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[MUSIC].

Let's compute the volume of a bead but first we have to figure out how we're

going to build the bead as a solid of revolution.

So start with the X, Y plane and I'm going to draw a circle of radius two in

the plane. So, its a circle of radius 2 if I take

this and rotate around the Y axis and I got a sphere, but I want to bead, so I am

going to imagine drilling out a hole around the Y axis.

So to do that we can draw these vertical lines here.

This is the line x equals 1. This is the line, x equals minus 1.

[NOISE] Okay, now if I just take the the region in here, and rotate that region

around the y axis, that gives me a bead. It's the same thing as giving me a

sphere. With a hole drilled out around the y

axis. We've got a choice.

We can attack this problem with shells or with washers.

If I want to do it with washers, since I'm rotating around the y axis, I'd be

cutting this into horizontal strips. Because if I take one of these horizontal

strips and rotate it around the y axis, that gives me a washer, and that looks

like here is my here's my big washer. And that's, that works.

There's nothing wrong with that, but it turns out that if you do it the way the

interval that you get is kind of yucky. But let's try it instead with shells.

If I go with shells, then I'd be cutting into vertical strips.

Parallel to the axis of rotation. And when I take one of those little

vertical strips and rotate it around. Well then exactly what I've got is, is

one of these shell pieces. Now we invoke the formula for volume of a

shell in terms of dx. So the formula is 2 pi, the radius of the

shell. Which if I think of just one of these

pieces say at x, the radius is x, since that's how far it is from the axis of

rotation. The height of the shell, which I haven't

figured out yet, times the thickness of the shell, which is then dx.

Now I need to determine the height of the shell.

So that big sphere has radius 2. And this orange curve is that circle with

radius 2. So I can write down that that orange

curve is y equals the square root of 2 squared minus x squared.

And, that's almost enough information to tell me the height of the shell.

Let me write down what the height of this shell is, then, the height of the shell

is well, how tall is this thing. Well here, from here to here, is the

square root of 2 squared minus x squared, and then it's the same distance, down to

the other side here. So, the height, is twice this quantity.

So, I can write that down; 2 Pi X times the height of the shell, which is 2 times

the square root of 2 squared, which is 4 minus X squared.

That's the height of the shell, and then the thickness of the shell dx.

Let's think about the endpoints of integration.

This purple line, right, was the line x equals 1.

And the orange circle, right, has radius 2.

So, x can take values between 1 and 2, and that tells me what my end points of

integration should be. So, this is the volume of just one of

those shells, and I'm going to integrate that x goes from 1 to 2, to get all of

the shells, and add them all up. Finally, we integrate it's 2 pi times the

integral from 1 to 2, and I'll write 2x, the square root of 4 minus x squared dx.

And this is really where shells shine, I set this up now, and you can see I can

make a u substitution here. So I'll set u equal 4 minus x squared, du

in that case is minus 2x dx, put a minus sign there and a minus sign there.

Now I've got du square root of u, so this integral becomes minus 2 pi.

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Square of u d u and I can change the endpoints, u goes from when x is 1, u is

3. When x is 2, u is 0, and this is negative

from 3 to 0, so I can rewrite this as 2 pi the integral from 0 to 3 of the square

root of u du. Now, that's not hard to do.

I can rewrite that integrand as 2 pi the interval of u to the 1/2 du from 0 to 3.

And I can easily anti differentiate that using the power rule.

So this is 2 pi, u to the 3 halves over 3 halves evaluated at 0 and 3.

Now, when I plug in zero I just get zero, so my answer is whatever I get when I

plug in three. So the volume is 2 pi times 3 to the 3

halves power over 3 halves. Now, instead of writing 3 to the 3 halves

power, I could write this as 2 pi over 3 halves times 3 times the square root of

3. Right, that's the same as 3 to the 3

halves power. But look, this 3 and this 3 cancel and

I'm dividing by a half here, so this ends up being 4 pi times the square root of 3.

This is the volume of our bead. We did it, but is the answer reasonable?

Well to think about how reasonable our calculation for the volume of the bead

was. We could think about how we build the

bead, alright we build this thing by starting with the sphere of radius two

and then drilling out a tube of radius one.

So if I start with the sphere of radius two and I subtract even more volume

right, this cylinder of height four extends above the top of this sphere and

below the bottom of the sphere. So this is even more material than I

removed to get the bead. Well that means that the volume of this

sphere minus the volume of the cylinder's got to be even a little bit less than the

volume of the bead, right? Because this is a little bit more,

material than the material that we removed to get the bead.

Well how big are these three things? This thing here is a 4 3rds pie radius

cubed thats the volume of a sphere minus this thing here is pie r squared times

height is the volume of a cylinder. I could simplify this a bit this is a 4

times 8 so that's 32 3rd pi minus 4 pi, and I can simplify that a little bit

further even, that's 20 3rd pi. And what do we get to the volume of the

big bead we calculated that to be 4 pi square root of 3.

And yeah, this quanitiy here is just a bit smaller than this quantity here.

So, it seems like our answer's reasonable.