[MUSIC].
We've done a bunch of area calculations at this point, so let's move up a
dimension and think about volume. Let me ask really loudly.
What is volume? Of course, what is area?
Right? Area is the thing that integral's
calculate. So instead of trying to give a definition
of volume, lets just focus on how were going to compute volume, whatever it is.
And, were going to comput volume by integrating the volume of little tiny
pieces that we cut up our objects into. Let's work through an example.
Well, here's a cone, I want to compute the volume of this cone.
And just for concreteness, let's suppose that the cone has a height of 2 units,
and a radius of 1 unit, and I want to compute its volume.
I want to set this thing up as an integration problem, I want to cut this
thing up into slices, and add up the volumes of the little slices.
So, imagine that I take this, and I just slice it up into a bunch of really thin
pieces. Now, what's the volume of a slice?
Well, that slice, is practically a little tiny cylinder.
So what's the volume of one of these very thin cylinders?
Well let's suppose the cylinder has some radius r, and I'm going to call its
height, an its height should be really thin, so I'll call its height dx.
And in that case the volume of the cylinder is, what's the area of the top?
It's phi r squared times its height, which is dx.
So this here will be our formula for the volume of one of these thin cylinders.
And once again, we're seeing the importance of the differentials.
Right? We're using the differentials to write
down an equation for the volume of the thin cylinder.
Well any how, lets put this together in interval.
But to set up the integral to calculate the volume of this cone, I first have to
figure out the volume of one of the slices that make up this cone.
To do that, let's turn this cone on it's side, and let's imagine that this is the
point 0, 0. So let's make this the Y axis, and the X
axis will head in that direction. Then, what do I have here?
Well, this line, is the line y equals x over 2.
Now, I'll use that to figure out the radius of one of those slices.
Pick some value of X, and let's think about one of the slices, one of these
thin cylinders that I'm using to build up the big cone.
I don't know the radius of that that particular cylinder.
Well, it's this distance here, and I know the Y coordinate up there, the Y
coordinate is x over 2, so the radius of this cylinder is x over 2.
I can write down the volume of that disk. The volume's Phi times the radius, which
is x over 2 square, times the thickness of that disk, which is DX.
And now I'm integrating this from where to where?
Well, X goes from 0 all the way down to 2.
So this'll be the integral from x equals 0 to 2.
And this integral will calculate the volume of my cone.
And now we use the fundamental theorem of calculus.
I can simplify this a bit, I've got, some constants here I can pull out.
Right? I can pull out the phi, and I can pull
out 1/2 squared. So this is phi over 4 times the integral
from 0 to 2, of just x squared dx. Now what's an anti-derivative for x
squared? Well, x to the 3rd over 3, and then I'm
integrating that, I'm evaluating that at 0 and at 2, and taking the difference.
So when I plug in zero don't get anything, but when I plug in 2, I get 8,
that's 2 cubed over 3, and then minus 0, so this is the volume of the cone.
Which, I can simplified somewhat, I can write this as 2 times phi over 3.
We did it! Well, I mean, you could have just looked
up this formula at the back of some calculus textbook.
But that's beside the point. Right?
The point here, is that we're teaching you the secrets of volume.
You can now verify that the formulas in the back of the textbook are correct, and
that's part of the fun of mathematics. We're not just giving you information,
we're giving you information, and the tools with which to verify that that
information is correct.