Video Transcript
In this video, we will learn how to
find volumes of composite solids consisting of two or more regular solids. Let’s begin by reminding ourselves
what a composite solid is.
A composite solid can be formed
from two or more single solids. Here, we have a cone and a cylinder
with the same radius. We could join them together by
placing the cone on top of the cylinder, and this would create a composite
solid. We can also form composite solids
by removing one solid from another. For example, if we had two
cylinders of different diameters with the same height, we could remove the smaller
cylinder from the larger cylinder to create a solid a little like a tube.
Let’s review, first of all, how we
can find the volume of a single solid. Remember, the volume is a measure
of how much space there is inside a three-dimensional shape. If the solid whose volume we’re
looking to find is a prism, meaning it has a constant cross section. Then we can find its volume by
finding the area of the constant cross section and then multiplying this by the
depth or the height of the prism. For example, in the case of a
cylinder, the constant cross section is a circle. So, we can find its area using the
formula 𝜋𝑟 squared, where 𝑟 is the radius of the circle, and then multiply this
by the height of the cylinder, ℎ, in order to find its volume.
For other solids which aren’t
prisms we need to remember the individual formulae for calculating their
volumes. For example, in the case of a cone
with a radius of 𝑟 units and a height of ℎ units, its volume can be found using the
formula one-third 𝜋𝑟 squared ℎ. The volume of a cone is simply
one-third of the volume of the cylinder with the same dimensions. In the case of a pyramid, we found
the volume using the formula one-third multiplied by base area multiplied by the
perpendicular height. Again, the volume is one-third of
the volume of the prism, with the same base area and height. Now that we’ve recapped some of the
key formulae we’ll need, let’s consider some examples.
Benjamin made a cardboard house at
school. The lower part of the house is a
rectangular prism and the upper part is a triangular prism. Find the volume of the house.
We’re told, and we can see in the
diagram, that this composite solid is made up of two prisms: a rectangular prism or
a cuboid and a triangular prism. The total volume will, therefore,
be the sum of these two individual volumes. Let’s, therefore, consider each of
these prisms separately, starting with the rectangular prism. The volume of a rectangular prism
or cuboid is found by multiplying its three dimensions together, its length
multiplied by width multiplied by height.
And we can see all three of these
measurements on the figure. They are 45 centimeters, 17
centimeters, and 20 centimeters. So, we have 45 multiplied by 17
multiplied by 20. If we don’t have a calculator, then
we’d need to use either the column or grid method to perform this
multiplication. But if we do have a calculator, we
can use it to tell us that the volume of this rectangular prism is 15,300.
Now, for the triangular prism, we
know that the volume of any prism is its cross-sectional area multiplied by the
depth of the prism. The cross section of this prism is
a triangle with a base of 45 centimeters and a perpendicular height of 18
centimeters. The depth of the prism is 17
centimeters. So, we have 45 multiplied by 18
over two for the cross-sectional area. That’s base times perpendicular
height over two. And then, we multiply by 17, which
is the depth of the prism. If we’re doing this calculation by
hand, we can cancel a factor of two first of all, giving 45 multiplied by nine
multiplied by 17, which is equal to 6,885.
To find the total volume of the
house then, we add these two individual volumes together, giving 22,185. As this is a volume, the units will
be cubic units. And as the units for the lengths
were centimeters, the units for the volume are cubic centimeters. So, our answer to the problem is
that the volume of the house Benjamin made is 22,185 cubic centimeters.
Let’s now consider a second example
in which the composite solid has been formed by removing one solid from another.
A roll of paper towels has the
given dimensions. Determine, to the nearest
hundredth, the volume of the roll.
This roll of paper towels is a
composite solid, which has been formed by removing a smaller cylinder from a larger
one. We can, therefore, find its volume
by subtracting the volume of the small cylinder from the volume of the larger
cylinder. We know that to find the volume of
a cylinder, we use the formula 𝜋𝑟 squared ℎ. That’s the area of the circular
cross section multiplied by the height of the cylinder. The height of each of our cylinders
is the same. It’s 30 centimeters. But we need to be a little bit
careful when determining the radii because the measurements we’ve been given are
actually the diameters and not the radius of each circle.
That’s okay, though, because we
know that the radius is simply half the diameter. So, for the larger cylinder, the
radius is half the diameter of 16 centimeters. That’s eight centimeters. And for the smaller cylinder, the
radius is half the diameter of four centimeters, which is two centimeters. So, we can calculate each volume by
substituting the relevant values of 𝑟 and ℎ into our formula. For the large cylinder, we have
that the volume is equal to 𝜋 multiplied by eight squared multiplied by 30. And for the smaller cylinder, the
volume is 𝜋 multiplied by two squared multiplied by 30.
If we wish, we can factor by 30 𝜋,
giving that the volume is equal to 30 𝜋 multiplied by eight squared minus two
squared. Eight squared is 64 and two squared
is four. So, within the parentheses, we have
64 minus four, which is equal to 60, giving a calculation of 30𝜋 multiplied by
60. 30 multiplied by 60 is 1,800. So, if we wish, we could give our
answer in an exact form; it’s equal to 1,800𝜋. But the question asked us to
determine this volume to the nearest hundredth. So, we need to evaluate this on our
calculators. 1,800𝜋 is equal to
5,654.866776. And rounding this to the nearest
hundredth, we have 5654.87. The units for this volume will be
cubic centimeters. And so, we have our answer to the
problem. The volume of the roll of paper
towels to the nearest hundredth is 5654.87 cubic centimeters.
Let’s look at another example.
The figure shows the design of a
swimming pool. Work out, in cubic meters, the
volume of water needed to fill the swimming pool completely.
Now, this swimming pool is a
composite solid. And we need to look carefully at
the diagram to identify the individual solids it’s composed of. The shallow end of this swimming
pool is a cuboid or rectangular prism with dimensions of 10 meters, 15 meters, and
1.5 meters. The deep end of the swimming pool
is another cuboid with dimensions of 15 meters, four meters, and 15 meters. We know that for a cuboid, the
volume is equal to the length multiplied by the width multiplied by the height. So, we can determine the volume of
water needed to fill the deep end and the shallow end of the swimming pool. The volumes are 900 and 225 cubic
meters, respectively.
Now, where it gets a little
trickier is the sloping part in the middle of the pool. But if we look carefully, we can
see that this sloping part is in fact a prism. The cross section of this prism is
this face now highlighted in green, and it is a trapezium or trapezoid. We know that the volume of a prism
is the area of its constant cross section multiplied by the depth or height of the
prism. And we should also recall that the
area of a trapezoid is half the sum of the parallel sides, which we often refer to
as 𝑎 and 𝑏 multiplied by the perpendicular distance between them, which we often
refer to as ℎ.
Let’s look carefully at the diagram
to determine each of these values. The two parallel sides of our
trapezoid are the depths of the deep and shallow ends of the pool. They are four meters and 1.5
meters. The perpendicular distance between
these two sides can be found by subtracting the length of the shallow end and the
length of the deep end, which are 10 meters and 15 meters, from the total length of
the pool, which is 30 meters. That gives five meters. The depth of this prism, not to be
confused with the depth of the pool, is the same as the depth of the other two. It’s 15 meters.
So the volume of this trapezoid or
prism, it’s a half multiplied by four plus 1.5 multiplied by five, for the area of
the cross section, and then multiplied by 15, which is the depth of the prism. When evaluated, that gives 206.25,
and again the units are cubic meters. The total volume of water needed to
completely fill the swimming pool then is found by adding these three values
together. That’s 1,331.25 cubic meters.
A lot of the skill involved in this
question was in looking carefully at the diagram to identify the individual solids
that made up this composite solid and, of course, their individual dimensions. Using color, as I did in this
example, can really help with this.
Let’s now consider another
real-life example in which we model a tree as a composite solid.
By modeling the trunk of the tree
as a cylinder and the head of the tree as a sphere, ignoring any air between the
leaves and branches, work out an estimate for the volume of the tree seen in the
given figure. Give your answer in terms of
𝜋.
In this question, we’re working out
an estimate for the volume of the tree because in reality it won’t be a perfect
sphere and a perfect cylinder, but these are reasonable models. We need to recall two key
formulae. Firstly, the volume of a sphere is
calculated using the formula four-thirds 𝜋𝑟 cubed, where 𝑟 is the radius of the
sphere. And secondly, the volume of a
cylinder is 𝜋𝑟 squared ℎ, where 𝑟 is the radius of the circular base and ℎ is the
height of the cylinder.
Let’s consider the volume of the
leaves first of all, which we’re modeling as a sphere. From the diagram, we can see that
the diameter of this sphere is nine feet. The radius will be half of
this. So, the radius of this sphere is
4.5 or nine over two feet. We have then that the volume of the
leaves modeled as a sphere is four-thirds multiplied by 𝜋 multiplied by nine over
two cubed. To cube nine over two, we can cube
the numerator and denominator separately. Nine cubed is 729 and two cubed is
eight. We can then cancel a factor of four
from the numerator and denominator and also a factor of three, which leaves 243 over
two 𝜋. Notice that the question asked us
to give our answer in terms of 𝜋, so we won’t evaluate this.
Next, we consider the volume of the
tree trunk, which we’re modeling as a cylinder. We can see that the height of the
cylinder is eight feet and the diameter of its base is 1.5 feet. If we write this as a fraction,
it’s three over two. And so, the radius will be half of
this. We can double the denominator to
give a radius of three over four feet. The volume of the trunk is,
therefore, estimated as 𝜋 multiplied by three-quarters squared multiplied by
eight.
Again, we can square the numerator
and denominator of our fraction separately to give nine over 16. And then, we can cancel a factor of
eight from the numerator and denominator to give nine 𝜋 over two as the volume of
the tree trunk. The total volume of the tree then
can be found by adding these two values together, 243 over two 𝜋 plus nine over two
𝜋. 243 over two plus nine over two is
252 over two. And 252 divided by two is 126. So, by modeling this tree as a
composite solid composed of a cylinder and a sphere, we found an estimate for the
total volume of the tree to be 126𝜋 cubic feet.
Let’s consider one final example
also involving spheres.
The shape in the given figure
consists of a cylinder with a hemisphere attached to each end. Work out its volume, giving your
answer to two decimal places.
We’re told in the question, but we
can also see from the diagram, that this composite shape consists of a cylinder and
two hemispheres. We can see that these two
hemispheres are congruent because they each have a radius of three feet. So, the total volume will be equal
to the volume of the cylinder plus twice the volume of the hemisphere. Two identical hemispheres though
make a sphere. So, we can simplify slightly by
calculating the volume of the cylinder and the volume of a sphere.
Let’s consider the cylinder first
of all then. We know that its volume is
calculated using the formula 𝜋𝑟 squared ℎ. That’s the cross-sectional area
multiplied by the height of the cylinder. From the figure, we can see that
the height of the cylinder is 10 feet, but what about its radius? Well, it’s just the same as the
radius of the hemisphere on each end, so it’s three feet. The volume of the cylinder is,
therefore, 𝜋 multiplied by three squared multiplied by 10. That simplifies to 90𝜋. And we’ll keep our answer in terms
of 𝜋 for now.
For the two hemispheres, which
we’ve already said we can model as a single sphere, the volume is given by
four-thirds 𝜋𝑟 cubed. We, therefore, have four-thirds
multiplied by 𝜋 multiplied by three cubed. Three cubed is equal to 27. And we can then cancel a factor of
three from the numerator and denominator. We’re left with four multiplied by
𝜋 multiplied by nine, which is 36𝜋. The total volume of the shape in
the figure then is 90𝜋 for the volume of the cylinder plus 36𝜋 for the volume of
the sphere, or two hemispheres, which is 126𝜋.
This would be a perfectly
acceptable format for our answer, and indeed, it’s an exact value. But the question asked for the
answer to two decimal places. So, evaluating this on a
calculator, and we have 395.84067. Rounding appropriately and we have
our answer to the problem, the units of which will be cubic feet. The total volume of the shape in
the given figure to two decimal places is 395.84 cubic feet.
Let’s now summarize what we’ve seen
in this video. Firstly, a composite solid can be
formed in two ways, either by joining two or more solids together or by removing one
or more solids from another. In order to calculate the volume of
a composite solid, we need to find the volumes of the individual solids it’s
composed of and then either add them or subtract them. In order to work with composite
solids, we must, therefore, be comfortable calculating the volumes of individual
solids to include prisms, cones, pyramids, and spheres.
