### Video Transcript

Consider the region between the
curves π¦ equals five π₯ squared and π₯ squared plus π¦ squared equals two, for π¦
greater than or equal to zero. Find the volume of the solid of
revolution obtained by rotating this region about the π₯-axis, giving your answer to
two decimal places.

In this example, weβre looking to
find the volume of a solid formed by rotating the area between two curves about the
π₯-axis. So letβs begin by sketching this
area. π¦ equals five π₯ squared is a
quadratic. So its graph is a parabola with its
lowest point at zero. And the function π₯ squared plus π¦
squared equals two is the equation of a circle centered at zero with radius root
two.

The region specified is for
π¦-values greater than or equal to zero. So, in fact, weβre only concerned
with an area above the π₯-axis. And the region to be rotated about
the π₯-axis to form a solid is the shaded region between the two curves as shown in
the diagram. The solid obtained by rotating this
region might look a bit like a doughnut, but with an indent rather than a hole at
its center.

To find the volume of this solid of
revolution, we recall that the volume π of a solid generated by revolving the
region bounded by two curves, π¦ equals π of π₯ and π¦ equals π of π₯, where π is
greater than or equal to π on the interval π, π, about the π₯-axis is given by π
equals π times the definite integral with respect to π₯ from π₯ equals π to π₯
equals π of π of π₯ squared minus π of π₯ squared.

In our case, within the specified
region, the circle with radius root two is greater than five π₯ squared apart from
at the points of intersection, where theyβre equal. And what weβre actually doing in
our integration is summing the areas of the infinitely many cross-sectional vertical
disks or washers in between where the two curves meet.

So, now for the function π, if we
rearrange π₯ squared plus π¦ squared equals two to make π¦ the subject, we have π¦
equals the square root of two minus π₯ squared, which is π of π₯. π of π₯ is then five π₯
squared.

Now, the limits of integration π
and π are the π₯-values where the two curves intersect. And we can find these values in
various ways, perhaps using a calculator or graphing software. Weβll do this here by equating our
two functions and solving for π₯. This means we need to solve 25π₯ to
the fourth power plus π₯ squared minus two equals zero for π₯. Substituting π’ equals π₯ squared,
we can solve the resulting quadratic for π’. And now making some space for our
calculation, using the quadratic formula or otherwise, we have π’ equal to negative
one plus or minus the square root of 201 all over 50.

Now, since this is equal to π₯
squared, we can only use the positive result. And evaluating this, we find π₯
squared equals 0.26354 and so on. Next, taking the square root on
both sides, we have π₯ equal to positive or negative 0.51337 to five decimal
places. So these values are our limits of
integration π and π. And with our functions π of π₯ and
π of π₯, we have the integral as shown.

To make things a little easier to
manage, we can note that our solid is symmetric about π₯ is zero and the
π¦π§-plane. So we can take our lower bound to
be zero and double the result. Now, taking the squares of our
functions, our integrand is two minus π₯ squared minus 25π₯ to the fourth power. Integrating each term separately,
we then have two π times two π₯ minus π₯ cubed over three minus 25 over five π₯ to
the fifth power, which is five π₯ to the fifth power, all evaluated between π₯ is
zero and 0.51337.

Substituting our limits of
integration in and evaluating, where for π₯ is equal to zero each term is zero, and
we restrict the value of π to two decimal places for reasons of space, we have two
π times 0.80335 and so on. This evaluates to 5.0476 and so
on. And so the volume of the solid of
revolution formed by rotating the given region about the π₯-axis is 5.05 cubic units
to two decimal places.