Video Transcript
Find the volume of the solid
obtained by rotating the region bounded by the curves 𝑥 equals six minus five
𝑦 squared and 𝑥 equals 𝑦 to the fourth power about the 𝑦-axis.
In this example, we’re looking
to find the volume of the solid obtained by rotating a region bounded by two
curves about the 𝑦-axis. And so, we recall that the
volume obtained by rotating a region about the 𝑦-axis, whose cross-sectional
area is given by the function 𝐴 of 𝑦, is the definite integral between 𝑐 and
𝑑 of 𝐴 of 𝑦 with respect to 𝑦. This is sometimes written,
alternatively, as the definite integral between 𝑐 and 𝑑 of 𝜋 times 𝑥 squared
d𝑦. So, to help us picture what’s
happening, we’re going to begin by sketching the area bounded by the two
curves.
The graph of 𝑥 equal six minus
five 𝑦 squared looks a little something like this. And 𝑥 equals 𝑦 to the fourth
power looks as shown. And so, this is the region
we’re going to be rotating about the 𝑦-axis. By either solving the equations
𝑥 equals 𝑦 to the fourth power and 𝑥 equals six minus five 𝑦 squared
simultaneously or using graphing software or a calculator, we find these curves
intersect at the points where 𝑦 equals one and 𝑦 equals negative one. Then, when we rotate this
region about the 𝑦-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 𝐴 of 𝑦
is 𝜋 times six minus five 𝑦 squared squared minus 𝜋 times 𝑦 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 60𝑦 squared plus 25𝑦 to the fourth power minus 𝑦 to the eighth
power. Then, we integrate term by
term.
The integral of 36 is 36𝑦. When we integrate negative 60𝑦
squared, we get negative 60𝑦 cubed divided by three, which simplifies to
negative 20𝑦 cubed. The integral of 25𝑦 to the
fourth power is 25𝑦 to the fifth power divided by five, which simplifies to
five 𝑦 to the fifth power. And finally, the integral
negative 𝑦 to the eighth power is negative 𝑦 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 𝑦-axis is 376 over nine
𝜋 cubic units.
