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When we were in school they told us that the Surface Area of a sphere = $4\pi r^2$

Now, when I try to derive it using only high school level mathematics, I am unable to do so. Please help.

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But you do know Cavalieri, right? –  J. M. Oct 17 '11 at 14:27
    
@J.M. Frankly, I hadn't heard about Cavalieri before you mentioned it. Plus this is a line from the Wikipedia article " Today Cavalieri's principle is seen as an early step towards integral calculus and while it is used in some forms, such as its generalization in Fubini's theorem, results using Cavalieri's principle can often be shown more directly via integration." Can such a method be avoided entirely? If I remember rightly my teacher had given me a convincing explanation using elementary concepts. –  Green Noob Oct 17 '11 at 14:42
    
I see, you mentioned "high school level", and it is now clear to me that I'm no longer that young... –  J. M. Oct 17 '11 at 14:55

4 Answers 4

up vote 10 down vote accepted

Imagine a vertical cylinder enclosing the sphere, with height $2r$, radius $r$, and open ends. This cylinder has surface area $4\pi r^2$. The trick is to show that if you slice the cylinder and the sphere into infinitesimally thin horizontal rings, then at a given height, the surface area of the spherical ring equals the surface area of the cylindrical ring. Thus the total surface areas are equal.

Suppose the cylindrical ring has has height $\delta h$, and therefore area $2\pi r \times \delta h$. If the ring is at a height $r\sin\theta$ above the equator of the sphere, with $-\pi < \theta < \pi$, then the spherical ring has radius $r\cos\theta$, but its surface is at an angle $\theta$ from the vertical. So its area is $2\pi r \cos \theta \times \delta h/\cos \theta$, which is the same as the cylindrical ring.

This really needs some nice pictures, but I have no skill in that direction.

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A rigorous version of the above argument, and more, was given by Archimedes more than two thousand years ago, in On the Sphere and the Cylinder. –  André Nicolas Oct 17 '11 at 15:08
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Archimedes knew how to draw pictures, too. –  TonyK Oct 17 '11 at 18:11
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Here are some figures : mathcentral.uregina.ca/QQ/database/QQ.09.99/wilkie1.html –  leonbloy Oct 18 '11 at 1:13
    
Perhaps the picture from this answer could be close to what you want. –  Martin Sleziak Nov 21 '12 at 13:17

In F. G. - M., Cours de Géométrie Élémentaire, 1917, the surface area of a sphere is proved in a sequence of theorems.

In short as follows.

enter image description here

I. The lateral surface area of a regular pyramidal frustum with two parallel bases with perimeters $p=ns$ and $p^{\prime }=ns^{\prime }$ and $n$ trapezoidal lateral faces with apothem $a$ is

$$S_{P}=n\frac{s+s^{\prime }}{2}a=\frac{p+p^{\prime }}{2}a,$$

where $s$ and $s^{\prime }$ are the lengths of the sides of the bases regular polygons.

II. The lateral surface area of a conical frustum is

$$S_{C}=\lim_{n\rightarrow \infty }S_{P}=2\pi R^{\prime \prime }l=2\pi zh,$$

where $z=\overline{EG}$, with $EG\perp AB$.

III. The surface generated by a regular polygonal line rotating around a diameter which does not cross it has an area given by

$$S_{m}=2\pi a^{\prime }h,$$

where $a^{\prime }$ is the apothem.

IV. The lateral surface area of the portion of a sphere limited by two planes is

$$S_{F}=\lim_{m\rightarrow \infty }S_{m}=2\pi Rh.$$

V. The surface area of a sphere is

$$S=2\pi Rh=2\pi R\times 2R=4\pi R^{2}.$$

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What do you mean by integration? If one finds a method in Euclidean geometry for showing that the volume $V$ of a sphere of radius $r$ is $\frac43 \pi r^3$, do you consider that "integration"?

Suppose you know that formula for the volume of a sphere. If the radius increases from $r$ to $r+dr$, where $dr$ is an infinitely small increment, then the corresponding infinitely small change in volume $dV$ is $\frac43\pi(r+dr)^3 - \frac43\pi r^3$. But $$ \frac{\frac43\pi(r+dr)^3 - \frac43\pi r^3}{dr} = \frac{dV}{dr} = 4\pi r^2. $$ If you multiply the surface area $A$ of the sphere by the infinitely small thickness $dr$ of the atmosphere surrounding it, you get $A\;dr$, but you also get $dV$.

Hence $A$ must be $4\pi r^2$.

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How do you derive the formula for the volume? Surely that is a harder problem than finding the formula for the surface area? If so, this approach gets you nothing. –  TonyK Oct 17 '11 at 20:36
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I see no reason to think that "surely that is a harder problem than finding the formula for the surface area". In a secondary-school geometry course, I saw it done by applying Cavalieri's principle to two bodies, one of which was the sphere. The other was that portion of the interior of the smallest right circular cylinder that encloses the sphere, that is outside of the largest right circular cone whose vertex is the center of the sphere and that is contained within the cylinder. –  Michael Hardy Oct 17 '11 at 20:41
    
And it's worth understanding why the fact that one of these is the derivative with respect to $r$ of the other is not a freak coincidence. –  Michael Hardy Oct 17 '11 at 20:42
    
The way you describe it, it still looks like a harder problem. –  TonyK Oct 17 '11 at 20:46
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@TonyK: All I can say is that I disagree. A lot of subjectivity may be involved in assessing which is more difficult. –  Michael Hardy Oct 18 '11 at 1:28

The volume of a rotated graph of a function $f$ around the $x$-axis segment $[a,b]$ is $V=\pi \int_a^b f^2(x)dx$.

You could describe the unit sphere as the rotated graph of $f(x)=\sqrt{1-x^2}$ around $[-1,1]$. Then the volume of the sphere is $V=4\pi /3$. By a scaling argument, the general formula can be derived.

For the area of the sphere, my primary school teacher said that it could be thought of a pyramid with base the surface of the sphere. Then applying the volume formula: $Volume=(Base Area)\times Height / 3$ it follows that

$$ 4\pi R^3/3=A\times R/3$$ Then $A= 4\pi R^2$. The last part is not rigorous at all, but it works. :)

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