Some of you may have seen in school that the

surface area of a sphere is 4pi*R^2, a suspiciously suggestive formula given that it’s an clean

multiple of pi*R^2, the area of a circle with the same radius. But have you ever wondered

why is this true? And I don’t just mean proving this 4pi*R^2 formula, I mean viscerally

feeling a connection between this surface area, and these four circles. How lovely would it be if there was some shift

in perspective that showed 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 paper

around a sphere doesn’t really work. Nevertheless, I’d like to show you two ways of thinking

about this surface area connecting it in a satisfying way to these circles. The first

is a classic, one of the true gems of geometry all students should experience. 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 four-fold

relation is not unique to spheres, but is instead one specific instance of a much more

general fact for all convex shapes in 3d. Starting with a birds 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 the sphere. Or rather,

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

With that, we can unwrap that label to understand it as a simple rectangle. The width of this rectangle comes from the

cylinder’s circumference, so it’s 2*pi*R, and the height comes from the height of the

sphere, which is 2R. This already gives the formula, 4pi*R^2, but in the spirit of mathematical

playfulness it’s nice to see how four circles with radius R fit into this. The idea is that

you can unwrap each circle into a triangle, without changing its area, and fit these nicely

onto our unfolded cylinder label. More on that in a bit. The more pressing question is why on earth

the sphere can be related to the cylinder. This animation is already suggestive of how

this works. The idea is to approximate the area of the sphere with many tiny rectangles

covering it, and to show how if you project those little 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 on the cylinder, quite surprisingly, ends up

having the same area as the original rectangle. But why should that be? Well, 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. On the

one hand, as this rectangle is projected outward, its width will get scaled up. For rectangles

towards the poles, that length is scaled quite a bit, since they’re projected over a longer

distance. For those closer to the equator, less so But on the other hand, because these rectangles

are at a slant with respect to the z-direction, during this projection the height of each

such rectangle will get scaled down. Think about holding some flat object and looking

at its shadow. As you reorient that object, the shadow looks more or less squished for

some angles. Those rectangles towards the poles are quite slanted in this way, so their

height gets squished a lot. For those closer to the equator, less so. It will turn out that these two effects, of

stretching the width and squishing the height, cancel each other out perfectly. Already as a rough sketch, wouldn’t you

agree this is a very pretty way of reasoning? Of course, the meat here comes from showing

why these two competing effects on each rectangle cancel out perfectly. In some ways, the details

fleshing this out are just as pretty as the zoomed out structure of the full argument. Let me go ahead and cut away half the sphere

so we get get a better look. For any mathematical problem solving it never hurts to start by

giving names to things. Let’s say the radius of the sphere is R. Focus on one specific

rectangle, and let’s call the distance between our rectangle and the z-axis is d. You could

complain that this distance d is a little ambiguous depending on which point of the

rectangle you’re going from, but for tinier and tinier rectangles that ambiguity will

be negligible. And tinier and tinier is when this approximation-with-rectangles gets closer

to the true surface area anyway. To choose an arbitrary standard let’s say d is the

distance from the bottom of the rectangle. To think about projecting out to the cylinder,

picture two similar triangles. This first one shares its base with the base of the rectangle

on the sphere, and has a tip at the same height on the z-axis a distance d-away. The second

is a scaled up version of this, scaled so that it just barely reaches the cylinder,

meaning its long side now has length R. So the ratio of their bases, which is how much

our rectangle’s width gets stretched out, is R/d. What about the height? How precisely does

that get scaled down as we project? Again, let’s slice a cross section here. In fact,

why don’t we go ahead and completely focus our view to this 2d cross section. To think about the projection, let’s make

a little right triangle like this, where what was the height of our spherical rectangle

is the hypotenuse, and its projection is one of the legs. Pro tip, anytime you’re doing

geometry with circles or spheres, keep at the forefront of your mind that anything tangent

to the circle is perpendicular to the radius drawn to that point of tangency. It’s crazy

how helpful that one little fact can be. Once we draw that radial line, together with the

distance d we have another right triangle. Often in geometry, I like to imagine tweaking

the parameters of a setup and imagining how the relevant shapes change; this helps to

make guesses about what relations there are. In this case, you might predict that the two

triangles I’ve drawn are similar to each other, since their shapes change in concert

with each other. This is indeed true, but as always, don’t take my word for it, see

if you can justify this for yourself. Again, it never hurts to give more names to

things. Maybe call this angle alpha and this one beta. Since this is a right triangle,

you know that alpha + beta + 90 degrees=180 degrees. Now zoom in to our little triangle,

and see if we can figure out its angles. You have 90 degrees + beta + (some angle) forming

a straight line. So that little angle must be alpha. This lets us fill in a few more

values, revealing that this little triangle has the same angles, alpha and beta, as the

big one. So they are indeed similar. Deep in the weeds it’s sometimes easy to

forget why we’re doing this. We want to know how much the height of our sphere-rectangle

gets squished during this projection, which is the ratio of this hypotenuse to the leg

on the right. By the similarity with the big triangle, that ratio is R/d. So indeed, as this rectangle gets projected

outward onto the cylinder, the effect of stretching out the width is perfectly canceled out by

how much the height gets squished due to the slant. As a fun sidenote, you might notice that it

looks like the projected rectangle is a 90 degree rotation of the original. This would

not be true in general, but by a lovely coincidence, the way I’m parametrizing the sphere results

in rectangles where the ratio of the width the the height starts out as d to R. So for

this very specific case, rescaling the width by R/d and the height by d/R actually does

have the same effect as a 90 degree rotation. This lends itself to a rather bizarre way

to animate the relation, where instead of projecting each rectangular piece, you rotate

each one and rearrange them to make the cylinder. Now, if you’re really thinking critically,

you might still not be satisfied that this shows that the surface area of the sphere

equals the area of this cylinder label since these little rectangles only approximate the

relevant areas. Well, the idea is that this approximation gets closer and closer to the

true value for finer and finer coverings. Since for any specific covering, the sphere

rectangles have the same area as the cylinder rectangles, whatever values each of these

two series of approximations are approaching must actually be the same. I mean, as you get really aggressively philosophical

about what we even mean by surface area, these sorts of rectangular approximations and not

just aids in our problem-solving toolbox, they end up serving as a way of rigorously

defining the area of smooth curved surfaces. This kind of reasoning is essentially calculus,

just stated without any of the jargon. In fact, I think neat geometric arguments like

this, which require no background in calculus to understand, can serve as a great way to

tee things up for new calculus students so that they have the core ideas before seeing

the definitions which make them precise, rather than the other way around.

Unfold circle So as I said before, if you’re itching to

see a direct connection to four circles, one nice way is to unwrap these circles into triangles.

If this is something you haven’t seen before, I go into much more detail about why this

works in the first video of the calculus series. The basic idea is to relate thin concentric

rings of the circle with horizontal slices of this triangle. Because that circumference

of each such ring increases linearly in proportion to the radius, always 2pi times that radius,

when you unwrap them all and line them up, their ends will form a straight line, giving

you a triangle with a base of 2pi*R, and a height of R, as opposed to some other curved

shape. And four of these unwrapped circles fit into

our rectangle, which is in some sense an unwrapped version of the sphere’s surface.

Second proof Nevertheless, you might wonder if there’s

a way than this to relate the sphere directly to a circle with the same radius, rather than

going through this intermediary of the cylinder. I do have a proof for you to this effect,

leveraging a little trigonometry, though I have to admit I still think the comparison

to the cylinder wins out on elegance. I’m a big believer that the best way to

really learn math is to do problems yourself, which is a bit hypocritical coming from a

channel essentially consisting of lectures. So I’m going to try something a little different

here and present the proof as a heavily guided sequence of exercises. Yes, I know that’s

less fun and it means you have to pull out some paper to do some work, but I guarantee

you’ll get more out of it this way. The approach will be to cut the sphere into

many rings parallel to the xy plane, and to compare the area of these rings to the area

of their shadows on the xy plane. All the shadows of the rings from the northern hemisphere

make up a circle with the same radius as the sphere, right? The main idea will be to show

a correspondence between these ring shadows, and every other ring on the sphere. Challenge

mode here is to pause now and see if you can predict how that might go. We’ll label each one of these rings based

on the angle theta between a line from the sphere’s center to the ring and the z-axis.

So theta ranges from 0 to 180 degrees, which is to say from 0 to pi radians. And let’s

call the change in angle from one ring to the next d-theta, which means the thickness

of one of these rings with be the radius, R, times d-theta. Alright, structured exercise time. We’ll

ease in with a warm-up Question #1: What is the circumference of

this ring at the inner edge, in terms of R and theta? Go ahead and multiply your answer

the thickness R*d-theta to get an approximation for this ring’s area; and approximation

that gets better and better as you chop up the sphere more and more finely. At this point, if you know your calculus,

you could integrate. But our goal is not just to find the answer, it’s to feel the connection

between the sphere its shadow. So… Question #2: What is the area of the shadow

of one of these rings on the xy-plane? Again, expressed in terms of R, theta and d-theta. Question #3: Each of these ring shadows has

precisely half the area of one of these rings on the sphere. It’s not the one at angle

theta straight above it, but another one. Which one?

(As a hint, you might want to reference some trig identities)

Question #4: I said in the outset there is a correspondence between all the shadows from

the northern hemisphere, which make up a circle with radius R, and every other ring on the

sphere. Use your answer to the last question to spell out exactly what that correspondence

is. Question #5: Bring it on home, why does this

imply that the area of the circle is exactly ¼ the surface area of the sphere, particularly

as we consider thinner and thinner rings? If you want answers or hints, I’m quite

sure people in the comments and on reddit will have them waiting for you. And finally, I’d be remiss not to make a

brief mention of the fact that the surface area of a sphere is a specific instance of

a much more general fact: If you take any convex shape, and look at the average area

of all its shadows, averaged over all possible orientations in 3d space, the surface area

of the solid will be precisely 4 times that average shadow area. As to why this is true, I’ll leave those

details for another day. Hey, given the time of year I thought I’d

take a moment to let you know about some new additions to the 3blue1brown store. Aside

from the usual fare, like shirts, mugs and posters, there are now some Fourier Series

socks, which show certain periodic functions graphed on a cylinder, the way all periodic

functions wish they were graphed. And, by popular demand, there are the plushie

pi creatures, both ordinary and extra plushified. I’ll admit that I was initially skeptical

when people asked about them, because, you know, what would you do with one, exactly?

But after getting them, and seeing the pictures people would send, what I realized is they

basically serve the same function as a flag. Just instead of representing loyalty to a

country, or even to the channel per se, it’s to math, and the idea that math has some personality

to it, more so than it often gets credit for.