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Current time:0:00Total duration:9:01

AP.CALC:

CHA‑5 (EU)

, CHA‑5.C (LO)

, CHA‑5.C.1 (EK)

Over here I've drawn
part of the graph of y is equal to x squared. And what we're going
to do is use our powers of definite integrals to find
volumes instead of just areas. So let's review what
we're doing when we take just a regular
definite integral. So if we take the definite
integral between, say, 0 and 2 of x squared dx,
what does that represent? Well, let's look
at our endpoints. So this is x is equal to 0. Let's say that this right
over here is x is equal to 2. What we're doing is
for each x, we're finding a little dx around
it-- so this right over here is a little dx. And we're multiplying that
dx times our function, times x squared. So what we're doing is we're
multiplying this width times this height right over here. The height right over
here is x squared. And we're getting the area
of this little rectangle. And the integral
sign is literally the sum of all of these
rectangles for all of the x's between x is equal to
0 and x is equal to 2. But the limit of that as these
dx's get smaller and smaller and smaller, get
infinitely small, but not being equal to 0. And we have an infinite
number of them. That's the whole power
of the definite integral. And so you can imagine, as these
dx's get smaller and smaller and smaller, these rectangles
get narrower and narrower and narrower, and we
have more of them, we are getting a better
and better approximation of the area under the
curve until, at the limit, we are getting the
area under the curve. Now we're going to apply
that same idea, not to find the area
under this curve, but to find the
volume if we were to rotate this curve
around the x-axis. So this is going to stretch our
powers of visualization here. So let's think
about what happens when we rotate this
thing around the x-axis. So if we were to rotate
it, and I'll look at it and say that we're looking it
a little bit from the right. So we get kind of a base that
looks something like this. So this is my best
attempt to draw it. So you have a base that
looks something like that. And then the rest
of the function, if we just think
about between 0 and 2, it looks like one
of those pieces from-- I don't know if you
ever played the game Sorry-- or it looks like a little
bit of a weird hat. So it looks like this, and let
me shade it in a little bit so it looks something like that. And just so that
we're making sure we can visualize this
thing that's being rotated. We care about the entire
volume of the thing. Let me draw it from a
few different angles. So if I drew from the top, it
would look something like this. It'll become a
little more obvious that it looks
something like a hat. It would point up like this,
and it goes down like that. It would look
something like that. So in this angle, we're not
seeing the bottom of it. And if you were to
just orient yourself, the axes, in this
case, look like this. So this is the y-axis. And the x-axis goes right
inside of this thing and then pops out
the other side. And if this thing
was transparent, then you could
see the back side. It would look
something like that. The x-axis, if you
could see through it, would pop the base
right over there, would go right through
the base right over there. And it'd come out
on the other side. So this is one orientation
for the same thing. You could visualize it
from different angles. So let's think about how we
can take the volume of it. Well, instead of thinking
about the area of each of these rectangles, what
happens if we rotate each of these rectangles
around the x-axis? So let's do it. So let's take each of these. Let's say you have this
dx right over here, and you rotate it
around the x-axis. So if you were rotate this
thing around the x-axis-- so I'm trying my best
to-- around the x-axis, you rotate it. What do you end up with? Well, you get something that
looks kind of like a coin, like a disk, like a
quarter of some kind. And let me draw it out here. So that same disk out here
would look something like this. And it has a depth of dx. So how can we find the
volume of that disk? Let me redraw it out here, too. It's really important to
visualize this stuff properly. So this is my x-axis. My disk looks
something like this. My best attempt at the x-axis
sits it right over there. It comes out of the center. And then this is the
surface of my disk. And then this right over
here is my depth dx. So that looks pretty good. And then let me just
shade it in a little bit to give you a little
bit of the depth. So how can we find
the volume of this? Well, like any disk
or cylinder, you just have to think about what
the area of this face is and then multiply
it times the depth. So what's the area of this base? Well, we know that
the area of a circle is equal to pi r squared. So if we know the
radius of this face, we can figure out
the area of the face. Well, what's the radius? Well, the radius
is just the height of that original rectangle. And for any x, the
height over here is going to be equal to f of x. And in this case, f
of x is x squared. So over here, our radius
is equal to x squared. So the area of the
face for a particular x is going to be equal to
pi times f of x squared. In this case, f
of x is x squared. Now, what's our
volume going to be? Well, our volume is going to be
our area times the depth here. It's going to be that
times the depth, times dx. So the volume of this thing
right over here-- so the volume just of this coin, I
guess you could call it, is going to be equal to. So my volume is going to
be equal to my area times dx, which is equal to pi
times x squared squared. So it's equal to pi--
x squared squared is x to the fourth--
pi x to the fourth dx. Now, this expression
right over here, this gave us the volume
just of one of those disks. But what we want is the
volume of this entire hat, or this entire bugle
or cone-looking, or I guess you could say the
front-of-a-trumpet-looking thing. So how could we do that? Well, the exact same technique. What happens if we were to take
the sum of all of these things? So let's do that, take the
sum of all of these things. And I'll switch to one color--
pi times x to the fourth dx. We're going take the sum
of all of these things from x is equal to 0 to 2. Those are the boundaries
that we started off with. I just defined them arbitrarily. We could do this, really,
for any two x values-- between x is equal
to 0 and x equals 2. And we're going to take the
sum of the volumes of all of these coins. But the limit-- as the depths
get smaller and smaller and smaller and we have
more and more and more coins, at the limit,
we're actually going to get the volume
of our cone or our bugle or whatever we want to call it. So if we just evaluate
this definite integral, we have our volume. So let's see if we can do that. And now this is just taking
a standard definite integral. So this is going to be
equal to-- and I encourage you to try it out
before I do it. So we can take the pi out. So it's going to be equal
to pi times the integral from 0 to 2 of x
to the fourth dx. I don't like that color. Now, the antiderivative
of x to the fourth is x to the fifth over 5. So this is going to be equal to
pi times x to the fifth over 5. And we're going
to go from 0 to 2. So this is going to be
equal to pi times this thing evaluated at 2. Let's see. 2 to the third is 8. 2 to the fourth is 16. 2 to the fifth is-- let
me just write it down. 2 to the fifth over 5 minus
0 to the fifth over 5. And this is going
to be equal to-- 2 to the fifth is 32, so it's
going to be equal to pi times 32/5 minus-- well,
this is just 0-- minus 0, which is
equal to 32 pi over 5. And we're done. We were able to
figure out the volume of this kind of wacky shape.