### Video Transcript

In this lesson, weβll learn how to
use integration to find the volume of a solid with a variable cross section. Weβll use whatβs called a slicing
method and begin by reminding ourselves of some of the ideas weβll need to use. Weβll then discuss how and in what
circumstances the method is used. And then, weβll look at some
examples.

We know that area is a measure of
the space covered by a two-dimensional shape or region. For example, for a rectangle with
side lengths π and π, the area is π times π. For a circle with radius π, the
area is ππ squared. And for a trapezoid with side
lengths π and π and height β, the area is β over two times π plus π. And remember that area is measured
in units squared.

Volume is a measure we use for the
amount of three-dimensional space taken up by a solid. For example, for a cuboid with
sides π, π, and β, the volume is π times π times β. For a sphere with radius π, the
volume is four over three ππ cubed. And for a circular cylinder with
radius π and height β, the volume is ππ squared times β. And remember that volume is
measured in units cubed.

We know also that if we have two
curves π of π₯ and π of π₯, and we want to find the area between the two curves
for values of π₯ between π and π, then the area is the integral between π and π
of π of π₯ minus π of π₯ with respect to π₯. Weβre going to extend this idea and
use definite integrals to find volumes of three-dimensional solids.

We do this by slicing the solid
into an infinite number of cross sections, using integration to sum the areas of the
cross sections or slices. So, for the volume of a
three-dimensional shape, the volume is equal to a definite integral of the area of
the cross sections with respect to π₯. Before we start integrating,
however, letβs have a closer look at how this might work geometrically.

Here, we have a cylinder with
semicircular ends that are parallel to each other and perpendicular to the
sides. The sides are of length β. If we take a cross section from
anywhere along β, the area of each cross section is the same. The volume is equal to the
cross-sectional area times the length β. This works fine if the area of the
cross section is constant, so we have a formula for the area. But if the area of the cross
sections of our solid are not constant, we can use the slicing method and a definite
integral to calculate the volume. Letβs see how this works.

Suppose we have a circle defined by
π₯ squared plus π¦ squared is equal to π squared with the centre at the origin. The circle is to be the base of our
solid. Now, suppose the cross sections are
equilateral triangles with one edge accord across the circle. The cross sections are
perpendicular to the π₯-axis. To find the volume of the solid
generated by these triangles, where the base is the circle, we let π΄ of π₯ be the
area of the triangle crossing the π₯-axis at the point π₯. So, that is, the value of π₯
changes, so will the area of the triangle.

If we split our solid into
cross-sectional slabs, where each slap has a width of Ξπ₯, if we label our slab π
π, then the volume of this slab is approximately equal to the area of the cross
section at π₯ is equal to π₯ π star times Ξπ₯. Where π₯ π star is a sample point
in between π₯ π minus one and π₯ π. An estimate for the total volume of
a solid is, therefore, the sum of the volumes of these π cross-sectional slabs. This is just an estimate, of
course. And as the number of
cross-sectional slabs π gets larger, and Ξπ₯ gets smaller, the slices get thinner
and thinner.

We define the volume of the whole
solid, therefore, as the limit of this sum as π tends to infinity. And we recognise this as the limit
of the Riemann sum, which is a definite integral. We can now define the volume for a
solid π between π₯ is equal to π and π₯ is equal π with cross-sectional area π΄
of π₯ perpendicular to the π₯-axis where π΄ is continuous as the integral between π
and π of π΄ of π₯ with respect to π₯.

Itβs important to remember that
this is the volume of a solid where the cross-sectional area is perpendicular to the
π₯-axis. If the cross sections were
perpendicular to the π¦-axis, our limits would change to π¦-values, and our areas
would be our functions of π¦ so that the integral itself would be with respect to
π¦. Before beginning an example, letβs
make a note of our strategy.

When finding volumes of solids by
slicing, the first thing to do is to determine the base of the solid. We then determine the shape of the
cross sections and try and sketch a picture of what we have. We then need to find a formula for
the area of the cross section. And remember, if the cross section
is perpendicular to the π₯-axis, then the area will be a function of π₯. If the cross section is
perpendicular to the π¦-axis, then the area will be a function of π¦. And similarly, if the cross section
is perpendicular to the π§-axis, then the area will be a function of π§. Then, for the volume of the solid,
we integrate the area of the cross section over an appropriate interval. The interval of the integration
will be defined by the base of the solid. Now, letβs look at an example.

Use the slicing method to find the
volume of the solid whose base is the region inside the circle π₯ squared plus π¦
squared is equal to two centred at the origin, if the cross sections are equilateral
triangles perpendicular to the π₯-axis.

Weβre asked to find the volume of a
solid whose base is the region inside the circle π₯ squared plus π¦ squared is equal
to two centred at the origin where the cross sections are equilateral triangles
perpendicular to the π₯-axis. Letβs first sketch our base and our
cross sections. Our base is the circle π₯ squared
plus π¦ squared is equal to two. This means the radius π squared is
equal to two, which means that the radius π is equal to plus or minus the square
root of two.

Our cross sections are equilateral
triangles perpendicular to the π₯-axis. The peak of our solid will be
traced out by the apexes of these triangles. Remember, weβll want to calculate
the volume of the solid. And thatβs equal to the definite
integral of the areas of the cross sections between limits defined by the boundaries
of the base. Our first task is then to calculate
the area of the triangles.

Remember that the area of a
triangle is half the base times the height. In our case, the base is two π¦
since itβs split in half by the π₯-axis. And since this is an equilateral
triangle, each side has length two π¦. Now, remember that because the
cross sections are perpendicular to the π₯-axis, we need to find the area as a
function of π₯. In order to do that, weβll first
find the height β in terms of π¦. And using the equation for our
base, we can find the area as a function of π₯.

To find the height β, letβs use the
Pythagorean theorem. This says, for a right angle
triangle with sides π, π, π, π squared plus π squared is equal to π
squared. In our case, this means that β
squared plus π¦ squared is equal to two π¦ all squared. Thatβs equal to four π¦
squared. So, β squared is four π¦ squared
minus π¦ squared, which is three π¦ squared, which gives us that β, since itβs a
length, is the positive square root of three times π¦.

Now, using our formula for the area
of a triangle, we have the area of our cross-sectional triangle is equal to half
times two π¦ times the square root of three times π¦. Cancelling our twos, that gives us
the area is equal to the square root of three times π¦ squared. Now, from the equation of our
circle, we have π₯ squared plus π¦ squared is equal to two. This means that π¦ squared is two
minus π₯ squared. So, if we substitute this into our
area, we get areas of function of π₯ so that our area is the square root of three
times two minus π₯ squared.

Remember that the volume is the
integral of the area where the limits are defined by the boundaries of our base. In our case, then, the limits are
negative root two to root two. So, our volume is the integral
between negative root two and root two of the square root of three times two minus
π₯ squared with respect to π₯. As root three is a constant, we can
take this outside. So, our integral becomes root three
times the integral between negative root two and root two of two minus π₯ squared
with respect to π₯.

We can also use the symmetry of the
circle to make this a little bit easier. By the symmetry of the circle, we
can change the lower limit to zero and multiply the integral by two. So, we now have the volume equal to
two times the square root of three times the integral between zero and the square
root of two times two minus π₯ squared with respect to π₯. The integral of two with respect to
π₯ is two π₯ evaluated between zero and square root of two. And the integral of π₯ squared with
respect to π₯ is one over three π₯ cubed evaluated between the limits zero and root
two.

Our volume is, therefore, two times
the square root of three times two π₯ minus π₯ cubed over three evaluated between
zero and root two. That gives us two times the square
root of three times two root two minus root two cubed over three minus zero minus
zero. Evaluating this, we get eight times
the square root of six over three.

The volume of the solid whose base
is the region inside the circle π₯ squared plus π¦ squared is two centred at the
origin where the cross sections are equilateral triangles perpendicular to the
π₯-axis is eight times root six over three units cubed. We found this volume by finding the
areas of the cross-sectional triangles in terms of π₯. We then integrated this area over
an interval defined by the base.

Letβs try this on a different
example.

Use the slicing method to find the
volume of the solid whose base is the region under the parabola π¦ is equal to four
minus π₯ squared in the first quadrant and whose cross-sectional slices are squares
perpendicular to the π₯-axis with one edge in the π₯π¦-plane.

We want to find the volume of a
solid whose base is the region under the parabola π¦ is equal to four minus π₯
squared. Weβre looking in the first
quadrant. And where the solid has
cross-sectional slices which are squares perpendicular to the π₯-axis with one edge
in the π₯π¦-plane. The first thing we can do is draw
our parabola π¦ is equal to four minus π₯ squared. And π¦ is equal to zero, our
equation, gives us π₯ squared is equal to four. So, our parabola crosses the
π₯-axis at π₯ is equal to negative two and two. When π₯ is equal to zero, π¦ is
equal to four. So, our parabola intersects the
π¦-axis at π¦ is equal to four.

The base of our solid is the region
under the parabola in the first quadrant. The cross sections of our solid are
squares which are perpendicular to the π₯-axis with one edge in the π₯π¦-plane. To find our volume, we integrate
the areas of the cross section where the limits of integration are defined by our
parabola, the lower limit is zero, and the upper limit is two.

We know that the area of a square
with side length π is equal to π squared. In our case, the length of the
sides of each square is π¦. And the value of π¦ depends on
where the square meets the π₯-axis. The area of each square is,
therefore, π¦ squared. And since π¦ is equal to four minus
π₯ squared, we have the area of the cross section equal to four minus π₯ squared all
squared. So, we have the area of the cross
section as a function of π₯. Our volume is, then, the integral
between zero and two of four minus π₯ squared all squared with respect to π₯.

To make the integral a little
easier, we can expand the bracket. This gives us the integral between
zero and two of 16 minus eight π₯ squared plus π₯ to the power of four with respect
to π₯. The integral of 16 with respect to
π₯ between zero and two is 16π₯ between zero and two. The integral of negative eight π₯
squared with respect to π₯ is negative eight times π₯ to the power of three divided
by three between zero and two. And the integral of π₯ to the power
of four is π₯ to the five divided by five evaluated between zero and two.

Evaluating this, we have 16 times
two minus eight times eight over three plus two to the five over five minus
zero. Evaluating this gives us 256
divided by 15. The volume of the solid whose base
is the region under the parabola π¦ is equal to four minus π₯ squared in the first
quadrant where the cross-sectional slices are squares perpendicular to the π₯-axis
with one edge in the π₯π¦-plane is equal to 256 divided by 15 units cubed.

In this example, our slices were
squares perpendicular to the π₯-axis with one edge in the π₯π¦-plane. Letβs look at one more example
where, in this case, the cross sections are perpendicular to the π¦-axis.

Use the slicing method to find the
volume of the solid whose base is the region under the parabola π¦ is equal to nine
minus π₯ squared and above the π₯-axis and where slices perpendicular to the π¦-axis
are squares.

Weβre asked to find the volume of a
solid whose base is defined by the parabola π¦ is equal to nine minus π₯ squared and
above the π₯-axis. The slices of the solid are
perpendicular to the π¦-axis and are squares. Letβs first draw the base region of
our solid. Our parabola is π¦ is equal to nine
minus π₯ squared so that when π¦ is equal to zero, π₯ squared is equal to nine so
that π₯ is equal to positive or negative three. When π₯ is equal to zero, π¦ is
equal to nine. And this is where our function
crosses the π¦-axis.

The region under this curve and
above the π₯-axis forms the base of our solid. Our cross sections are squares. And these are perpendicular to the
π¦-axis. The length of the sides of each
square is two π₯. And the value of π₯ depends on
where the square crosses the π¦-axis. The area of a cross-sectional
square is, therefore, two π₯ times two π₯, which is four π₯ squared. The volume of our solid, however,
is an integral along the π¦-axis between π¦ is zero and nine of the area of the
cross-sectional squares as a function of π¦. So, we need to convert our area
into a function of π¦.

We know that π¦ is equal to nine
minus π₯ squared, which gives us π₯ squared equal to nine minus π¦. Our area as a function of π₯ is
four π₯ squared so that four π₯ squared is equal to nine minus π¦ times four. In other words, four π₯ squared is
equal to four times nine minus π¦ so that our area, in fact, is four times nine
minus π¦. And we can now work out our volume,
which is the integral between zero and nine of four times nine minus π¦ with respect
to π¦.

We can take the four outside since
this is a constant multiple. We know that the integral of nine
with respect to π¦ is nine times π¦. And we know that the integral of
negative π¦ with respect to π¦ is negative π¦ squared over two. So, then, our volume is four times
nine π¦ minus π¦ squared over two evaluated between zero and nine. This is four times nine times nine
minus nine squared over two minus zero. That gives us four times 81 minus
40.5, which is 162. The volume of the solid whose base
is the region under the parabola π¦ is nine minus π₯ squared and above the π₯-axis
where slices are perpendicular to the π¦-axis and are squares is, therefore, 162
units cubed.

Now, letβs remind ourselves of the
key points in finding the volume of a solid by the slicing method. The first thing we need to do is
determine the base of our solid. We then determine the shape of the
cross section of the solid and then sketch the cross section and the base. We then find a formula for the area
of our cross section. If the cross section is
perpendicular to the π₯-axis, this will be a function of π₯. If the cross section is
perpendicular to the π¦-axis, the area will be a function of π¦. And if the cross section is
perpendicular to the π§-axis, then the area will be a function of π§.

To find our volume we then
integrate the area of the cross section over an interval defined by the base
region. If our cross sections are
perpendicular to the π₯-axis, then our volume is equal to the integral between π₯ is
π and π₯ is π of the area, which is a function of π₯ with respect to π₯. If the cross sections are
perpendicular to the π¦-axis, then our limits change to π¦-values, our area is a
function of π¦, and we integrate with respect to π¦. If our cross sections are
perpendicular to the π§-axis, then our limits become π§ values, our function area is
a function of π§, and we integrate with respect to π§. So, the key in finding the volume
is to determine the base of the solid, the shape of the cross section, and where
that cross section sits with respect to the axes.