### Video Transcript

The region bounded by the curves 𝑥
equals three root 𝑦, 𝑥 equals zero, and 𝑦 equals three is rotated about the
𝑦-axis. Find the volume of the resulting
solid.

So, the first thing I’ve done is
actually drawing a sketch of our function so that 𝑥 equals three root 𝑦. Okay, so now, what else can we put
on our sketch? So what I’ve now added to our
sketch is 𝑥 equals zero and 𝑦 equals three. And I’ve added these on because
actually these are two of the lines that our region is gonna be bounded by alongside
the curve.

Well, if we look back at the
question, the question is asking us to actually rotate our region about the
𝑦-axis. So if we do this, we’re actually
gonna create a 3D solid which is actually gonna look a little bit like this. Well, what I’ll do is to draw it
from another angle so we can really understand the shape that we’re looking for. So inside our sketch, we can see a
rough idea of what we’re looking at from the side.

Okay, great, so we now know what
we’re looking for. But how would we go about finding
it? Well, we’re gonna start by looking
at a small cross section of our region. And if we take this small cross
section and we rotate it like we are going to with the whole region, the more we
actually get is a disc which is why this is sometimes called the disc method.

Okay, so we’ve got our disc. Let’s think about actually what
dimensions of our disc are going to be. Well, first of all, we know that
our radius is gonna be equal to three root 𝑦. And we know that because our small
region that we’re looking at actually goes from the 𝑦-axis to the function
itself. So therefore, we know the distance
between those two is going to be three root 𝑦 or the value of the function at that
point.

Okay, great, so we’ve got our
radius, while our depth is going to be 𝑑𝑦 cause what it’s gonna be is a very small
amount of 𝑦. So we actually denote that using
𝑑𝑦. Okay, great, so now we’ve got our
dimensions, what we’re going do now? Well, first of all, we can actually
say “well, let’s work out the area of our cross section” because we’ve actually got
a circle here. So therefore, the area is gonna be
equal to 𝜋𝑟 squared which will give us 𝜋 multiplied by three root 𝑦 all squared
and this gives us 𝜋 times nine 𝑦.

Okay, great, we found the area. So now what we need to is actually
find the volume of our disc. So therefore, we can say that the
volume is just gonna be equal to 𝜋 nine 𝑦 multiplied by 𝑑𝑦. And that’s because the area of our
cross section is 𝜋 nine 𝑦 and our depth is 𝑑𝑦. So therefore, we found our
volume. Okay, great, but this is still just
a volume of our disc. How do we find the volume of the
whole region?

So in order to find the volume of
the whole solid, what we’re actually gonna use is the definite integral. And what definite integral does is
it gives us a sum of an infinite number of our small discs. And we’re saying “infinite number”
because if it was a finite number of discs, then it’ll just be an estimate. But because it’s an infinite
number, we know that we’re gonna get the full volume of our solid.

As you also may have noticed as
I’ve written now a definite integral, I’ve actually put the limits three and
zero. And that’s because these are the
limits of our region cause if you have a look it’s where 𝑦 is equal to three and 𝑦
is equal to zero, our region is bound between those points. So therefore, we can say that the
volume of the resulting solid is gonna be found using the definite integral of 𝜋
nine 𝑦 with the limits three and zero.

And just to remind us how we
actually find the value of a definite integral, we can say that if we have a
definite integral with the limits 𝑏 and 𝑎, then we can say that actually the value
of this is gonna be equal to our integral with the upper limit 𝑏 substituted in for
𝑥 minus the integral with the value of our lower limit 𝑎 substituted in for
𝑥.

Okay, great, so now we know how to
do that, let’s get on and find the value of our definite integral. The first thing I’ve actually done
is actually taking the 𝜋 outside of our integration. And that’s because 𝜋 won’t affect
the integration itself cause it’s just a coefficient. So then, we actually integrate our
function. So we integrate nine 𝑦 which gives
us nine 𝑦 squared over two.

And just to remind us how we
actually did that, we actually raised the exponent by one. So I added one to one which gave us
two. So we’ve got nine 𝑦 squared and
then we divide by the new exponent.

Okay, now, we move on to our next
step which is actually substituting in our upper and lower limits. Okay, so we’re gonna get 𝜋
multiplied by then we’ve got nine times three squared over two minus nine multiplied
by zero squared over two. I won’t necessarily usually write
this because we know the answer of this part is gonna be equal to zero. But I just put that in there to
show the complete method. So now, what we’re gonna do is
actually calculate this and find our value. And this is all equal to 81 over
two 𝜋.

So therefore, we can say that if
the region bounded by the curves 𝑥 equals three root 𝑦, 𝑥 equals zero, and 𝑦
equals three is rotated about the 𝑦-axis, the volume of the resulting solid is
going to be equal to 81 over two 𝜋.