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Concept of volume, the volume of the prism and the cylinder
Similarly to the area, which assigns a measure to planar objects, we might need to define some measure which describes their size. This can affect their value, aesthetic role or usage. One of the important information about a diamond or a pearl is its size. When creating a workplace, its size depends on the number of people that will use it.
The volume of solids can be defined similarly to the area of polygons. First we define the volume of polyhedra.
The volume of a polyhedron is a positive number describing the polyhedron, which satisfies the following:
- The volume of the unit cube is \latex{ 1 }.
- The volume of congruent polyhedra is equal.
- If a polyhedron is divided into sub-polyhedra, then the total volume of the parts equals the volume of the original polyhedron.
It can be shown that for any polyhedron, one can assign a uniquely determined positive number satisfying the conditions, therefore every polyhedra has a uniquely determined volume.
It can also be shown that the above mapping is uniquely determined even for solids with curved surface. Thus cylindrical solids and cones also have volume. The proof of this statement is quite hard (it requires advanced mathematical tools), so for now we have to accept it without proof.
The volume of the cuboid
Determining the volume of a polyhedron can be viewed as a comparison to the volume of the unit cube. At a start, let us determine the volume of the cuboid.
THEOREM: The volume of the cuboid equals to the product of the lengths of the edges meeting in a vertex:
\latex{V=a\times b\times c.}
Proof
The statement of the theorem is clearly true if the cuboid can be divided into unit cubes, that is, if its edges are all integers in length. (Figure 40)
If this is not the case, than we need more refined tools to complete the proof, but the statement is still valid.
If this is not the case, than we need more refined tools to complete the proof, but the statement is still valid.
\latex{ a }
\latex{ b }
\latex{ c }
Figure 40
The surface of a cuboid is equal to the sum of the areas of its bounding faces (Figure 41). This sum, according to the net of the cuboid is
\latex{A=2\times (ab+ac+bc).}
In the special case of the cube where the edges are equal in length: \latex{a = b = c,} the volume is
\latex{ ac }
\latex{ ab }
\latex{ ab }
\latex{ ac }
\latex{ a }
\latex{ a }
\latex{ bc }
\latex{ bc }
\latex{ c }
\latex{ c }
\latex{ b }
\latex{ b }
\latex{ b }
\latex{ b }
\latex{ b }
\latex{ c }
\latex{ c }
\latex{ b }
\latex{ b }
\latex{ b }
Figure 41
\latex{V=a^{3}.}
The size of the surface is
\latex{A=6\times a^{2}.}
Example 1
We are to construct a cuboid using \latex{ 30 } congruent cubes. How should we proceed if we want to use as little paint to colour the resulting solid as possible?
Solution
According the condition, we have to try to minimize the surface of the resulting cuboid. Let us think of the cubes as unit cubes, then the volume of the cuboid is \latex{ 30 }.
Examine in how many ways \latex{ 30 } can be expressed as the product of \latex{ 3 } positive integers:
Examine in how many ways \latex{ 30 } can be expressed as the product of \latex{ 3 } positive integers:
\latex{30=1\times1\times 30=1\times 2\times 15=1\times 5\times 6=2\times 3\times 5.}
The possibilities are shown in Figure 42.
\latex{ 1 }
\latex{ 2 }
\latex{ 6 }
\latex{ 5 }
\latex{ 5 }
\latex{ 3 }
\latex{ 10 }
\latex{ 15 }
\latex{ 30 }
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
Figure 42
According to the formula \latex{A=2\times (ab+ac+bc)} of the surface, the surface areas of the constructions are as follows:
\latex{A_{1}=2\times (1+30+30)=122;}
\latex{A_{2}=2\times (2+15+30)=94;}
\latex{A_{3}=2\times (3+10+30)=86;}
\latex{A_{4}=2\times (5+6+30)=82;}
\latex{A_{5}=2\times (6+10+15)=62.}
It can be concluded that the surface is minimal if the edges of the cuboid are \latex{a=2,b=3,c=5.}
Example 2
In a cuboid, the ratio of the edges meeting in one vertex is \latex{ 4:5:20 }. The length of its diagonal is \latex{ 42 }. Determine its volume and surface.
Solution
Let the length of the edges be denoted by \latex{ a }, \latex{ b } and\latex{ c }. (Figure 43)
By using the Pythagorean theorem, for the diagonal \latex{ AC } of the base face we have
By using the Pythagorean theorem, for the diagonal \latex{ AC } of the base face we have
\latex{AC^{2}=AB^{2}+BC^{2}=a^{2}+b^{2}.}
The space diagonal \latex{ AD } can be expressed using the right angled triangle \latex{ ACD }:
\latex{AD^{2}=AC^{2}+CD^{2}=a^{2}+b^{2}+c^{2}}
This means that the square of length of the space diagonal equals the sum of the squares of the edges. Using the given ratio, let us denote the lengths of the edges by \latex{4x,5x} and \latex{20x}. Substituting into the equation for the space diagonal gives
\latex{42^{2}=(4x)^{2}+(5x)^{2}+(20x)^{2},}
\latex{1764=441x^{2},}
\latex{x^{2}=4.}
\latex{1764=441x^{2},}
\latex{x^{2}=4.}
The solution satisfying the conditions of the problem is \latex{x=2}, which means that the edges of the cuboid are \latex{a=8,b=10,c=40.}
Therefore the volume is
Therefore the volume is
\latex{V=a\times b\times c=3,200.}
The surface area is
\latex{A=2\times (ab+ac+bc)=1,600.}
\latex{ A }
\latex{ a }
\latex{ B }
\latex{ b }
\latex{ C }
\latex{ c }
\latex{ D }
Figure 43
The volume of the prism
For a cuboid, if we think of the face with edges \latex{ a } and \latex{ b } as base, then its altitude is \latex{ c }. Then the volume of the cuboid can be viewed as the product of the area of its base, \latex{ ab }, and its altitude, \latex{ c }. This equality is true for every prism as well.
THEOREM: The volume of a prism is the product of the area of its base and the corresponding altitude:
\latex{V=A_{base}\times m.}
This statement holds for right prism, for oblique prism and for every cylindrical solids as well. The proof of this fact is not easy, the principles needed to verify are beyond the level of this book; although we will now present one of them as it will prove useful in the following.
At the start of the \latex{ 17 }th century, Italian mathematician Cavalieri stated a proposition concerning the volumes of different types of solids. This has since been called Cavalieri's principle.
CAVALIERI'S PRINCIPLE: If, for two solids, there exists a plane such that the area of the intersection of one solid with any of the planes parallel to the first plane equals the area of the intersection of the same plane with the other solid, then the two solids have equal volume. (Figure 44)
\latex{ t }
\latex{ t }
\latex{ T }
\latex{ T }
\latex{ m }
\latex{ m }
Figure 44
We shall note that we would need extra conditions to state the most general form of the principle, but this form is enough for determine the volume of the solid we are interested in.
Example 3
The volume of two prisms, one with a base of an equilateral triangle and the other with a base of a square are equal, as well as the length of the edges of their bases. Find the ratio of their altitudes.
Solution
Let us denote the length of the edges of the bases by \latex{ a }, the altitude of the triangular prism by \latex{m_{1}} and the altitude of the square prism by \latex{m_{2}.}
The volumes of the two solids are, respectively,
The volumes of the two solids are, respectively,
\latex{V_{1}=t_{1}\times m_{1}=\frac{a^{2}\times \sqrt{3} }{4}\times m_{1}, V_{2}=t_{2}\times m_{2}=a^{2}\times m_{2}.}
Since the volumes are equal, it follows that
\latex{\frac{a^{2}\times \sqrt{3} }{4}\times m_{1}=a^{2}\times m_{2}.}
From this, we can express the ratio of the altitudes:
\latex{\frac{m_{1} }{m_{2} }=\frac{a^{2} }{\frac{a^{2}\times \sqrt{3} }{4} }=\frac{4}{\sqrt{3} }=\frac{4\times \sqrt{3} }{3}.}
The volume of the cylinder
The volume of cylinders can be determined with the help of the prism: it holds that any cylinder can be approximated with arbitrary precision using an appropriately chosen prism. Thus the volume of a cylinder with radius \latex{ r } and altitude \latex{ m } is (Figure 45):
\latex{V=A_{base}\times m}, that is,
\latex{V=r^{2}\times \Pi \times m.}
\latex{ r }
\latex{ m }
Figure 45
When determining the surface area, observe that the lateral surface is a rectangle with sides equal to the circumference of the base disc and the altitude in length (Figure 46); the latter of which is also the generatrix a of the right circular cylinder:
\latex{A=2\times A_{base circle}+A_{generatrix}=2\times r^{2}\times \Pi +2\times r\times \Pi \times a,} from which
\latex{A=2\times r\times \Pi \times (r+a).}
Example 4
The radius of a cylinder, with altitude being \latex{ 2\, m }, is \latex{ 1\, m }. The cylinder is filled with water up to the \latex{\frac{1}{4} } of the diagonal in a lying position. How high will the water be if we erect the cylinder?
\latex{ r }
\latex{ r }
\latex{ a }
Figure 46
Solution
The volume of the water in the lying cylinder is equal to the volume of a prism with a segment-shaped base and an altitude of \latex{ 2\, m }. (Figure 47)
\latex{ d }
\latex{ m_1 }
\latex{ \frac{1}{4}d }
Figure 47
The area of the segment, and thus the area of the base, is the difference of the area of a sector and a triangle (Figure 48). The central angle of the
sector can be found using triangle OAC: \latex{\cos \alpha =\frac{\frac{r}{2} }{r}=\frac{1}{2},} therefore \latex{\alpha =60°}, and the central angle is \latex{2\alpha =120°}. Thus the area of the sector is one third of the area of a full disc. The area of the triangle OAB equals the area of an equilateral triangle with sides of length \latex{ r }, therefore
sector can be found using triangle OAC: \latex{\cos \alpha =\frac{\frac{r}{2} }{r}=\frac{1}{2},} therefore \latex{\alpha =60°}, and the central angle is \latex{2\alpha =120°}. Thus the area of the sector is one third of the area of a full disc. The area of the triangle OAB equals the area of an equilateral triangle with sides of length \latex{ r }, therefore
\latex{A_{segment}=\frac{r^{2}\times \Pi }{3}-\frac{r^{2}\times \sqrt{3} }{4}\approx 0.61m^{2}.}
\latex{\frac{r}{2}}
\latex{\alpha}
\latex{ O }
\latex{ r }
\latex{ A }
\latex{ C }
\latex{ B }
Figure 48
This means that the volume of the water in the cylinder is
\latex{V=A_{segment}\times m\approx 0.61\times 2=1.22m^{3}.}
In a standing position the base will be a disc with radius \latex{ r }. Let the height of the water in this case be \latex{m_{1}.} Using the previously computed volume,
\latex{V=r^{2}\times \Pi \times m_{1},}
\latex{m_{1}=\frac{V}{r^{2}\times \Pi }\approx \frac{1.22}{\Pi }\approx 0.39m.}
\latex{m_{1}=\frac{V}{r^{2}\times \Pi }\approx \frac{1.22}{\Pi }\approx 0.39m.}
The height of the water will be \latex{ 0.39\, m }.
Exercises
{{exercise_number}}. We would like to build a cuboid using \latex{ 36 } unit cubes. How many different ways are there? Which ones have the smallest and the largest surface area?
{{exercise_number}}. The ratio of the edges of a cuboid is \latex{3 : 4 : 5,} the length of its space diagonal is \latex{ 20 }. Determine the lengths of its edges. What is the volume and surface area of the cuboid? Determine the angle between the space diagonal and the shortest edge, and the angle between the space diagonal and the longest edge.
{{exercise_number}}. The ratio of the edges meeting in one vertex of a cuboid is \latex{2 : 3 : 4,} its volume is \latex{192\, cm^{3}.} Determine the lengths of its edges and its surface area.
{{exercise_number}}. The ratio of the edges of a cuboid with surface area \latex{1,300\, cm^{2}} is \latex{2 : 3 : 4.} Determine the lengths of its edges and its volume.
{{exercise_number}}. The edge of the base of a right prism is \latex{ 10\, cm } long, its altitude is \latex{ 20\, cm }. Find its volume and surface area if its base is
- an equilateral triangle;
- a regular pentagon;
- a regular hexagon;
- a regular \latex{ 10 }-gon.
{{exercise_number}}. Determine the volume and surface area of the rotational cylinder with the following data given:
- \latex{r=5 cm, m=10 cm;}
- \latex{r=10 cm, A_{lateral surface}=2,000 cm^{2};}
- \latex{m=10 cm, A_{lateral surface}=1,500 cm^{2}.}
{{exercise_number}}. A cylinder is cut out of a cube with sides of length \latex{ 10\, cm }. At least how many percent is the waste?
{{exercise_number}}. Take the inscribed and circumscribed cube of a sphere. The difference in the edge lengths of the cubes is \latex{ 10\, cm }. What are the volumes and surface areas of the two cubes?
{{exercise_number}}. A rectangle with sides \latex{a= 5 cm} and \latex{b=8 cm } is first rotated around its shorter, then around its longer edge. Find the volumes and surface areas of the resulting solids.
{{exercise_number}}. Place a plane on one of the edges of the base of a cube with edge length being \latex{ 10\, cm } such that the angle between the plane and the base of the cube is \latex{ 30º }. What are the volumes of the two solids the cube is cut into by the plane? Determine the surface areas of the two solids.
{{exercise_number}}. The sum of the lengths of the three edges meeting in a vertex of a cuboid is \latex{ 14 }, while the sum of their squares is \latex{ 84 }. The length of one edge is the geometric mean of the other two. Determine the volume and surface area of the cuboid.