We want to find the volume of the solid
obtained by rotating the
region in the first quadrant
bounded by y equals x
squared graphed here in blue,
y equals zero which would be the x-axis,
and x equals two, this vertical line here
rotated about the y-axis.
So this is the bounded region.
If we rotate this region about the y-axis,
it would produce this solid here
and our goal is to find
the volume of this solid.
Because our function is a function of x
and we have a vertical axis of rotation,
we'll be using the shell method
to find the volume given
by this formula here
where the volume is equal to 2 pi
times the integral of p of x times h of x
integrated with respect to x from a to b
where p of x is the distance
from the axis of rotation
and h of x would be the height of a shell.
To help set up the integral,
it's always helpful
to sketch a representative rectangle
where if we were to rotate the rectangle
about the y-axis, it
would give us one shell
of the volume.
So let's use this rectangle here
to help set up our integral.
If we rotate this about the y-axis,
if would give us one shell of the volume.
To help illustrate this,
here are several shells
that would give us the approximate volume
of our solid.
As the number of shells
approaches infinity,
the volume of the shells approaches
the volume of the solid which
gives us this integral here.
So looking at this rectangle,
notice how the width would be delta x
which is why we're
integrating with respect to x.
And then p of x, the distance
from the axis of rotation,
would be this distance here.
Notice how this distance would just be x.
So p of x equals x.
And then h of x, the
height of this rectangle,
would be this length here
which would just be the function value
given by y equals x squared.
So h of x equals x squared.
And now we have all
the information we need
to set up our integral.
The volume is equal to two pi
times the integral of, again,
p of x times h of x which would just be
x times x squared
integrated with respect to x
from zero to two.
So notice how our
integrand is just x cubed.
Let's evaluate this on the next slide.
And now we'll find the antiderivative.
The antiderivative of x to the third
would be x to the fourth divided by four.
So when x is two, we'd
have two to the fourth
which is 16.
16 divided by four would be four.
And then when x is zero,
this would be zero.
So the volume is two pi
times four cubic units
or eight pi cubic units.
Let's also look at our
decimal approximation.
So eight pi is approximately 25.1327.
So we have both the exact
and the approximate volume of our solid.
And because this is volume,
these would be cubic units.
So again, this would be the volume
of the solid here generated by rotating
this bounded region about the y-axis.
And we used the shell
method to find the volume.
I hope you found this helpful.
