4. Volume and Surface Area of Cylinders, Cones, and Spheres
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Chapter 10
4.
Volume and Surface Area of Cylinders, Cones, and Spheres
Calculating the volume and surface area of cylinders, cones, and spheres is crucial in various fields such as engineering, architecture, and manufacturing. These shapes frequently appear in real-world applications, and understanding their properties enables accurate measurements and efficient problem-solving. By mastering these calculations, individuals can apply them to determine material requirements, design structures, and solve practical challenges. The process involves using specific formulas for each shape, considering their unique properties, such as the radius and height for cylinders and cones, and the radius for spheres and hemispheres.
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Lesson Settings & Tools
| | 14 Theory slides |
| | 9 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Volume and Surface Area of Cylinders, Cones, and Spheres
Slide of 14
In this lesson, several popular 3D figures like cylinders, cones, and spheres will be explored. To better understand their characteristics, their volume and surface area formulas will be introduced.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Challenge
Relating the Volumes of Cones and Cylinders
The year is about to end and New Year's Eve parties are right around the corner. Tiffaniqua and her grandmother decide to bake chewy chocolate chip cookies for their gathering. As a part of their baking steps, Tiffaniqua fills a cone-shaped cup with flour and pours it into a cylindrical cup.
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Discussion
Cylinder
The first 3D figure of this lesson to breakdown is the cylinder. Here, the method of how to calculate the volume and surface area of a cylinder will be understood.
Rule
Volume of a Cylinder
Note that the surface area of a cylinder consists of the two equal circular areas and one rectangular lateral face.
Rule
Surface Area of a Cylinder
Consider a cylinder of height h and radius r.
The surface area of this cylinder is given by the following formula.
S=2π rh+2π r^2
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Example
Finding the Volume and Surface Area of a Cylinder
Tiffaniqua loves fireworks. Her and her grandmother team up to buy a box full of them to celebrate the new year.
a Calculate the volume of the cylindrical-shaped firework. Round the answer to the nearest integer.
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b Calculate the surface area of the cylindrical-shaped firework. Round the answer to the nearest integer.
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Hint
a The volume of a cylinder is the product of π, the square of its radius, and its height.
b The surface area of a cylinder is the sum of the areas of the base and top circles and the lateral face.
Solution
a It is a given that the firework is in the shape of a cylinder. This means we can use the formula for the volume of a cylinder to determine its volume.
It is known that h= 18 and r= 4. Substitute these two values into the formula.
The volume of a firework is about 905 cubic centimeters. b It is now time to calculate the surface area of a cylindrical firework. Notice that the surface area of a cylinder consists of the areas of the base and top circles and the lateral face.
Recall the surface area formula as the sum of these three areas. SA=2π rh +2 π r^2 Once again, substitute h= 18 and r= 4 into the surface area formula.
The surface area of a cylindrical firework is about 553 square centimeters.
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Pop Quiz
Practice Finding the Surface Area and Volume of a Cylinder
Find the volume or surface area of the cylinder. The radius and height are given in centimeters.
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Discussion
Cone
The second 3D figure of this lesson to understand is the cone. One by one, the formulas of the volume and surface area of a cone will be considered.
Rule
Volume of a Cone
The volume of a cone is one third the product of its base area and its height.
The base area B is the area of the circle and the height h is measured perpendicular to the base.
V = 1/3 B h
Since the base is a circle, its area depends on its radius. Therefore, the base area can also be expressed in terms of the radius r.
V = 1/3 π r^2 h
Note that the surface area of a right cone consists of a circular base and a curved lateral face.
Rule
Surface Area of a Cone
Consider a right cone with radius r and slant height l.
The surface area of a right cone is the sum of the base area and the lateral area. The area of the base is given by π r^2 and the lateral area is π rl.
SA=π r^2 +π r l
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Example
Finding the Volume and Surface Area of a Cone
During her winter vacation, Tiffaniqua wants to design a paper Christmas tree that spins! The tree will be made in a conical shape. She plans to use 136-centimeter sticks to make the framework of the lateral face. In that case, the radius of the base will be 64 centimeters.
a Help her calculate the volume of the conical Christmas tree. Round the answer to the nearest cubic centimeter.
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b Help her calculate the surface area of the conical Christmas tree. Round the answer to the nearest square centimeter.
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Hint
a The volume of a cone is one-third of the product of the area of its base and height. Use the Pythagorean Theorem to find the height of the cone.
b The surface area of a cone is the sum of the area of its base and the area of its lateral face.
Solution
a The volume of a cone is one-third of the product of the area of its base and height. The base is already known, therefore, start by calculating the height of the conical Christmas tree. Recall that the distance from the vertex of the cone to its base is the height of the cone. The slant height of the cone is 136 centimeters and a radius is 64 centimeters.
The height, radius, and slant height of the cone form a right triangle, △ ABC. Use the Pythagorean Theorem to find the height of the cone, AB.
AC^2 = AB^2 + BC^2
SubstituteII
AC= 136, BC= 64
136^2 = AB^2 + 64^2
▼
Solve for AB
CalcPow
Calculate power
18 496 = AB^2 + 4096
SubEqn
LHS-4096=RHS-4096
14 400 = AB^2
RearrangeEqn
Rearrange equation
AB^2 = 14 400
SqrtEqn
sqrt(LHS)=sqrt(RHS)
sqrt(AB^2) = sqrt(14 400)
SqrtPowToNumber
sqrt(a^2)=a
AB = 120
Since a negative root does not make sense for a side length, consider only the positive root. The height of the cone is 120 centimeters. Now, recall the volume formula of the cone. V=1/3π r^2 h Here, r is the radius and h is the height of the cone. Substitute r= 64 and h= 120 into the volume formula.
V=1/3π r^2 h
SubstituteII
r= 64, h= 120
V=1/3π ( 64)^2 ( 120)
▼
Evaluate right-hand side
CalcPow
Calculate power
V=1/3π(4096)(120)
Multiply
Multiply
V=1/3 * π * 491 520
MoveRightFacToNumOne
1/b* a = a/b
V=491 520π/3
CalcQuot
Calculate quotient
V=163 840π
UseCalc
Use a calculator
V=514 718.540364 ...
RoundInt
Round to nearest integer
V ≈ 514 719
The volume of the cone is about 514 719 cubic centimeters.
b This time, the surface area of the cone will be calculated. Recall that the surface area of a cone is the sum of the base area and lateral face.
SA=π r^2 + π r l Note that the radius of the base r is 64 centimeters and the slant height of the cone l is 136 centimeters. Substitute these two values into the formula.
The conical tree will be about 40 212 square centimeters. Note that this means Tiffaniqua needs to have at least 40 212 square centimeters green paper to design the moving Christmas tree.
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Pop Quiz
Practice Finding the Volume and Surface Area of a Cone
The applet shows right cones. The dimensions of the figure are given in decimeters. Use the given information to answer the question. If necessary, round the answer to one decimal place.
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Discussion
Sphere
The next 3D figure of the lesson is a perfectly rounded figure called the sphere.
Concept
Sphere
Now, how to calculate the volume and surface area of a sphere will be examined.
Rule
Volume of a Sphere
Rule
Surface Area of a Sphere
The surface area of a sphere with radius r is four times pi multiplied by the radius squared.
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Example
Finding the Volume and Radius of a Sphere
It snowed a lot before winter break. Tiffaniqua's school organized an inter-class snowperson competition on the last day before the holiday. The person who makes the biggest snowperson wins. 
Tiffaniqua and her class worked together. They decided to make two huge spherical snowballs. Then, they will put them on top of each other and shape the snowmperson. They plan that the head of the snowperson will have 1 meter of radius and the body will have 32 square meters surface area.
a Calculate the volume of a spherical snowball with a radius of 1 meter. Round the answer to the nearest cubic meter.
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b Calculate the radius of a spherical snowball with a surface area of 32 square meters. Round the answer to one decimal place.
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Hint
a Recall the formula for the volume of a sphere.
b Recall the formula for the surface area of a sphere.
Solution
a Start by recalling the formula for the volume of a sphere.
V=4/3π r^3 It is known that the radius of the spherical snowball will be 1 meter. Substitute r= 1 into the formula and calculate V.
V=4/3π r^3
Substitute
r= 1
V=4/3π ( 1)^3
BaseOne
1^a=1
V=4/3π * 1
IdPropMult
Identity Property of Multiplication
V=4/3π
UseCalc
Use a calculator
V=4.188790 ...
RoundInt
Round to nearest integer
V ≈ 4
They need approximately 4 cubic meters of snow to make a snowball with a radius of 1 meter.
b In the second scenario, they plan to make a snowball that has 32 square meters surface area. Recall the formula for the surface area of a sphere.
SA=4π r^2 Now, substitute SA=32 into the formula to find r.
SA=4π r^2
Substitute
SA= 32
32=4π r^2
▼
Solve for r
DivEqn
.LHS /4π.=.RHS /4π.
32/4π=r^2
RearrangeEqn
Rearrange equation
r^2=32/4π
SqrtEqn
sqrt(LHS)=sqrt(RHS)
sqrt(r^2)=sqrt(32/4π)
SqrtPowToNumber
sqrt(a^2)=a
r=sqrt(32/4π)
UseCalc
Use a calculator
r=1.595769...
RoundDec
Round to 1 decimal place(s)
r ≈ 1.6
The radius of the snowball will be 1.6 meters for this case. It looks more likely that a snowperson made from two snowballs with radii of 1 meter and 1.6 meters will be the greatest sized snowperson.
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Discussion
Hemisphere
The final figure is the half of a sphere which is called hemisphere.
Concept
Hemisphere
A hemisphere is a three-dimensional object formed by half of a sphere and a flat circular base. Any plane that goes through the center of a sphere divides it into two hemispheres.
The radius r of a hemisphere is the segment that connects the center O with any point on the hemisphere. The radius of a sphere is the same as the radius of any of its hemispheres. The volume of a hemisphere with radius r is half the volume of a sphere with radius r. Volume V=2/3π r^3 The surface area of a hemisphere with radius r is the circular flat area plus half the surface area of a sphere with radius r.
Surface Area 2 π r^2 + π r^2 =3π r^2
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Example
Finding the Surface Area and Radius of a Hemisphere
Tiffaniqua's father bought a big snow globe as a gift for her Christmas present.
a Since this is a present for Tiffaniqua, her father wants to use beautiful wrapping paper to make it even more special. The radius of the hemispherical part is 11 centimeters. Help him to find the surface area of this present to determine the amount of the wrapping paper. Round the answer to the nearest square centimeter.
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b Her father also has a box in the shape of a hemisphere to put the gift into it. The box has the volume of 3 cubic centimeters. Does the snow globe fit this box?
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Hint
a Recall the surface area formula of a hemisphere.
b Recall the volume formula of a hemisphere.
Solution
a The snow globe has a radius of 11 centimeters. Notice that the surface area of this hemispherical globe consists of a circular base and a half of the surface area of a sphere with the same radius, 11 centimeters. Recall the sum of these formulas.
SA=2π r^2 + π r^2=3 π r^2 Now, substitute r=11 centimeters into the formula and calculate it.
SA=3π r^2
Substitute
r= 11
SA=3 π ( 11)^2
CalcPow
Calculate power
SA=3 π * 121
Multiply
Multiply
SA=363 π
UseCalc
Use a calculator
SA=1140.398133 ...
RoundInt
Round to nearest integer
SA ≈ 1140
Tiffaniqua's father needs approximately 1140 square centimeters of paper to wrap the present.
b This time, the volume of the snow globe will be calculated. Recall that the volume of a hemisphere is half of the volume of sphere with the same radius.
Notice that the box is 3 cubic decimeters. Multiply it by 100 to convert cubic decimeters to cubic centimeters.
3 dm^3=3000 cm^3
Since the volume of the box is greater than 2788 cubic centimeters, the gift fits this box. The answer is yes.
V=2/3π r^3 Now, substitute r=11 into the volume formula.
V=2/3π r^3
Substitute
r= 11
V=2/3π ( 11)^3
CalcPow
Calculate power
V=2/3π (1331)
MoveRightFacToNum
a/c* b = a* b/c
V=2662π/3
UseCalc
Use a calculator
V=2787.639881 ...
RoundInt
Round to nearest integer
V ≈ 2788
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Pop Quiz
Finding the Volume or Surface Area of Different Spheres and Hemispheres
In the following applet, calculate either the volume or the surface area of the given sphere or hemisphere. The radius is given in centimeters. Round the answer to the nearest integer.
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Closure
Calculating the Volumes of Cones and Cylinders
The challenge presented at the beginning of this lesson said that Tiffaniqua and her grandmother filled a cone-shaped cup with flour and poured it into a cylindrical cup.
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Hint
The volume of a cylinder is equal to the product of its base area and height. The volume of a cone is equal to one third of the product of its base area and height.
Solution
It is a given that the cylindrical cup is filled up by the given cone-shaped cup after three pours. This means that the volume of the given cylinder is equal to three times the volume of the given cone.
V_(cylinder)= 3 * V_(cone)
Recall that the volume of a cylinder is equal to the product of its base area and height. The volume of a cone is equal to one-third of the product of its base area and height. Note that both have circular bases.
V_(cylinder)=π r_1^2 h_1 [0.5em] V_(cone)=1/3 π r_2^2 h_2
Notice that both circular bases have the same radius. That is, r_1= r_2= 8 centimeters. Now, substitute the radii, h_1= h, and h_2= 12 into the corresponding formulas. Then, multiply the volume of the cone by 3 and set it equal to the volume of the cylinder.
V_(cylinder)= 3 * V_(cone) ⇓ π ( 8)^2( h)= 3 * 1/3 π ( 8)^2( 12)
Simplify the obtained equation and solve it for h.
π ( 8)^2( h)= 3 * 1/3 π ( 8)^2( 12)
▼
Solve for h
MoveLeftFacToNumOne
a* 1/b= a/b
π (8)^2(h)=3/3 π (8)^2(12)
CalcQuot
Calculate quotient
π (8)^2(h)=1 π (8)^2(12)
IdPropMult
Identity Property of Multiplication
π (8)^2(h)=π (8)^2(12)
DivEqn
.LHS /(π 8^2).=.RHS /(π 8^2).
π (8)^2(h)/π (8)^2=π (8)^2(12)/π (8)^2
CancelCommonFac
Cancel out common factors
π (8)^2(h)/π (8)^2=π (8)^2(12)/π (8)^2
SimpQuot
Simplify quotient
h=12
The height of the cone-shaped cup is 12 centimeters. It can be concluded that if the volume of a cylinder is three times the volume of a cone and their radii are the same then their heights should also be the same.
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Volume and Surface Area of Cylinders, Cones, and Spheres
Exercise 3.1
Consider the given sphere and cylinder with the same radius r and height r.
Select the correct statements for the volume and surface areas of these figures.
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We will determine the correct statements. To do so, we will recall the formulas for the surface area and volume of the given figures one at a time. Let's start with the surface areas.
Surface Areas
Recall that the surface area of a cylinder consists of two circular bases and a rectangular lateral face.
| Sphere | Cylinder |
|---|---|
| 4π r^2 | 2π r^2 + 2π r h |
Note that in our case the height of the cylinder is equal to its radius r.
We ended with the same formula for the surface area of the sphere. Therefore, we can say that their surface areas are equal. Let's now examine their volume formulas.
Volumes
Recall that the volume of a cylinder is the product of its base area and height.
| Sphere | Cylinder |
|---|---|
| 4/3π r^3 | π r^2 h |
Once again, we will substitute h= r into the cylinder's volume formula. π r^2 h h= r π r^3 Note that the volume of the sphere is 43 times the volume of the cylinder. \begin{gathered} V_\text{Sphere}={\color{#009600}{\dfrac{4}{3}}} \cdot {\color{#FD9000}{V_\text{Cylinder}}} \end{gathered}
Conclusion
We can conclude that two of the given statements are correct.
- Their surface areas are equal.
- The volume of the sphere is 43 times the volume of the cylinder.
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Volume and Surface Area of Cylinders, Cones, and Spheres
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