[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.
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.