Name: The Napkin Ring Problem
Uploaded: 2020-03-28T10:17:49.000Z
Duration: 10 min 44 s
Description: Thousands of YouTube videos with English-Chinese subtitles! Now you can learn to understand native speakers, expand your vocabulary, and improve your pronunciation...

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Hey, Vsauce! Michael here! If you core a sphere; that is, remove a cylinder from it, you'll be left with a shape

called a Napkin ring because, well, it looks like a napkin ring!

It's a bizarre shape because if two Napkin rings have the same height, well

they'll have the same volume regardless of the size of the spheres they came from! (Cool)

This means that if you cut equally tall napkin rings from an orange and from the Earth,

well, one could be held in your hand. The other would have the circumference of our entire planet,

I mentioned this counterintuitive fact while making a Kendama with Adam Savage. Check that video out if you haven't, or better yet,

just come see us. We're bringing brain candy live to 24 new cities this fall. It's going to be busy, but right now

We're talking about balls and coring them!

I have here TWO napkin rings from very differently sized spheres; one is from a tiny ball,

just a little tomato that I've cored, so it's got a little hole in it right there.

The other Napkin ring is made from an orange,

but both Napkin rings have the same height. The tomato has a smaller

circumference than the orange which means less volume, but its ring is thicker which means more volume. Both of those effects

exactly cancel out. So these two napkin rings have identical volumes they take up the same amount of space

To see why the napkin ring problem is true,

let's discuss Cavalieri's principle. It states that for any two solids like these two cylinders I've built here

sandwiched between parallel planes if any other parallel plane

Intersects both in regions of equal area no matter where it's taken from

Well then the solids have the same volume. That's clearly true here these cylinders are built out of stacks of VSauce stickers

100 in each Stack so their volumes are the same

Skew one of them like this it shape will change, but it's volume hasn't it's still contained the same amount of stuff

And Cavalieri's principle ensures that they still have the same volume because any cross-section taken from down here up here in the middle anywhere

Will always give us a region of the same area as the other because those regions are always equal area circles.

Now let's Apply Cavalieri's principle to Napkin Rings

We can see that two napkin Rings with similar Heights have

identical volumes by showing that when cut by a plane the area of one's cross section

Always equals the area of the others now to do this

Notice that the area of the spheres cross section minus the area of the cylinders cross section gives us the area of the Napkin rings

Cross section. Depending on where we slice the Napkin ring the cross sections will have different areas

But they will always be the same as each other. Let's calculate the areas of these blue rings

first of all let's call the height of the Napkin ring h and the radius of the sphere they're cut from

Alright, perfect now a cross section of a sphere like this and a cross section of a cylinder like this are both circles

So their areas can be determined by using Pi times the radius squared

So if we want to find the area of the spheres cross section and subtract the area of the cylinders cross section

(I'll draw a picture of a cylinder here), all we need to do is take Pi

multiply it by the radius of the sphere cross section

Square that and then subtract Pi times the radius of the cylinder Squared, but what are their Radii?

Well, if this is the center of the sphere we can draw a line straight up to the corner of the cylinder down the side

of the cylinder and then connect to form a right triangle

The Pythagorean Theorem will really help us here it tells us that the length of one side squared plus the length of the other side

squared equals the length of the hypotenuse squared

Now this distance right here this side of the triangle what we want

It's the radius of the cylinder, so we'll call this the little r radius of the cylinder (beautiful little picture there)

this side length, which is just half the height of the cylinder, so the height of the cylinder divided by 2

Equals the hypotenuse squared the hypotenuse happens to be the radius of the sphere itself which is capital R

Perfect now let's solve for the radius of the cylinder. Which is what we want

We'll just subtract h over 2 squared from both sides that'll give us the radius of the cylinder

equaling the radius of the sphere squared minus

We can take the square root of both sides so that we wind up with the radius of the cylinder

equaling the square root of the radius of the sphere

minus 1/2 the height of the cylinder squared

Perfect ok now let's take a look at the area of a cross-section of the sphere now for this

let's draw a straight line from the center out to the edge of

Cross-section, and we'll go straight down and connect back up, hey look! Another right triangle

And notice that this distance now the side of the triangle down here is actually the radius

Cross-section up here. They're both equal so we even want to solve for this the radius of the circle that is the spheres cross-section

The sphere's Cross-Section squared plus this distance squared (which is y)

Equals the hypotenuse squared well, what do you know the hypotenuse is the radius of the sphere again (capital R)

Ok let's subtract y squared from both sides the radius of the spheres cross-section squared

equals the radius of the sphere squared minus

Y squared will take the square root of both sides and end up

learning that the radius of the sphere's cross-section equals the square root of

The radius of the sphere squared minus y squared. Y is the height that this

Cross-section is taken from above the equator the higher up we take these cross sections of the sphere the smaller their radii will be

Whereas the cylinders radius is always the same no matter where we cut from

Anyway, let's take these two Radii and plug them into our formula

Okay, the area of the cross section of the sphere is what we want first

Okay, that's just the square root of R squared minus y squared

Not too bad now the radius of the cylinder is the square root of R squared minus

Half the height of the cylinder squared now what you might notice is that we're taking the square root of something and then squaring it

But now let's distribute Pi to the terms inside the parentheses so pi times R squared gives us Pi R

Squared Pi times negative y squared gives us negative pi y squared

squared is negative pi R squared negative Pi times negative h over 2 squared is positive Pi

Squared. Great, now we can keep simplifying but what you might notice is that we have a pi r squared and a

minus Pi R squared, well, that equals 0 so these

Completely cancel each other out, but what we're left with are

Terms containing no mention of the spheres Radius whether the radius is large or small

Doesn't matter all you need to know to find the area of the cross section of a napkin ring