### Video Transcript

Determine the volume of the solid
generated by rotating the region bounded by the curve π¦ equals seven π₯ squared and
the line π¦ equals seven π₯ a complete revolution around the π₯-axis.

So what Iβve done to actually help
us visualize this question is Iβve actually sketched the first quadrant of both
graphs. So what we actually have here is a
graph of π¦ equals seven π₯ and also a graph of π¦ equals seven π₯ squared. So what weβre actually looking for
in this question is actually to determine the volume of the solid that we
generated.

So what weβre actually looking for
in this question is to determine the volume of the solid generated by rotating the
region bounded by our curve and our line one complete revolution around the
π₯-axis. So what Iβve actually done is
marked a very small section here so that we can take a look at it a section at a
time.

So Iβve actually taken the small
section that Iβve marked and actually rotated it around the π₯-axis would end up
with this shape that looks a bit like this. So actually, we can say that it
looks a bit like a washer. And thatβs why the method weβre
gonna use is actually called the washer method.

So if you actually take a look at
this region and the little sketch that Iβve got here is what it actually have is a
washer, which is a circle with a circular region cut from its centre. And as we actually want to find the
volume, the first thing we want to do is actually find the cross-sectional area.

So to find the area of this shape,
what we actually have are two radiuses: the radius of the full circle and the radius
of the small inner circle. So therefore, we can say that the
area of the washer is gonna be equal to the area of the larger circle, so ππ
squared with a capital π
, minus the area of the inner circle, so ππ squared with
a small π.

So we can actually see how this
applies to our question. And we can take a look again at the
graph. And what weβve got is our small π
is gonna be equal to seven π₯ squared because this is where the actual washer would
actually hit the curve. And we also know that the big π
is
actually gonna be equal to seven π₯ cause we can see that the outer radius is
actually where it touches the line π¦ equals seven π₯.

But this is all very useful. But however, what weβre actually
trying to find in this question is the volume. So how do we get about finding
that? Well, if we actually take a small
section of our washer and we look at it from the side, then what we can see is
obviously we found the cross-sectional area or a formula for that.

Well, what we need is the depth, if
weβre trying to work out the volume. And the depth is just gonna be ππ₯
because what ππ₯ means is a very small amount or a small change in π₯. So therefore, our volume is
actually gonna be equal to π΄ because thatβs our cross-sectional area of the washer
multiplied by ππ₯ because as we said thatβs the small depth that weβve seen in our
diagram.

So therefore, if we actually want
to think about this volume in terms of actually the functions that weβre gonna be
using, we can say that the π, our volume, is equal to π and then π multiplied by
π π₯ squared, so one of our functions squared that would be this function with the
big π
, minus ππ₯ squared, which will be our function with the small π, multiplied
by ππ₯.

This is great because actually what
weβve done here is found volume. However, itβs still only the volume
of very small section of our washer. And we want to find the volume of
the whole region. Well, actually, it leads us onto
this formula here, which is actually the volume formula for the whole region.

So we can say that the π is equal
to π multiplied by then the definite integral between the limits π and π of ππ₯
squared minus ππ₯ squared ππ₯. And we get this because actually
what it tells us is that itβs actually the sum of an infinite number of ππ₯ squared
minus ππ₯ squared ππ₯ is between the limits π and π. And it has to be an infinite number
because actually if we had a finite number, then itβll just be an estimate. But while weβre using a definite
integral, we find the exact volume of the whole region.

Okay, so fab, weβve actually got
this and weβve also shown exactly where it comes from. So letβs get on and actually use it
to solve our problem. So first of all, what we actually
need to do is find the limits. And to do this, we actually need to
see where both of our functions meet. I will sketch where this point was
on the graph because we said at one. But letβs see how weβd actually
work it out.

So weβd actually put our functions
equal to each other. So we have seven π₯ squared equals
seven π₯. So first of all, what we do is we
actually subtract seven π₯ from each side. So we get seven π₯ squared minus
seven π₯ equals zero. And then, we actually divide each
side by seven. And we do this so that itβs gonna
be easier to factor. So when we divide both sides by
seven, we get π₯ squared minus π₯ is equal to zero.

And next, we actually factor to
find π₯. So we take π₯ out as a factor cause
itβs a factor of both terms. So we have π₯ multiplied by π₯
minus one is equal to zero. So therefore, we know now that our
limits are gonna be equal to zero or one. And this is because if we had π₯
equal to zero, then zero multiplied by anything is equal to zero. So that satisfies the equation. And if we had π₯ is equal to one,
then inside the parentheses, weβd have one minus one, which gave us zero. So again, we give our answer as
zero.

Okay, so great, so now we can
actually say that our volume is gonna be equal to π multiplied by the definite
integral between the limits one and zero. But then, which one of our
functions is going to come first when weβre looking for the definite integral? Well, to help us to decide, weβre
actually gonna look at something we looked at earlier, which was that π β so our
cross-sectional area β is equal to π big π
squared minus π small π squared.

So therefore, we can say that our
definite integral is gonna be of seven π₯ all squared because seven π₯ was our big
π
then minus seven π₯ squared all squared because that was our small π and then
ππ₯. So great, we actually got this sort
of formula here. But how do we actually work out the
value of it? How do we work out the value of a
definite integral?

Okay, so well to find this, we got
this general rule, which tells us that if weβre trying to find the value of a
definite integral between the limits π and π, then what we do is we actually
integrate our function. So we get capital πΉ π₯, which Iβll
put here to denote that. And then, we substitute in π,
which is our upper limit for our π₯-values. And then, what we do is to find the
value of that and then we subtract from it the value of our integral with π
substituted in, which is our lower limit instead of our π₯-values.

Okay, great, now we know what to
do. Letβs get on and do it and find our
volume. So now, what we need to do is
actually integrate our function that we have here. But before we do that, what I need
to do is actually expand the parentheses just to make it easier to do.

So weβve got volume is equal to π
multiplied by the definite integral between the limits one and zero of 49π₯ squared
because seven π₯ squared is 49π₯ squared minus 49π₯ to the power of four ππ₯. So then, when we actually integrate
that, what weβre gonna get is our volume is equal to π multiplied by then the
integral, which is 49π₯ cubed, over three minus 49π₯ to the power of five over five
and again between our limits one and zero.

And just a quick recap of how we
actually integrated, we look at our first term and see how we integrated that. But what we do is we have 49 then
multiplied by π₯. And what we do is we increase the
exponent by one, so two plus one, so 49π₯ cubed. So that gave us the numerator. And what we do is we divide by the
new exponent. So we divide by two plus one. So we divide by three. So therefore, we get 49π₯ cubed
over three.

So now what we need to do is move
on to the next step. Well, the next step is actually
weβre gonna substitute our limits in for π₯. So this lead us to our next line,
which I wouldnβt actually normally write it out in full. But Iβve just done that just to
show exactly whatβs happening.

So weβve got π multiplied by then
we got 49 multiplied by one all cubed over three minus 49 multiplied by one to the
power of five over five and thatβs because thatβs our upper limit then subtract 49
times zero to the power of three over three minus 49 multiplied by zero to the power
of five over five because thatβs our lower limit. But of course, obviously, this
second parenthesis will actually gonna be equal to zero because of the fact that our
lower limit is zero.

So therefore, this is gonna leave
us with the volume is equal to π multiplied by 49 over three minus 49 over five
because obviously one cubed and one to the power of five just gives us one.

So now, our next step is to
actually make the denominators the same so that we can actually deal with these
fractions. So we can say that itβs equal to
245 over 15 minus 147 over 15 because obviously we multiplied the first fraction β
the numerator and the denominator β by five and the second fraction β numerator and
denominator β by three to make them both have a denominator of 15.

So therefore, what we can say is
that the volume of the solid generated by rotating the region bounded by the curve
π¦ equals seven π₯ squared and the line π¦ equals seven π₯ a complete revolution
around the π₯-axis is equal to 98 over 15π.