[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.