But why is a sphere's surface area four times its shadow?

Some of you may have seen in school that the surface area of a sphere is 4 pi R squared,

a suspiciously suggestive formula given that it's a clean multiple of the more

popular pi R squared, the area of a circle with the same radius.

But have you ever wondered why this is true?

And I don't just mean proving the 4 pi R squared formula,

I mean viscerally feeling to your bones a connection between this surface area

and these four circles.

How lovely would it be if there were some subtle shift in perspective that shows

how you could nicely and perfectly fit these four circles onto the sphere's surface?

Nothing can be quite that simple since the curvature of a sphere's

surface is different from the curvature of a flat plane,

which is why trying to fit, say, a piece of paper around the sphere just doesn't work.

Nevertheless, I would like to show you two separate ways of thinking about

the surface area that connect it in a satisfying way to these circles.

The first comes from a classic, one of the true gems of geometry

that I think all math students should experience the same way

all English students should read at least some Shakespeare.

The second line of reasoning is something of my own,

which draws a more direct line between the sphere and its shadow.

And lastly, I'll share why this fourfold relation is not unique to spheres,

but is instead one specific instance of a much more general

fact for all convex shapes in three dimensions.

Starting with the bird's eye view here, the idea for the first approach is to

show that the surface area of the sphere is the same as the area of a cylinder

with the same radius and the same height as that sphere, or rather,

a cylinder without the top and bottom, what you might call the label of that cylinder.

And once you have that, you can unwrap that label to understand it simply as a rectangle.

The width of this rectangle comes from the cylinder's circumference,

so it's 2 pi times R, and the height comes from the height of the sphere,

which is 2 times the radius.

And this already gives us the formula, 4 pi R squared when we multiply the two.

But in the spirit of mathematical playfulness,

it's nice to see how four circles with radius R can actually fit into this picture.

The idea will be to unwrap each circle into a triangle without changing its area,

and then to fit those nicely into the unfolded cylinder label.

More on that in a couple minutes.

The more pressing question is, why on earth should

the sphere be related to the cylinder in this way?

The way I'm animating it is already suggestive of how this might work.

The idea is to approximate the area of the sphere with many tiny rectangles covering it,

and to show how if you project these rectangles directly outward,

as if casting a shadow by little lights positioned on the z-axis,

pointing parallel to the xy-plane, the projection of each rectangle onto the cylinder,

surprisingly, ends up having the same area as the original rectangle.

But why should that be?

There are two competing effects at play here.

For one of these rectangles, let's call the side along the latitude lines its width,

and the side along the longitude lines its height.