5. Properties of Spheres
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Chapter 3
5.
Properties of Spheres
This lesson delves into the geometric properties of spheres, including how to calculate their volume and surface area. It also introduces advanced concepts like Cavalieri's Principle, which is used to prove the formula for the volume of a sphere. Practical applications range from determining the volume of a baseball to calculating the amount of leather needed for its surface. The lesson even explores intriguing problems like the Napkin Ring Problem, which examines the volume of two different spherical objects that have been hollowed out to the same height. This information is invaluable in various fields such as sports science, engineering, and even space exploration.
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| | 14 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Properties of Spheres
Slide of 14
Using the formulas for the volume of a cylinder or volume of a cone, the formula to find the volume of a sphere can be obtained. Even cooler, in this lesson, the formula for finding the surface area of a sphere will be developed.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Explore
Cone and Hemisphere Volumes
Consider a hemisphere and a cone that have the same radius r, and the height of the cone is also r. Use the cone to fill up the hemisphere with water.
What is the relationship between the volume of the full sphere and the volume of the given cone?
Discussion
Spheres
Discussion
Volume of a Sphere: Developing Its Formula
In the exploration, it was seen that it took two cones with radius r and height r to fill up a hemisphere of radius r. Therefore, the volume of the hemisphere is twice the volume of the cone.
Additionally, the volume of the sphere can be connected to the volume of a cylinder. To do so, remember that the volume of a cone is one-third the volume of a cylinder with same radius and height.
Finally, the volume of a cylinder with height r is given by the formula πr3. Substituting this expression into the last equation, an explicit formula to calculate the volume of a sphere is obtained.
Now, the formula for the volume of a sphere will be rigorously proven.
VHemisphere=2⋅VCone
Since the hemisphere is half of a sphere, the volume of a sphere is twice the volume of the hemisphere. Therefore, the volume of a sphere with radius r is four times the volume of a cone with radius and height r.
VCone=31⋅VCylinder
Consequently, the volume of a sphere is four-thirds the volume of a cylinder with radius and height r.
VSphere=34πr3
Discussion
Formula for the Volume of a Sphere
The volume of a sphere with radius r is four-thirds the product of pi and the radius cubed.
Proof
Cavalieri's Principle will be used to show that the formula for the volume of a sphere holds. For this purpose, consider a hemisphere and a right cylinder with a cone removed from its interior, each with the same radius and height.
Finding the Hemisphere's Cross-Sectional Area
Draw a right triangle with height x, base y, and hypotenuse r. Here, x is the distance between the center of the base of the hemisphere and the center of the cross sectional circle, y is the radius of the cross sectional circle, and r is the radius of the hemisphere.
r2=x2+y2
▼
Solve for y
y=±r2−x2
Finding the Cylinder's Cross-Sectional Area
The area of the cross-section of the cylinder can be found similarly. The cross-section's area is equal to the area between two circles. Since the height and the radius of the cylinder are equal, an isosceles right triangle can be formed inside the cylinder. Therefore, the radius of the smaller circle is also x.
Conclusion
It can be stated that both solids have the same cross-sectional area at every altitude.AH=π(r2−x2)=AC
Moreover, they have the same height. By Cavalieri's Principle, the hemisphere and the cylinder with a cone removed from its interior have the same volume. 
Vhemisphere=Vcylinder−Vcone
SubstituteValues
Substitute values
Vhemisphere=πr3−31πr3
▼
Simplify right-hand side
NumberToFrac
a=33⋅a
Vhemisphere=33πr3−31πr3
MoveRightFacToNum
ca⋅b=ca⋅b
Vhemisphere=33πr3−3πr3
SubFrac
Subtract fractions
Vhemisphere=32πr3
Vsphere=2⋅Vhemisphere=2⋅32πr3=34πr3
Example
Volume of a Water Tank
Last weekend, Tearrik installed a spherical water tank for his house.
If the tank has a radius of 2.15 feet, what is the maximum amount of water it can hold? Round the answer to two decimal places.
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Hint
Find the volume of the tank.
Solution
Finding the largest amount of water the tank can hold is the same as the finding its volume. Since it has a spherical shape, its volume is four-thirds of pi multiplied by the radius cubed.
V=34πr3
It is given that the radius of the tank is 2.15 feet. To find the volume, all that is needed is to substitute this value into the volume formula and solve.
Tearrik's tank can hold about 41.63 cubic feet of water.
Example
Earth's Radius
The volume of the Earth is approximately 259875159532 cubic miles.
Assuming the Earth is spherical, what is the Earth's radius? Round the answer to one decimal place.
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Hint
Solve the volume formula of the sphere for the radius and substitute the given volume.
Solution
Begin by writing the formula to find the volume of a sphere.
Next, to find the Earth's radius, substitute the given volume into the derived formula and simplify.
Consequently, the Earth has a radius of approximately 3958.8 miles.
V=34πr3
Since the volume is given and the radius is needed, solve the formula for r.
V=34πr3
▼
Solve for r
MultEqn
LHS⋅3=RHS⋅3
3V=4πr3
DivEqn
LHS/4π=RHS/4π
4π3V=r3
RearrangeEqn
Rearrange equation
r3=4π3V
RadicalEqn
3LHS=3RHS
3r3=34π3V
RootPowToNumber
3a3=a
r=34π3V
r=34π3V
Substitute
V=259875159532
r=34π3(259875159532)
UseCalc
Use a calculator
r=3958.755865…
RoundDec
Round to 1 decimal place(s)
r≈3958.8
Example
School Competition
Davontay needs to fill a cylindrical bucket in a school competition by throwing water balloons from a distance of 20 feet. Each balloon has a spherical shape with a radius of 2.5 inches. The radius of the bucket is 10 inches, with a height of 15 inches.
What is the minimum number of balloons needed to fill up the bucket?
Using this information, to find the volume of the bucket, r=10 and h=15 can be substituted into the formula for the volume of a cylinder.
Next, substitute r=2.5 into the formula for the volume of a sphere to find the volume of each balloon.
Now, to determine how many water balloons will fill up the bucket, divide the volume of the bucket by the volume of a balloon.
By multiplying the last equation by VS the equation VC=72VS is obtained. This indicates that 72 water balloons will fill up the bucket. Remember, what has been determined is the minimum number of balloons needed, but there could be a need for more if, for instance, Davontay misses some shots.
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Hint
Find the volume of the bucket and the volume of each balloon. Then, divide the volume of the bucket by the volume of a balloon.
Solution
To determine the minimum number of balloons needed to fill up the bucket, first solve for the volume of the bucket and the volume of each balloon. To do so, recall the formulas to find the volume of a cylinder and a sphere.
| Volume of a Cylinder | Volume of a Sphere |
|---|---|
| VC=πr2h | VS=34πr3 |
VSVC=362.5πin31500πin3
Simplifying the right-hand side will give the number of balloons needed.
VSVC=362.5πin31500πin3
▼
Simplify right-hand side
CrossCommonFac
Cross out common factors
VSVC=362.5πin31500πin3
CancelCommonFac
Cancel out common factors
VSVC=362.51500
DivByFracD
b/ca=ba⋅c
VSVC=62.53⋅1500
Multiply
Multiply
VSVC=62.54500
CalcQuot
Calculate quotient
VSVC=72
Example
Volume of a Pencil
Ramsha bought a standard pencil whose radius is 4 millimeters and the length, not including the eraser, is 180 millimeters. After a good sharpening, the tip turned into a 12 millimeter high cone.
Assuming the eraser is half of a sphere, what is the volume of the pencil? Round the answer to two decimal places.
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Hint
The pencil is made of a cone, a cylinder, and a hemisphere, all with the same radius. Therefore, the volume of the pencil equals the sum of the volumes of each solid.
Solution
The given pencil can be seen to consist of a cone, a cylinder, and half of a sphere — all with the same radius.
Consequently, the volume of the pencil equals the sum of the volumes of each of these solids.
In conclusion, the volume of the Ramsha's pencil is approximately 8779.7 cubed millimeters.
| Volume of a Cone | Volume of a Cylinder | Volume of a Hemisphere |
|---|---|---|
| V1=31πr2h | V2=πr2h | V3=32πr3 |
Volume of the Pencil's Tip
The tip of the pencil has a radius of 4 millimeters and a height of 12 millimeters. Substituting these values into the first formula will give the volume of the tip. The pencil's tip has a volume of 64π cubed millimeters.Volume of the Pencil's Body
The body of the pencil has a radius of 4 millimeters. To find the height of this cylinder, subtract the height of the pencil's tip from the original length of the pencil.180mm−12mm=168mm
Next, substitute r=4 and h=168 into the formula for a cylinder's volume.
It has been found that the pencil's body has a volume of 2688π cubed millimeters. Volume of the Pencil's Eraser
The eraser is a hemisphere with a radius of 4mm. Therefore, to find its volume, substitute r=4 into the hemisphere volume formula. Consequently, the pencil's eraser has a volume of 3128π cubed millimeters.Pencil's Volume
Finally, the volume of the pencil is equal to the sum of the volume of its parts.VP=V1+V2+V3
▼
Substitute values
VP=π(64+2688+3128)
VP=π(3192+38064+3128)
VP≈8779.7
Discussion
Formula for the Surface Area of a Sphere
Consider the fact that baseballs commonly have a radius of 1.43 inches. With this information, the volume of a baseball can be calculated, right? What about the surface area? How much leather is needed to make the lining of a baseball?
To answer these question, the following formula can be used.
Rule
Surface Area of a Sphere
The surface area of a sphere with radius r is four times pi multiplied by the radius squared.
Proof
Informal JustificationTo find the surface area of a sphere, this informal justification will use the areas and volumes of known figures. It includes approximating the ratio between a sphere's surface area and its volume. In considering a known figure, suppose that a sphere is filled with n congruent pyramids. The area of the base of each pyramid is B.
VpyramidVpyramid=31Bh⇓=31Br
The ratio of the area of the base of the pyramid to its volume can be obtained by dividing B by the formula for the volume.
VpyramidApyramid base
▼
Evaluate
SubstituteExpressions
Substitute expressions
31BrB
DivByFracD
b/ca=ba⋅c
Br3B
CrossCommonFac
Cross out common factors
Br3B
CancelCommonFac
Cancel out common factors
r3
n⋅Vpyramidn⋅Apyramid base
▼
Evaluate
CancelCommonFac
Cancel out common factors
n⋅Vpyramidn⋅Apyramid base
SimpQuot
Simplify quotient
VpyramidApyramid base
SubstituteValues
Substitute values
r3
SAsphereVsphere=n⋅Apyramid base=n⋅Vpyramid
Considering these relationships, the ratio of the surface area of the sphere to its volume is the same as the ratio of the areas of the bases of the pyramids to their volumes.
VsphereSAsphereVsphereSAsphere=n⋅Vpyramidn⋅Apyramid base⇓=r3
Since the formula for the volume of a sphere is known, it can be substituted in this ratio to find the formula for the surface area of the sphere.
VsphereSAsphere=r3
MultEqn
LHS⋅Vsphere=RHS⋅Vsphere
SAsphere=r3⋅Vsphere
▼
Evaluate
Substitute
Vsphere=34πr3
SAsphere=r3(34πr3)
MultFrac
Multiply fractions
SAsphere=r(3)3(4πr3)
ReduceFrac
ba=b/ra/r
SAsphere=33(4πr2)
ReduceFrac
ba=b/3a/3
SAsphere=4πr2
Example
Surface Area of a Baseball
Find the amount of leather needed to make the lining of a baseball that has a radius of 1.43 inches. Round the answer to one decimal place.
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Hint
Use the formula to find the surface area of a sphere.
Solution
The amount of leather needed to make the lining of a baseball is the same as the surface area of the baseball. Therefore, the following formula can be used.
Consequently, to make the lining of a baseball, about 25.7 square inches of leather are needed. 
S=4πr2
Next, substitute r=1.43 into the previous formula.
S=4πr2
▼
Substitute r for 1.43 and evaluate
Substitute
r=1.43
S=4π(1.43)2
CalcPow
Calculate power
S=4π(2.0449)
UseCalc
Use a calculator
S=25.696971…
RoundDec
Round to 1 decimal place(s)
S=25.7
Covering of the Baseball
Notice that the surface area of a sphere with radius r is four times the area of a circle with the same radius as the sphere. This relationship can be roughly seen in the covering of a baseball.
Pop Quiz
Finding the Volume or Surface Area of Different Spheres
In the following applet, calculate either the volume or the surface area of the given sphere and round the answer to two decimal places.
Illustration
Napkin Ring Problem
In talking about spheres, there is a fascinating case called the Napkin Ring Problem.
Consider two solids that have spherical shapes, for example, a soccer ball and the planet Earth. Each is represented in the diagram, along with its respective diameter.
The interesting fact about these two napkin rings is that, since they have the same height, they have exactly the same volume.
Why
Showing the Volumes are EqualTo show that the two napkin rings have the same volume, the Cavalieri's Principle will come into action. Start by considering a cross-section of each napkin ring at the same height.
The area of each cross-section is equal to the area of the outer circle minus the area of the inner circle. For a moment, focus all the attention on only one of the cross-sections. Let h be the height of the cylinder, r be the radius of the sphere, and y be the height above the center at which the cross-section was made. 
Having the two radii, the area of the cross-section can be found.
As might be seen, the area of the cross-section does not depend on the radius of the sphere, it depends only on the height of the napkin ring and the height at which the cross-section was made. Therefore, finding the area of the second cross-section will produce the same expression.
Consequently, since both napkin rings have the same height and the same cross-sectional area at every altitude, by Cavalieri's Principle, the two napkin rings have the same volume.
From the diagram, the radius of the inner circle is AB, which is equal to OD. Since △ODE is a right triangle, by the Pythagorean Theorem, OD can be found.
OD=r2−(2h)2=AB
Also, AC is the radius of the outer circle and applying the Pythagorean Theorem to the right triangle AOC, an expression for it can be deducted.
AC=r2−y2
ACross-Section=πAC2−πAB2
▼
Substitute values and simplify
SubstituteII
AC=r2−y2, AB=r2−(2h)2
ACross-Section=π(r2−y2)2−π⎝⎜⎛r2−(2h)2⎠⎟⎞2
PowSqrt
(a)2=a
ACross-Section=π(r2−y2)−π(r2−(2h)2)
FactorOut
Factor out π
ACross-Section=π(r2−y2−(r2−(2h)2))
Distr
Distribute -1
ACross-Section=π(r2−y2−r2+(2h)2)
SubTerms
Subtract terms
ACross-Section=π(-y2+(2h)2)
CommutativePropAdd
Commutative Property of Addition
ACross-Section=π((2h)2−y2)
Closure
Relationship Between Circles and Spheres
Briefly describe a circle and a sphere. Write the definitions side by side.
| Circle | Sphere |
|---|---|
| A circle is the set of all the points in a plane that are equidistant from a given point. | A sphere is the set of all points in the space that are equidistant from a given point called the center of the sphere. |
As can be seen, the two definitions are identical except for the fact that a circle is a two-dimensional figure while a sphere is three-dimensional. But this is not all. There is one more gnarly relationship. To see it, consider a circle and draw a line containing one diameter.
Find the surface area of the sphere. Write the answer in terms of π.
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The surface area of a sphere is 4 times π multiplied by the radius squared. S=4π r^2 From the diagram, we see that the radius is 4 centimeters. Therefore, we should substitute 4 for r in the formula and then evaluate the right-hand side.
The surface area of the given sphere is 64π square centimeters.
For this part, we are given the diameter of the sphere. To obtain the radius, we divide the diameter by 2. r=14/2 =7 m By substituting the radius into the formula for the surface area of a sphere, we can evaluate the surface area.
The surface area of the given sphere is 196π square meters.
For this last sphere, we been given the volume of the sphere as 36π ft^3. The volume of a sphere is four-thirds the product of π and the radius cubed. V=4/3π r^3 Let's substitute the volume of the sphere and solve the resulting equation for the radius.
The radius is 3 feet. Now we can calculate the surface area of the given sphere.
The surface area of the sphere is 36π square feet.
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Find the volume of the sphere. Write the answer in terms of π.
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{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.68333em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.22222em;\">V<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.8141079999999999em;vertical-align:0em;\"><\/span><span class=\"mord\"><span class=\"mord text\"><span class=\"mord Roboto-Regular\">m<\/span><\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.8141079999999999em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">3<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","answer":{"text":["288\\pi"]}}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.68333em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.22222em;\">V<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.8141079999999999em;vertical-align:0em;\"><\/span><span class=\"mord\"><span class=\"mord text\"><span class=\"mord Roboto-Regular\">mm<\/span><\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.8141079999999999em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\">3<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","answer":{"text":["972\\pi"]}}
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The volume of a sphere is 43 times π multiplied by the cube of the radius. V=4/3π r^3 Examining the diagram, we see that the radius of this sphere is 3 feet. Therefore, let's substitute 3 for r in the formula, and evaluate the right-hand side.
The volume of the sphere is 36π cubic feet.
This time, we have been given the diameter of the sphere. However, we need the radius, which is half the diameter. r=12/2 = 6 m Now we have the information we need to find the sphere's volume.
The volume of the sphere is 288π cubic meters.
This time, we know the sphere's surface area. The surface area of a sphere is 4 times π multiplied by the radius squared. S=4π r^2 Let's figure out the radius by substituting the given surface area for S and solving for r.
Since the radius is a distance, we disregard the negative option. Therefore, the radius is 9 millimeters. Now we can calculate the sphere's volume.
The volume of the sphere is 972π cubic millimeters.
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Baseball is a fun sport played by millions of people, mainly in the United States. Regulations state that the surface area of a baseball must be at least 25.8 square inches and at most 27.2 square inches.
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A baseball has the shape of a sphere. The surface area of a sphere is calculated as 4 times π multiplied by the radius squared. S=4π r^2 From the exercise, we know that the surface area must be at least 25.8 square inches. This measurement must then correspond to the least acceptable radius. Since the allowed surface area is 25.8 square inches, we can set S equal to 25.8 and solve for r.
The least possible radius is about 1.43 inches.
Note that this time we are asked to calculate the greatest acceptable volume. To calculate the volume of a sphere, we use the following formula. V=4/3π r^3 The greatest acceptable volume must correspond to the greatest acceptable surface area. Since we know this value, we can determine the greatest acceptable radius.
We will keep the radius in exact form for now. The greatest possible radius is sqrt(27.24π) inches. Let's substitute this into the formula for calculating a sphere's volume.
The greatest possible volume is about 13.34 cubic inches.
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Consider the following hemisphere and its circular base.
Calculate the following measure for the hemisphere. Express the answer in terms of π.
Volume
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Surface Area
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A hemisphere is half a sphere. This means that the volume of a hemisphere can be obtained by dividing the formula for calculating the volume of a sphere by 2. V_(HS)=43π r^3/2 From the diagram, we see that the radius of the hemisphere is 3 meters. With this information, we can calculate the volume of the hemisphere.
The volume is 18π cubic meters.
As in the previous section, we know that a hemisphere is half of a sphere. Therefore, one part of its surface area must be half the surface area of a sphere. However, when we cut a sphere in half, we actually create a second circular surface that becomes part of the total surface area. We get the following formula. S_(HS)=1/2(4π r^2)+ π r^2 Let's substitute the radius into this formula and evaluate.
The surface area of the hemisphere is 27π square meters.
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Properties of Spheres
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