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# Disk method for rotation around x-axis

Finding the volume of a solid of revolution that is defined between two functions. Created by Sal Khan.

Video transcript

Let's do some more volumes
of solids of revolutions. So let's say that
I have the graph y is equal to square root of x. So let's do it, so it
looks something like this. So that right over there is y is
equal to the square root of x. And let's say I also have
the graph of y equals x. So let's say y equals x
looks something like this. It looks just like that. y equals x. And what I care about
now is the solid I get if I were to rotate the
area between these two things around to the x-axis. So let's try to visualize it. So the outside is going to
be kind of a truffle shape, it's going to look
like a truffle shape. And then we hollow out
a cone inside of it. So I'll make my best
attempt to draw this shape. So it's going to look something
like-- so the outside is going to look
something like this. It's going to look
something like that. And we care about the interval. We care about the interval
between the points that they intersect. So between this point
and this point here. So the outside is going to
look something like this. So this is the base
of the truffle. That is the base of
the truffle, it's going to have this kind of
truffle shape on the outside. But I guess maybe we're
on some type of a diet, we don't want to eat
to the entire truffle. So we carve out a
cone on the inside. So the inside of
it is essentially hollow except for this
kind of shell part. So we carve out a
cone in the center. So we rotate it
around to the x-axis. Truffle on the outside, carved
out a cone on the inside. So what's going to be
the volume of that thing? So it's essentially, if we
take a slice of our figure, it's going to be, this
is going to be the wall. And we're essentially
going to take the volume of this
entire wall that we're rotating around the x-axis. So how do we do that? Well, it might dawn
on you that if we found the volume of the
truffle if it was not carved out, and then
subtract from that the volume of the cone, we
would essentially find out the volume of the
space in between the outside of the truffle and the
cone part of the truffle. So how would we do that? Well, so to find the
volume of the outer shape. So let me draw it over here. So actually let me
just draw it over here. If we think about the
volume of the outer shape, once again, we can
use the disk method. So at any given point in time
our radius for one of our disks is going to be equal
to the function. Let's rotate that disk around. Actually let me do it
in a different color, it's hard to see that
disk because it's in the same magenta. So this is a radius, and
let's rotate it around. So I'm rotating the disk around. This is our face of the disk,
that's our face of the disk. It's going to have
a depth of dx. We've seen this multiple times. It's going to have
a depth of dx. So the volume of
this disk is going to be our depth, dx, times
the area of the face. The area of the face is going to
be pi times the radius squared. The radius is going to
be equal to the value of the outer function, in this
case the square root of x. So it's going to be pi
times our radius squared, which is pi times the
square root of x squared. And so if we want
to find the volume of the entire outer
thimble, or truffle, or whatever we want to call it. Before we even carve
out the center, we just take a sum of a bunch
of these, a bunch of these disks that we've created. So that's one disk, we would
have another disk over here. Another disk over here. For each x we have another disk. And as we go, as x's
get larger and larger, the disks have a larger
and larger radius. So we're going to sum
up all of those disks. And we're going to take the
limit as each of those disks get infinitely thin, and we
have an infinite number of them. But we have to figure out our
boundaries of integration. So what are our
boundaries of integration? What are the two points right
over here where they intersect? Well we could just
set these two things to be equal to each other. If you just said x is
equal to square root of x, when does x equal
the square root of x? I mean you can square
both sides of this. You could say when
does x squared equal x? You could, well we
could keep it there. You could kind of
solve this, there's multiple ways you could do it. But you could solve this kind
of just thinking about it. If x is equal to 0, x
squared is equal to x. And you see that on the
graph right over here. x is equal to 0. And also 1 squared
is equal to 1. 1 is equal to the
square root of 1. You could have
done other things. You could say OK, x squared
minus x is equal to 0. You could factor out an x. You get x times x
minus 1 is equal to 0. And so either one of
these could be equal to 0, so x is equal to 0 or x
minus 1 is equal to 0. And then you get x equals
0 or x is equal to 1. Which gives us our
boundaries of integration. So this goes from x
equals 0 to x equals 1. And so for the
outside of our shape we can now figure
out the volume. But we're not done. We also need to
figure out the volume of the inside of our shape
that we're going to take out. So we're going to
subtract out that volume. So we're going to
subtract out of volume. Our x values, once again,
are going between 0 and 1. And so let's think
about those disks. So let's think about,
let's construct a disk on the inside, right over here. So if I construct a
disk on the inside. So now I'm carving out
the cone part of it. What is the area of the
face of one of those disks? Well it's going to be pi
times the radius squared. In this case, the
radius is going to be equal to the value of
this inner function, which is just x. And so this is just
y is equal to x. And then we're going to
multiply it times the depth, times the depth of
each of these disks. And each of these disks are
going to have a depth of dx. If you imagine a quarter that
has an infinitely-thin depth right over here. So it's going to be dx. And so the volume our, kind of
our truffle with a cone carved out, is going to
be this integral minus this integral
right over here. And we could evaluate
it just like that. Or we could even say,
OK we could factor out a pi out of both of them. There are actually,
there's multiple ways that we could write it. But let's just
evaluate it like this, and then I'll generalize
it in the next video. So this is going to be equal
to the definite integral from 0 to 1. You take the pi outside. Square root of x
squared is going to be x dx minus the integral,
we can factor the pi out. From 0 to 1 of x squared dx. And we could say
this is going to be equal to pi times the
antiderivative of x, which is just x squared over
2 evaluated from 0 to 1, minus pi, times the
antiderivative of x squared, which is x to the third over
3 evaluated from 0 to 1. This expression is
equal to-- and I'm going to arbitrarily
switch colors just because the green's getting
monotonous-- pi times 1 squared over 2 minus 0 squared over 2. I could write squared. 1 squared over 2 minus 0
squared over 2, minus pi times 1 to the third over 3 minus
0 to the third over 3. And so we get, this
is equal to-- let me do it in that
same blue color-- so this is this simplified. This is just 0 right over here. This is 1 squared over
2, which is just 1/2. So it's just pi over 2, 1/2
times pi minus-- well this is just 0, this is
1/3, minus pi over 3. And then to simplify
this, it's just really subtracting fractions. So we can find a
common denominator. Common denominator is 6. This is going to be 3 pi over 6. This is 3 pi over 6
minus 2 pi over 6. pi over 3 is 2 pi over 6,
pi over 2 is 3 pi over 6. And we end up with, we
end up with 3 of something minus 2 of something, you
end up with 1 of something. We end up with 1 pi over 6. And we are done. We were able to find the
volume of that wacky kind of gutted-out truffle.