrate theta from 0 to pi and phi from 0 to 2 pi. The result is the surface area, 4 pi R^2. Does that help?

Hi przyk, That works, and it's an interesting exercise in basic calculus to visualise why it works... but it's just shifting the problem, because it requires knowing the volume formula. Hi miex, I don't know what calculus you've done, so forgive me if this is too basic. You can derive the surface area formula by cutting the sphere into slices, approximate each slice by a cylinder, get the surface area around each slice, and add them all up. For example: Cut a sphere of radius into 10 slices, and expand the slices into cylinders: Please Register or Log in to view the hidden image! The radius (r) of each slice depends on the radius of the sphere (R) and the distance of the slice from the middle (h): r = √(R²-h²) The surface area (a) around the edge of each slice is the circumference times the thickness (Δh): a = 2πr.x =2.πRΔh√(R²-h²) For a sphere of radius 5, the numbers look like this: <table border><tr><th>Slice</th><th>h</th><th>r</th><th>a</th></tr><tr><td>1</td><td>4</td><td>3</td><td>18.8</td></tr><tr><td>2</td><td>3</td><td>4</td><td>25.1</td></tr><tr><td>3</td><td>2</td><td>4.6</td><td>28.8</td></tr><tr><td>4</td><td>1</td><td>4.9</td><td>30.8</td></tr><tr><td>5</td><td>0</td><td>5</td><td>31.4</td></tr><tr><td>6</td><td>0</td><td>5</td><td>31.4</td></tr><tr><td>7</td><td>1</td><td>4.9</td><td>30.8</td></tr><tr><td>8</td><td>2</td><td>4.6</td><td>28.8</td></tr><tr><td>9</td><td>3</td><td>4</td><td>25.1</td></tr><tr><td>10</td><td>4</td><td>3</td><td>18.8</td></tr><tr><th colspan=3 align=right>Total:</th><th>270</th></tr></table> Of course, this is only an approximation. You'd get a better approximation by using 100 slices, or a thousand, or a million... but that's kind of fhard to add up, and you really need infinite slices anyway. That's where calculus comes in. It gives us the tools to find a formula for the actual area, by finding the limit that the area of the slices approaches as you take more and more slices. Is that enough to get you going?

I know this doesn't answer your question, but... Apparently Archemedies determined the volume of a sphere using a very novel (pre-calculus) concept. He used simple geometric relations and rates of change in shape to break up the problem. He compared these portions of the sphere with cylenders and cones who's volume formula was already known at the time. And then applied these pieces on an imaginary scale to show the relation in mass between them. It was a sort of mechanical calculus solution. You can see the steps here: http://www.cut-the-knot.org/pythagoras/Archimedes.shtml

This doesn't work, and I'm not completely sure why (I have an inkling, but I'm interested in the formalism). Integrating the outer surface area of cylinders like this to get the surface area of a sphere doesn't work. It results in an area of 4π²r, instead of 4πr². Why? What's the criteria for ensuring that some particular segmentation will result in the correct integral?

Just in case anyone isn't aware: That only works because variations in the radius of a sphere are in a direction that is perpendicular to the surface. It won't work for all volumes.

Correct. Not correct. First of all, look at your last equation. Neglecting the numerical factors, you have R (a length) times h (a second length) times √(R²-h²) (a third length). That product has units of volume, not area. But there's an even more fundamental problem at work here. The differential cylinder of revolution will not have an area of 2πrx. It should be 2π times r times the differential arc length of the circle. Let your circle be situated at the origin. Then its equation is x<sup>2</sup>+y<sup>2</sup>=R<sup>2</sup>. The surface area dA of a single slice is then: dA=2πyds dA=2π√(R²-h²)√(1+(y')<sup>2</sup>) dA=2π√(R<sup>2</sup>-x<sup>2</sup>)√(1+x<sup>2</sup>/(R<sup>2</sup>-x<sup>2</sup>)) Combine the terms under the second radical, then multiply the radicals together. You should find that x drops out of the integrand entirely. dA=2πR Now integrate that from x=-R to x=R, and you will have A=4πR<sup>2</sup>. edited to fix a number of typos

Yeah, I figured out how to fix it, but that wasn't the problem. The problem was that I was thinking that it shouldn't matter, that as the segment size reduced to zero, the area would be close enough. It clicked completely while I was composing this reply. As &Deltax; approaches zero, the ratio of the cylindrical section area to the actual spherical section area doesn't approach 1. But the frustrum area (or its approximation) does. I hadn't thought the limit through properly.

Well, I think what you should do is stick with the easy way, unless you really need to know how it works, in which case I can't help you, since I'm just a high school student in AP Calculus AB and all I've done is derivatives, but I think its best to stick with what works.