Surface Area of a Sphere without double integral

[cwbullivant]

Is it possible to come up with a derivation of the surface area of a sphere without using a double integral? Most of the ones I've found seem to involve double integrals;

For example, this was given as the "simplest" explanation in a thread from 2005:

S=\iint dS=R^{2}\int_{0}^{2\pi}d\varphi \int_{0}^{\pi} d\vartheta \ \sin\vartheta

I was thinking about using shell integration for it, but as I recall, shell integration and solids of revolution deal only in volumes, not surface areas (This was by far my weakest area of Calc II, FWIW).

I'm going to be doing double integrals fairly soon, but I wanted to know if there was a more simplistic method so I wouldn't have to wait until then.

 

[Vorde]

You can do surface areas in single variable calculus, but they are just as ugly as they are in multivariable. The reason is that instead of using dx you have to use ds where ds = arc length = $\sqrt{1+ (\frac{dy}{dx})^2} dx$

The general formula for surface area is $SA = \int 2 \pi y ds$. So a formula for surface area of a unit sphere would be $\int_{-1}^{1}{2 \pi y \sqrt{1+ (\frac{dy}{dx})^2}} dx$ where y is the half circle of radius 1.

 

[SteamKing]

You might try the First Theorem of Pappus applied to a semi-circular arc.

 

[Vorde]

You might try the First Theorem of Pappus applied to a semi-circular arc.

Wow, that's a super cool theorem. Never knew about that.

 

[Vargo]

Pappus' theorem is a cool way.

Another way goes like this. For a sphere, let V(r) be the volume, and S(r) be surface area (as a function of radius).

Then V(r) = V(1)r^3 and V'(r) = S(r). Therefore, S(r) = 3V(1)r^2.

So if you can calculate the volume of a sphere (via the method of disks for example), then you can get the surface area this way.