We want to use the washer method
to determine the volume of the solid formed
when the region bounded by y equals x
and y equals x cubed is rotated about
the y axis over the interval, x greater than
or equal to zero and less than or equal to one.
So because we have rotation about the y axis,
or a vertical axis, we'll be using
this formula here to determine the volume
using the washer method, where the volume
equals pi times the integral from c to d
of the square of big r of y minus
the square of little r of y integrated
with respect to y, where big r of y
is the outer radius and little r of y
is the inner radius.
Notice how here because we have rotation
about the y axis, or actually for
any vertical axis of rotation,
we'll be integrating with respect of y,
and therefore we must express big r of y
and little r of why as functions of y
even though they're given as functions of x.
So for the first step, let's go ahead and graph
y equals x and y equals x cubed
to determine the bounded region over
this interval and then look at the solid
if we rotate this region about the y axis.
So in green we have the graph of y equals x cubed
and in blue we have the graph of y equals x,
so this yellow region is the bounded region.
The axis of rotation is the y axis,
so rotating this region about the y axis,
which will give us this solid here,
so our goal is to find the volume of the solid.
We can also tell from the graph
these two functions intersect at the origin,
which is the point zero comma zero,
as well as the point one comma one.
The problem gave us the closed interval for x
from zero to one, which showed us how
the interval for y would be the same
from zero to one, and that's important,
because remember, we are integrating with respect to y.
To help us set up the integral
and find big r of y and little r of y,
it's always helpful to sketch a representative rectangle,
which is shown here in red, where if we go
to rotate this rectangle about the y axis,
it would give us one washer of volume of our solid.
Notice how the thickness or width of this rectangle
would be delta y, once again indicating
we'll integrate with respect to y.
Big r of y is the outer radius,
which would be the distance from the y axis
to the corresponding point on y equals x cubed,
which means big r of y would be this distance.
The inner radius would be the distance
from the y axis to the corresponding point
on the blue path given by y equals x,
so this would be little r of y.
So because big r of why is a distance
from the y axis to the corresponding point
on the function y equals x cubed,
because we need to express the outer radius
as a function of y, we need to solve this equation for x,
so if take the cube root of both sides of the equation,
or raise both sides of the equation to the one third power,
we can say x equals y to the one third,
which means big r of y is equal to y raised
to the power of one third.
Now for little r of y, again, it's the horizontal distance
from the y axis to the corresponding point on y equals x.
Well if y equals x, of course we can write
x equals y, and therefore little r of y is equal to y,
which means that volume v is equal to pi
times the integral, we're integrating
with respect to y, the interval for y
over this bounded region would be from zero to one
along the y axis, so it limits integration
here from zero to one, and then we have
the square of big r of y, which would be
y to the one third squared minus little r of y squared,
which would be just y squared,
integrated with respect to y.
Now that we have integral, let's evaluate
this on the previous slide.
We'd have pi times integral from zero to one of,
here we have y to the two thirds minus y squared,
now we'll find the anti-derivative.
The anti-derivative of y to the two thirds
would be y to the five thirds divided by five thirds
or three fifths, y to the five thirds.
Remember we obtained the five thirds
as the exponent because we added one to two thirds.
Then we have minus y to the third divided by three
or minus one third y to the third.
Now we need to find big f of b minus big f of a,
so we'll have pi times, when y is equal to one,
we just have three fifths times one to the five thirds
minus one third times one to the third,
and then when y is zero, both terms would be zero.
Simplifying we just have pi times
the quantity three fifths minus one third.
Here the common denominator would be 15,
so you multiply three fifths by three over three,
and one third by five over five,
so here we have nine 15ths minus five 15ths,
which is four 15ths, so the volume is equal
to four pi divided by 15 cubic units.
Of course we can also get our decimal approximation
if we wanted, four pi divided by 15
is approximately zero point eight three seven eight.
And again, of course this would be cubic units.
So here's the exact volume and here's
the approximate volume, which again would be
the volume of this solid here,
formed by rotating this bounded region about the y axis.
I hope you found this helpful.