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,
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,
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
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.