Video Transcript
Find the volume of the region
formed by the curve 𝑥 equals the square root of 𝑦, the 𝑥-axis, and the line 𝑥
equals two when revolved around the 𝑦-axis.
To start, let’s sketch what we
know. On our grid, we have a curve for 𝑥
equals the square root of 𝑦. But in order to make the sketch, I
square both sides of the equation. The square root of 𝑦 squared just
equals 𝑦. So we know that we’re looking at a
curve that looks something like this. It’s also bound by the 𝑥-axis and
the line 𝑥 equals two. And the space between these three
lines is the region that will be revolved around the 𝑦-axis.
If we sketch a 3D representation of
what would happen, we would get something like this, where this pink space is going
to represent the volume of our region 𝑅 revolved around the 𝑦-axis. The curve 𝑦 equals 𝑥 squared
creates a cavity inside a cylinder. We’ll need to take the volume of
this cylinder, subtract the volume of the cavity in the center, which will give us
the volume of our region 𝑅 revolved around the 𝑦-axis. But how should we go about doing
that?
To solve this, we’ll use something
called the disc method or sometimes called the ring method. We can find the volume of a solid
by taking the definite integral from 𝑎 to 𝑏 of 𝑎 of 𝑦 with respect to 𝑦. In this case, 𝑎 of 𝑦 is the
cross-sectional area. We’re going to take the area with
respect to 𝑦 of this region because we’re revolving the region around the
𝑦-axis. Since we want to find first the
area of the larger cylinder, we consider a cross-section, which equals 𝜋 times the
radius squared. This is the line 𝑥 equals two. And that means, at all points, the
radius is going to be two.
In order to find the volume, we’ll
need to integrate from 𝑎 to 𝑏 𝜋 times the radius two squared with respect to
𝑦. And what should be our definite
integral? We know the shape is bound by the
𝑥-axis. So we’ll start at zero. And the height of the cylinder can
be found by finding the intersection of 𝑥 equals two and 𝑥 equals the square root
of 𝑦. If we plug in two for 𝑥, square
both sides of the equation, we’ll see that 𝑦 equals four at this intersection. That intersection is two, four. And we know that the height here is
from zero to four. And so we’ll integrate from zero to
four.
We know that two squared equals
four. And that means we’ll need to
integrate four 𝜋 with respect to 𝑦. The integral becomes four 𝜋 𝑦,
from four to zero. And so our volume will be four
times 𝜋 times four minus zero. The volume is 16𝜋. Remember, this is the volume for
our larger cylinder. And we need to repeat this process
with that inside cavity.
Here’s our cavity shape. We can go ahead and say that our
definite integral will be the same for both of these pieces. So we’re going to integrate from
zero to four if we consider a cross-sectional area of this shape. And the radius is going to be the
distance from the 𝑦-axis to our function. That means the radius will be equal
to the square root of 𝑦. And we need to square that value
and then integrate with respect to 𝑦.
The volume for our interior cavity
can be found by taking the definite integral from zero to four of 𝜋 times the
square root of 𝑦 squared with respect to 𝑦. The square root of 𝑦 squared is
just 𝑦. We need to take the integral of 𝜋
times 𝑦, which will be 𝜋 times 𝑦 squared divided by two, from zero to four. That volume will be equal to 𝜋
times four squared divided by two minus zero. Four squared equals 16. 16 divided by two equals eight. The volume is eight 𝜋. The interior cavity’s volume would
be eight 𝜋. And that means the volume of the
region we’re interested in will be 16𝜋 minus eight 𝜋, which is eight 𝜋.
The volume of the region formed by
the curve 𝑥 equals the square root of 𝑦, the 𝑥-axis, and the line 𝑥 equals two
when revolved around the 𝑦-axis is eight 𝜋.
