### Video Transcript

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.