### Video Transcript

In this video, weβre going to learn
how to calculate the volume of a solid created by rotating a region between either a
curve and an axis or between two curves about an axis by 360 degrees. Weβll learn how to find these
volumes using calculus methods called the disc and washer methods. And itβs therefore important that
youβre confident in applying the various methods for integrating polynomial
functions before accessing this video.

Suppose we have a curve given by
the equation π¦ equals π of π₯. Now, imagine that we rotate the
part of the curve between the vertical lines π₯ equals π and π₯ equals π about the
π₯-axis by 360 degrees. The curve would map out the surface
of a solid as it rotated. In this case, it might look a
little like the curved surface of a vase, for instance. The solid created is called a solid
of revolution, for fairly obvious reasons. And so, if the function π¦ equals
π of π₯ gave us a horizontal line, the solid would be a cylinder. If the function gave us a
semicircle, then the solid would be a sphere.

Now, we know how to find the
volumes of these three-dimensional shapes. But how do we calculate the volume
of any solid of revolution? We could estimate the volume by
splitting the region up in two cylinders or discs. This is a little bit like a
three-dimensional version of finding a right Riemann sum. We could, for example, split the
region into four subintervals and find the volume of each cylinder created. And, of course, the formula weβd
use here is the formula for the volume of a cylinder. Itβs the area of its cross section
multiplied by its perpendicular length.

Weβll let the function π΄ of π₯ be
the area of the cross section of each cylinder, and the width of each of our
subintervals, and therefore the length of each cylinder, can be Ξπ₯. This means an estimate for the
total volume created by the solid of revolution is the sum of the volume of each of
our discs. Itβs the sum of π΄ π₯ π times Ξπ₯
for values of π between one and four. We can more generally say that for
a solid to split into π discs, its volume will be approximately equal to the sum of
π΄ π₯ π star times Ξπ₯ for values of π between one and π, where π₯ π star is a
sample point in the subinterval from π₯ π minus one to π₯ π.

Now, of course, this is only an
estimate. But it follows that as we increase
the number of subintervals, the cylinders get shorter and shorter, and the estimate
for the volume gets closer and closer to the actual volume of the solid of
revolution. As the number of subintervals
approaches infinty, the sum approaches the exact volume of the solid of
revolution. And we therefore define the volume
as the limit of the sums as π approaches β. By recognising the limit of a
Riemann sum as a definite integral, we obtain the following definition.

We take the solid bounded between
the lines π₯ equals π and π₯ equals π. If the cross-sectional area of this
solid in the plane through π₯ and perpendicular to the π₯-axis is π΄ of π₯ where π΄
is a continuous function, then the volume is equal to the limit as π approaches β
of the sum from π equals one to π of π΄ of π₯ π star times Ξπ₯. But thatβs also equal to the
definite integral between π and π of π΄ of π₯ with respect to π₯. Weβre now going to have a look at
the application of this formula.

Consider the region bounded by
the curves π¦ equals π₯ plus four, π¦ equals zero, π₯ equals zero, and π₯ equals
three. Determine the volume of the
solid of revolution created by rotating this region about the π₯-axis.

Remember, to find the volume of
a solid of revolution for a region rotated about the π₯-axis, we use the
following definition. For the solid that lies between
the vertical lines π₯ equals π and π₯ equals π, whose cross-sectional area in
the plane through π₯ and perpendicular to the π₯-axis is π΄ of π₯ for a
continuous function π΄, the volume of this solid is given by the definite
integral between π and π of π΄ of π₯ with respect to π₯.

So, weβll begin by defining the
various elements in our question. The solid is bounded by the
vertical lines π₯ equals zero and π₯ equals three, so weβre going to let π be
equal to zero and π be equal to three. The region is also bounded by
the curves π¦ equals π₯ plus four and π¦ equals zero. And so, the region itself might
look a little something like this. Rotating this region about the
π₯-axis, and we obtain a three-dimensional shape as shown. The cross-sectional shape of
our solid will be a circle. And the area of each circle
will be given by π times radius squared.

Now, the radius of each circle
will be the value of the function at that point. And so, the function for its
area, π΄ of π₯, is π times π₯ plus four all squared. Our volume is therefore equal
to the definite integral between zero and three of π times π₯ plus four all
squared with respect to π₯. Since π is a constant, we can
take this outside of our integral and rewrite the volume as π times the
definite integral between zero and three of π₯ plus four all squared dπ₯. Now, we have two choices
here. We could use integration by
substitution or we could distribute the parentheses to evaluate our definite
integral. Letβs use integration by
substitution.

We let π’ be equal to π₯ plus
four. Thatβs the inner part of our
composite function. Differentiating π’ with respect
to π₯, and we simply get one. Now, whilst dπ’ by dπ₯ is not a
fraction, we treat it a little like one. And this means we can
alternatively write this as dπ’ equals dπ₯. We then replace π₯ plus four
with π’ and dπ₯ with dπ’. We are going to need to do
something with our limits, though. We use our substitution; this
is π’ equals π₯ plus four. Our lower limit is when π₯ is
equal to zero. So, π’ is equal to zero plus
four, which is equal to four. Our upper limit is when π₯ is
equal to three. So, π’ is equal to three plus
four, which is seven.

And weβre now ready to evaluate
the volume, π times the definite integral between four and seven of π’ squared
with respect to π’. We know that we can integrate a
polynomial term whose exponent is not equal to negative one by adding one to the
exponent and then dividing by that new value. So, we have π times π’ cubed
over three between four and seven. Thatβs π times seven cubed
over three minus four cubed over three, which becomes π times 279 over
three. 279 divided by three is 93. And so, we find the volume
obtained when we rotate our region about the π₯-axis to be 93π cubic units.

Notice how we use the formula for
area of a circle to help us evaluate our volume. We said that the radius of each
circle was given by the value of the function at that point. So, for that reason, the formula
for the volume of a solid of revolution obtained by rotating an area about the
π₯-axis is sometimes written as the definite integral between π and π of π times
π¦ squared dπ₯. These definitions are
interchangeable, but the latter can be much nicer to work with. And interestingly, we can also use
a slightly amended version of this formula to help us calculate the area of a solid
of revolution obtained by rotating a curve about the π¦-axis.

This time we look to interchange
the roles of π₯ and π¦. We want to give the equation as π₯
equals π of π¦ rather than π¦ equals π of π₯. Similarly, the limits must be given
in terms of π¦ as π¦ equals π and π¦ equals π. Then, our formula for the volume
becomes either the definite integral between π and π of π΄ of π¦ with respect to
π¦ where, of course, π΄ of π¦ is our function that describes the area of the cross
section of our solid, or the definite integral between π and π of π times π₯
squared dπ¦. Letβs now have a look at an
application of this latter formula.

Find the volume of the solid
generated by turning, through a complete revolution about the π¦-axis, the
region bounded by the curve nine π₯ minus π¦ equals zero and the lines π₯ equals
zero, π¦ equals negative nine, and π¦ equals zero.

Remember, when we rotate a
region bounded by a curve π₯ equals some function of π¦ and the horizontal lines
π¦ equals π and π¦ equals π about the π¦-axis, we use the formula the definite
integral between π and π of π times π₯ squared dπ¦. In our case, the horizontal
lines weβre interested in are given by π¦ equals negative nine and π¦ equals
zero. So, weβre going to let π be
equal to negative nine and π be equal to zero. The region weβre interested in
is bounded by the curve nine π₯ minus π¦ equals zero and π₯ equals zero. Now, we said π₯ needs to be
some function of π¦. So, we make π₯ the subject and
we find that our equation is π₯ equals π¦ over nine. Thatβs this region. And it looks a little something
like this when we rotate it about the π¦-axis.

Substituting everything we know
into our formula for the volume, and we find itβs equal to the definite integral
between negative nine and zero of π times π¦ over nine squared dπ¦. We take out a constant factor
of π and we distribute our parentheses. And we see, our integrand is
now π¦ squared over 81. And in fact at this stage, it
might also be sensible to take out this common factor of 81. Then, we know that to integrate
a polynomial term whose exponent is not equal to negative one, we add one to
that exponent and then divide by the new value. So, the integral of π¦ squared
is π¦ cubed over three.

Then, our volume is π over 81
times zero cubed over three minus negative nine cubed over three. And, of course, zero cubed over
three is zero. We might then choose to write
negative nine as negative nine times negative nine squared or negative nine
times 81. And this means we can now
simplify by dividing through by 81. And then, negative negative
nine divided by three is just three. And so, we found that the
volume of the solid generated by rotating our region about the π¦-axis is three
π cubic units.

In our next example, weβll consider
how we can apply these techniques in something called the washer method.

Find the volume of the solid
obtained by rotating the region bounded by the curves π₯ equals six minus five
π¦ squared and π₯ equals π¦ to the fourth power about the π¦-axis.

In this example, weβre looking
to find the volume of the solid obtained by rotating a region bounded by two
curves about the π¦-axis. And so, we recall that the
volume obtained by rotating a region about the π¦-axis, whose cross-sectional
area is given by the function π΄ of π¦, is the definite integral between π and
π of π΄ of π¦ with respect to π¦. This is sometimes written,
alternatively, as the definite integral between π and π of π times π₯ squared
dπ¦. So, to help us picture whatβs
happening, weβre going to begin by sketching the area bounded by the two
curves.

The graph of π₯ equal six minus
five π¦ squared looks a little something like this. And π₯ equals π¦ to the fourth
power looks as shown. And so, this is the region
weβre going to be rotating about the π¦-axis. By either solving the equations
π₯ equals π¦ to the fourth power and π₯ equals six minus five π¦ squared
simultaneously or using graphing software or a calculator, we find these curves
intersect at the points where π¦ equals one and π¦ equals negative one. Then, when we rotate this
region about the π¦-axis, we finally get this rather unusual doughnut shape. In fact, we call this a
ring.

The cross-sectional area of the
ring will be equal to the area of the outer circle minus the area of the inner
circle. Now, since the area of a circle
is given by π times radius squared and the radius of each circle is given by
the value of the function at that point, the area of the cross section π΄ of π¦
is π times six minus five π¦ squared squared minus π times π¦ to the fourth
power squared. Armed with this information, we
find that the volume is as shown. We take out a constant factor
of π and distribute our parentheses. And our integrand becomes 36
minus 60π¦ squared plus 25π¦ to the fourth power minus π¦ to the eighth
power. Then, we integrate term by
term.

The integral of 36 is 36π¦. When we integrate negative 60π¦
squared, we get negative 60π¦ cubed divided by three, which simplifies to
negative 20π¦ cubed. The integral of 25π¦ to the
fourth power is 25π¦ to the fifth power divided by five, which simplifies to
five π¦ to the fifth power. And finally, the integral
negative π¦ to the eighth power is negative π¦ to the ninth power over nine. We then substitute one and
negative one into this expression. And then, we calculate 36 minus
20 plus five minus a ninth minus negative 36 plus 20 minus five plus
one-ninth. That gives us 376 over
nine. And so, we find that the volume
of the solid obtained by rotating our region about the π¦-axis is 376 over nine
π cubic units.

In our very final example, weβre
going to look at how to find the volume obtained by rotating a solid about a line
parallel to an axis.

Consider the region bounded by
the curves π¦ equals π₯ cubed, π¦ equals zero, and π₯ equals two. Find the volume of the solid
obtained by rotating this region about π₯ equals three.

We know that when we rotate a
region bounded by a curve and the horizontal lines π¦ equals π and π¦ equals π
about the π¦-axis, we use the formula the definite integral between π and π of
π΄ of π¦ with respect to π¦, where π΄ of π¦ is the function that describes the
cross-sectional area of the shape. So, it might help to draw a
little sketch. We have the curves π¦ equals π₯
cubed, π¦ equals zero, and a vertical line at π₯ equals two. And so, that gives us the
region weβre looking to rotate. This time though, weβre not
rotating about the π¦-axis itself but about a line parallel to it. Itβs the line with equation π₯
equals three.

And so, rotating it about this
line, weβd end up with the shape shown. We call this a ring or a
washer. And we can say that the
cross-sectional area of the ring will be given by the area of the outer, the
larger circle minus the area of the inner circle. The area of a circle is π
times radius squared. And weβre going to rewrite the
equation for our curve as π₯ in terms of π¦, so thatβs π₯ equals the cube root
of π¦. And we can say that the radius
of the large circle would be the difference between the value of the function π₯
equals the cube root of π¦ and the function π₯ equals three. So, thatβs π times the cube
root of π¦ minus three all squared.

Similarly, the radius of our
smaller circle will be the difference between three and two. And so, the cross-sectional
area or the function for the cross-sectional area is π times the cube root of
π¦ minus three squared minus π times three minus two squared. We write the cube root of π¦ as
π¦ to the power of one-third. And we simplify three minus two
to be equal to one. The limits of our definite
integral are found by considering the equations of the horizontal lines that
bound our region. We know the lower of these is
π¦ is equal to zero. The upper limit is the point of
intersection between the curve π¦ equals π₯ cubed and π₯ equals two. So, thatβs eight.

We take out a constant factor
of π. And then, we distribute our
parentheses. We end up with an integrand of
π¦ to the power of two-thirds minus six π¦ to the power of one-third plus nine
minus one. And, of course, nine minus one
is simply eight. We then integrate by raising
the exponent of each term by one and then dividing by that new value. So, we get π times
three-fifths of π¦ to the power of five over three minus 18 over four times π¦
to the power of four over three plus eight π¦. We then substitute our limits
π¦ equals zero and π¦ equals eight in. Finally, we simplify. And we find that the volume of
the solid obtained by rotating our region about the line π₯ equals three is 56π
over five cubic units.

In this video, weβve learned that
we can find the volume of a solid of revolution obtained by rotating a region about
the π₯-axis using one of two formulae. The first is π is equal to the
definite integral between π and π of π΄ of π₯ with respect to π₯, where π΄ of π₯
is a continuous function that describes the cross-sectional area of the solid. Alternatively, we use the formula
the definite integral between π and π of π times π¦ squared dπ₯. For regions rotated about the
π¦-axis, we use the formulae π equals the definite integral between π and π of π΄
of π¦ with respect to π¦. Or, the definite integral between
π and π of π times π₯ squared dπ¦. Finally, we saw that we can even
use a washer method to find a volume created by rotating a region between two curves
about an axis or a line parallel to an axis.