It’s Professor Dave, let’s calculate volume.
We first talked about the concept of area
when we learned geometry, going over all the
formulas for the area of different polygons.
Then when we got to calculus, we learned about
integration as a new technique for calculating
area where curvature is involved.
We learned how to take some region between
a curve and an axis, or between two curves,
and find its area by integration, essentially
finding a length and sending it across the region.
But in geometry we also learned about three
dimensional figures, like cubes and prisms
and pyramids.
We described the volumes of these objects,
or the amount of three dimensional space that
they contain, and we had useful formulas to
calculate this as well.
But once again, when curvature gets involved,
all of these formulas are useless.
Luckily, just the way we run a line segment
across a two dimensional region to calculate
its area, we can run a plane region, or cross
section, across a three dimensional region
to calculate its volume.
This is also integration, just with an extra dimension.
Instead of adding up tiny rectangles under
a curve in the limit of infinitely thin rectangles,
we are adding up infinitely thin cross sections,
which we can call disks.
Since each of these disks is a two dimensional
area, taking the integral of an area function
will give us volume.
For example, take this sphere.
We can think of this as a series of circular
disks of varying radii.
If we place the center of the sphere at the
origin, and we run the disks through the shape
horizontally, then we can take the radius
of each circle as having the value of y, because
it will extend y units from the x-axis to
give a point on the sphere.
We know that the formula for the area of a
circle is pi r squared, and we just said that
the radius is equal to y, so that’s pi y squared.
But we want this in terms of the radius of
the sphere, because that is unchanging.
Let’s make a triangle like this, where the
distance from the center of the sphere to
the center of this circle is x, the distance
from the center of the circle to the surface
of the sphere is y as we said, and then the
hypotenuse is equal to the radius of the sphere,
since this point is on the surface of the sphere.
So to get the changing radius of the disk
in terms of the radius of the sphere, we use
the pythagorean theorem, and we get y squared
equals r squared minus x squared.
So let’s plug this new version of y squared
into our function for the area of the circle,
and we get pi times the quantity of r squared
minus x squared.
Now let’s integrate this with respect to
x, from negative r to r, as that interval
spans the shape.
We can pull out pi for starters.
Then we can notice that the volume of this
half is equal to the volume of this half,
so we can just integrate from zero to r if
we double our result, by putting a two here.
Now we get the antiderivative, remembering
that r is a constant, as the radius of the
sphere never changes.
That gives us (r squared x) minus (x cubed
over three).
Let’s evaluate for r, and we get two pi
times the quantity r cubed minus r cubed over three.
These combine to give two thirds r cubed,
leaving us with four thirds pi r cubed.
Evaluating for zero gives us zero, so this
is our final answer, and we will see that
we have just derived the formula for the volume
of a sphere.
So integration will give us a three dimensional
volume by adding up infinitely many two-dimensional
cross sections, just like the way it gives
us a two dimensional area by adding up infinitely
many one dimensional lengths.
But what are these three dimensional objects, and how can we represent them on the coordinate plane?
Well often times, these shapes will be produced
by taking some function and rotating it in
a third spatial dimension around the x axis.
Take something like root x, from zero to one.
Let’s fill in this area between the curve
and the x axis so we can see it.
Now imagine grabbing this little piece with
your hand and rotating it 360 degrees around
the x axis, so that it never leaves the axis,
and we see that it traces a three dimensional
shape in doing so.
This is called a solid of revolution, because
we obtain it by revolving a region about a line.
How can we get the volume of this shape?
Well as we said, we just add up all the cross
sections by integration.
So what is the formula for the area of one
of these circular cross sections?
Well the radius is root x, because that’s
how far it is from the x axis to the function,
so pi r squared gives us pi x.
That means that this is what we have to integrate,
and luckily it’s as easy as it gets.
We pull the pi out front, and x becomes x
squared over two.
Evaluating at one gives us pi over two, and
since the lower limit is zero, that’s our answer.
So while the act of defining the integrand
is definitely an extra layer of reasoning,
nothing has changed about the way we integrate.
In fact, when Newton and Leibniz were discovering
calculus, there was no one around to hand
them integrands, because no one even knew
what an integrand was.
It was questions like the ones that we are
considering now that led them to discover
calculus, so the fact that we are now finding
our own integrands, shouldn’t come as a
total surprise.
Unfortunately this technique does get a little trickier.
While rotating one function around an axis
gives us one type of solid, if we take the
area enclosed between two functions and rotate
that around an axis, that can give us a different
kind of solid whose cross sections are not
circles, but rather things called washers.
A washer is like a ring, and it will have
two radii, the one from the axis of rotation
to where the solid starts, which we call the
inner radius, and one from the axis of rotation
to where the solid ends, or the outer radius.
Just like the circle in the previous example,
whose radius changes as it moves through the
solid, this washer will also change as it
moves through the solid, and both the inner
and outer radii change according to these
two functions.
Let’s take the functions y equals x squared
and y equals x.
There is an enclosed area here, between x
equals zero and one, that we can imagine is
filled in.
We already learned how to calculate this area
by integration, but now let’s rotate this
plane section around the x axis.
That will give us this solid of revolution.
Now the inner radius will be equal to x squared,
because that’s the function that’s closer
to the axis, and the outer radius will be
equal to x, because that’s the function
that’s further away.
That means that the area of the washer that
is a cross section of this solid will be equal
to pi x squared, or the larger circle, minus
pi (x squared) squared, or pi x to the fourth,
which is the smaller circle.
Factor out a pi, and we have pi times the
quantity x squared minus x to the fourth.
That was the hard part, figuring out exactly
what it is that we need to integrate.
The easy part is the integration itself.
Let’s just integrate this area function
from zero to one.
We can pull the pi out of the integral, and
we get (x cubed over three) minus (x to the
fifth over five).
For x equals one, we get one third minus one
fifth, so to subtract we need a common denominator,
which means we get five fifteenths minus three
fifteenths.
When x is zero, everything is zero, so we
are left with two pi over fifteen as the volume.
The interesting thing about this technique
is that we could rotate the same area around
a different axis and get a different shape.
If we rotated this around y equals two, we
would get this, and the radii would be totally
different.
Or we could rotate it around the y axis instead,
which means we will need to things in terms
of y instead of x.
In each case, we just use critical thinking
to figure out the radii of the washer, by
finding these distances in terms of the relevant
variable, and then we integrate.
So as we can see, at its heart, calculating
the volume of solids of revolution by integration
just involves simple integrals that we have
already learned.
However, it is the case that we have to use
some critical thinking.
We must be able to look at a shape and figure
out what the formula will be for the area
of any cross section, and it might not always
be perfectly obvious what that is.
We may have to subtract an inner empty area
from a larger area, as with the washers, which
means finding expressions for the inner and
outer radius.
We may have to construct figures and solve
for missing lengths.
But once we have the formula for the area
of the cross section, we just integrate it
over the specified interval, and that’s
all there is to it.
For a little more practice, let’s check
comprehension.
