### Video Transcript

Let π· be the shaded region bounded
by the graph of π¦ equals two times the natural log of one over π₯ and the line π¦
equals three minus π₯, as shown. i) Find the area of π·. ii) Find the volume of the
solid generated when π· is rotated about the horizontal line π¦ equals negative
five. iii) Write, but do not evaluate, an integral expression that can be used to
find the volume of the solid generated when π· is rotated about the π¦-axis.

Remember we can find the area π΄ of
a region bounded by the curves π¦ equals π of π₯ and π¦ equals π of π₯ and the
lines π₯ equals π and π₯ equals π, where π and π are continuous and π of π₯ is
greater than or equal to π of π₯ for all π₯ in some closed interval π and π, by
evaluating the integral between π and π of π of π₯ minus π of π₯ with respect to
π₯.

In our case, π and π will be the
π₯-coordinates of the points of intersection of the two lines. And since in the interval weβre
looking at the graph of π¦ equals three minus π₯ sits above the graph of π¦ equals
two times the natural log of one over π₯, we let π of π₯ be equal to three minus π₯
and π of π₯ be equal to two times the natural log of one over π₯. To find the π₯-coordinates of the
points of intersection of our two lines, we need to solve them simultaneously. In other words, weβll find the
solutions to the equation three minus π₯ equals two times the natural log of one
over π₯.

We could use our graphical
calculators to perform this step. Alternatively, we could use our
graphical calculators to plot the graphs of π¦ equals three minus π₯ and π¦ equals
two times the natural log of one over π₯ and find the π₯-coordinates for where they
intersect. Either way, we find the π₯-values
weβre interested in are 0.25325 correct to five decimal places and 6.84788. And weβve chosen here five decimal
places because we want to be as accurate as possible.

And we now see that we can find the
area of π· by evaluating the integral between 0.25325 and 6.84788 of three minus π₯
minus two times the natural log of one over π₯ with respect to π₯. When we evaluate this using our
calculators, we get 10.22539 and so on. Correct to three decimal places
then, weβll say that the area of π· is 10.225 square units.

For part two, weβre looking to find
the volume. This time we use a formula which is
sometimes referred to as the washer method. It says that the volume formed by
rotating the difference between two functions π of π₯ and π of π₯ about the
π₯-axis, where once again π of π₯ is greater than or equal to π of π₯, is the
integral evaluated between π and π of π times π of π₯ squared minus π of π₯
squared with respect to π₯. We do need to be a little bit
careful here that weβre not rotating π· about the π₯-axis. Instead, weβre rotating it about a
line parallel to the π₯-axis, π¦ equals negative five.

In order to deal with this, weβre
going to subtract negative five from each of our equations. This is essentially going to shift
the lines up by five units. And in doing so, weβll then be
actually rotating about the π₯-axis rather than the line π¦ equals negative
five. And this makes a lot of sense since
when we shift π¦ equals negative five up by five units, we do get the π₯-axis. Since π is a constant, we can take
that outside our integral sign and we can see that the volume is going to be given
by π times the integral evaluated between 0.25325 and 6.84788 of three minus π₯
minus negative five squared minus two times the natural log of one over π₯ minus
negative five squared. If we evaluate this on our
graphical calculator, we get 264.19837 and so on. Correct to three decimal places,
thatβs 264.198 cubic units.

And for part three, weβre looking
to find another volume. This time though instead of
rotating about a line parallel to the π₯-axis, weβre rotating about the π¦-axis. In this case then, we interchange
the roles of π₯ and π¦. This is sometimes written as the
volume is equal to the integral evaluated between π and π, where the limits are
now given in terms of π¦ of π times π₯ squared with respect to π¦. Weβre going to rearrange both of
our equations so theyβre in terms of π¦.

π¦ equals three minus π₯ can be
written as π₯ equals three minus π¦. And when we rearrange π¦ equals two
times the natural log of one over π₯, we get π₯ equals π to the negative π¦ over
two. And now, we see that the volume of
the solid generator when π· is rotated about the π¦-axis will be given by π times
the integral evaluated between some new limits π and π of three minus π¦ squared
minus π to the negative π¦ over two squared with respect to π¦.

But what are π and π? Weβll use the equation π¦ equals
three minus π₯ to establish the corresponding values of π¦ when π₯ is equal to
0.25325 and 6.84788. Our value for π will be given by
π¦ equals three minus 0.25325 which is 2.747 correct to three decimal places. Our value of π will be given by
three minus 6.84788 which is negative 3.848 correct to three decimal places. And this means we can find the
volume of the solid generated when π· is rotated about the π¦-axis by evaluating π
times the integral between negative 3.848 and 2.747 of three minus π¦ squared minus
π to the negative π¦ over two squared with respect to π¦.