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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Lesson Settings & Tools
| | 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.
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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.
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Discussion
Spheres
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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. V_(Hemisphere) = 2* V_(Cone) 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.
V_(Sphere) = 4/3π r^3
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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.
Now, consider a plane that cuts the solids at a height x and is parallel to the bases of the solids.
The area of each cross-section will be calculated one at a time.
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.
Using the Pythagorean Theorem, an expression for y can be found.
r^2=x^2+y^2
▼
Solve for y
SubEqn
LHS-x^2=RHS-x^2
r^2-x^2=y^2
RearrangeEqn
Rearrange equation
y^2=r^2-x^2
SqrtEqn
sqrt(LHS)=sqrt(RHS)
y=±sqrt(r^2-x^2)
Since y is a distance, only the principal root is considered. Therefore, the area A_H of the circular cross-section can be found using the formula for area of a circle. For consistency, y will be used in place of r in the standard formula.
A_H=π y^2
Substitute
y= sqrt(r^2-x^2)
A_H=π * ( sqrt(r^2-x^2))^2
PowSqrt
( sqrt(a) )^2 = a
A_H=π(r^2-x^2)
This equation gives the area of the cross-section of the hemisphere at altitude x.
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.
Now that the radii of the circles are known, the area A_C of the cross-section can be calculated. It is the difference between the area A_G of the greater circle and the area A_S of the smaller circle.
A_C = A_G - A_S
SubstituteII
A_G= π r^2, A_S= π x^2
A_C = π r^2 - π x^2
FactorOut
Factor out π
A_C = π (r^2-x^2)
The area of the cross-section of the cylinder at altitude x can be found by using the above equation.
Conclusion
It can be stated that both solids have the same cross-sectional area at every altitude. A_H = π(r^2 -x^2)= A_C 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.
If the volume of the cone is subtracted from the volume of the cylinder, the volume of the hemisphere can be found.
V_(hemisphere) = V_(cylinder) - V_(cone)
SubstituteValues
Substitute values
V_(hemisphere) = π r^3 - 1/3π r^3
▼
Simplify right-hand side
NumberToFrac
a = 3* a/3
V_(hemisphere) = 3 π r^3/3 - 1/3π r^3
MoveRightFacToNum
a/c* b = a* b/c
V_(hemisphere) = 3 π r^3/3 - π r^3/3
SubFrac
Subtract fractions
V_(hemisphere) = 2 π r^3/3
Finally, by multiplying the volume of the hemisphere by 2, the formula for the volume of a sphere will be obtained.
V_(sphere) & = 2* V_(hemisphere) [0.7em] & = 2*2/3π r^3 [0.7em] & = 4/3π r^3
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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 = 4/3π r^3
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.
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Example
Earth's Radius
The volume of the Earth is approximately 259 875 159 532 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.
V = 4/3π r^3
Since the volume is given and the radius is needed, solve the formula for r.
V = 4/3π r^3
▼
Solve for r
MultEqn
LHS * 3=RHS* 3
3V = 4π r^3
DivEqn
.LHS /4π.=.RHS /4π.
3V/4π = r^3
RearrangeEqn
Rearrange equation
r^3 = 3V/4π
RadicalEqn
sqrt(LHS)=sqrt(RHS)
sqrt(r^3) = sqrt(3V/4π)
RootPowToNumber
sqrt(a^3)=a
r = sqrt(3V/4π)
Next, to find the Earth's radius, substitute the given volume into the derived formula and simplify.
r = sqrt(3V/4π)
Substitute
V= 259 875 159 532
r = sqrt(3( 259 875 159 532)/4π)
UseCalc
Use a calculator
r = 3958.755 865...
RoundDec
Round to 1 decimal place(s)
r ≈ 3958.8
Consequently, the Earth has a radius of approximately 3958.8 miles.
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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.
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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 |
|---|---|
| V_C = π r^2h | V_S = 4/3π r^3 |
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. V_C/V_S = 1500π in^3/62.53π in^3 Simplifying the right-hand side will give the number of balloons needed.
V_C/V_S = 1500π in^3/62.53π in^3
▼
Simplify right-hand side
CrossCommonFac
Cross out common factors
V_C/V_S = 1500π in^3/62.53π in^3
CancelCommonFac
Cancel out common factors
V_C/V_S = 1500/62.53
DivByFracD
a/b/c= a * c/b
V_C/V_S = 3* 1500/62.5
Multiply
Multiply
V_C/V_S = 4500/62.5
CalcQuot
Calculate quotient
V_C/V_S = 72
By multiplying the last equation by V_S the equation V_C = 72V_S 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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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.
| Volume of a Cone | Volume of a Cylinder | Volume of a Hemisphere |
|---|---|---|
| V_1 = 1/3π r^2h | V_2 = π r^2h | V_3 = 2/3π r^3 |
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. 180 mm - 12 mm = 168 mm 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 4 mm. Therefore, to find its volume, substitute r=4 into the hemisphere volume formula.
Consequently, the pencil's eraser has a volume of 1283π cubed millimeters.
Pencil's Volume
Finally, the volume of the pencil is equal to the sum of the volume of its parts.
V_P = V_1 + V_2 + V_3
▼
Substitute values
V_P = π(64 + 2688 + 128/3)
V_P = π(192/3 + 8064/3 + 128/3)
V_P ≈ 8779.7
In conclusion, the volume of the Ramsha's pencil is approximately 8779.7 cubed millimeters.
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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.
The formula to find the volume of a pyramid uses the height and the area of its base. Here, since the base of the pyramid lies on the surface of the sphere, the height of the pyramid is equal to the radius r of the sphere. V_(pyramid) &= 1/3Bh &⇓ V_(pyramid) &= 1/3Br 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.
A_(pyramid base)/V_(pyramid)
▼
Evaluate
SubstituteExpressions
Substitute expressions
B/1/3Br
DivByFracD
a/b/c= a * c/b
3B/Br
CrossCommonFac
Cross out common factors
3B/Br
CancelCommonFac
Cancel out common factors
3/r
This ratio is equal to the ratio of the areas of the bases of n congruent pyramids to their volumes.
n* A_(pyramid base)/n* V_(pyramid)
▼
Evaluate
CancelCommonFac
Cancel out common factors
n* A_(pyramid base)/n* V_(pyramid)
SimpQuot
Simplify quotient
A_(pyramid base)/V_(pyramid)
SubstituteValues
Substitute values
3/r
Since these pyramids fill the entire sphere, the sum of all the pyramid's base areas equals the surface area of the sphere. Additionally, the sum of the volumes of all the pyramids approximately equals the volume of the sphere. \begin{aligned} SA_{\text{sphere}} &= {\color{#0000FF}{n}}\cdot A_{\text{pyramid base}} \\ V_\text{sphere} &= {\color{#0000FF}{n}}\cdot V_{\text{pyramid}} \end{aligned} 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. SA_(sphere)/V_(sphere) &= n* A_(pyramid base)/n* V_(pyramid) [0.5em] &⇓ SA_(sphere)/V_(sphere) &= 3/r 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.
SA_(sphere)/V_(sphere) = 3/r
MultEqn
LHS * V_(sphere)=RHS* V_(sphere)
SA_(sphere)= 3/r * V_(sphere)
▼
Evaluate
Substitute
V_(sphere)= 4/3π r^3
SA_(sphere)= 3/r ( 4/3π r^3 )
MultFrac
Multiply fractions
SA_(sphere)= 3(4π r^3)/r(3)
ReduceFrac
a/b=.a /r./.b /r.
SA_(sphere)= 3(4π r^2)/3
ReduceFrac
a/b=.a /3./.b /3.
SA_(sphere)= 4π r^2
As stated previously, this is an informal justification for this formula and not a formal proof.
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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.
S = 4π r^2
Next, substitute r=1.43 into the previous formula.
S = 4π r^2
▼
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 = 25.696 971...
RoundDec
Round to 1 decimal place(s)
S = 25.7
Consequently, to make the lining of a baseball, about 25.7 square inches of leather are needed.
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.
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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.
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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.
Imagine each of these objects is cut with a horizontal plane 9 centimeters above the center of the sphere. Then, imagine making a second cut, this time 9 centimeters below the center of each sphere. Both the soccer ball and the Earth will be left with their own respective solid with a circular base. Additionally, the heights of each solid are the same.
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.
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 = sqrt(r^2 - (h/2)^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 = sqrt(r^2 - y^2)
Having the two radii, the area of the cross-section can be found.
A_(Cross-Section) = π AC^2 - π AB^2
▼
Substitute values and simplify
SubstituteII
AC= sqrt(r^2 - y^2), AB= sqrt(r^2 - (h/2)^2)
A_(Cross-Section) = π ( sqrt(r^2 - y^2))^2 - π( sqrt(r^2 - (h/2)^2))^2
PowSqrt
( sqrt(a) )^2 = a
A_(Cross-Section) = π(r^2 - y^2) - π(r^2 - (h/2)^2)
FactorOut
Factor out π
A_(Cross-Section) = π( r^2 - y^2 - (r^2 - (h/2)^2 ))
Distr
Distribute -1
A_(Cross-Section) = π( r^2 - y^2 - r^2 + (h/2)^2 )
SubTerms
Subtract terms
A_(Cross-Section) = π(- y^2 + (h/2)^2 )
CommutativePropAdd
Commutative Property of Addition
A_(Cross-Section) = π((h/2)^2 - y^2 )
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.
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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.
Next, rotate the circle about this line to see what figure is formed.
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Properties of Spheres
Exercise 3.1
A spherical lune is the region we find between two great circles of a sphere.
Find a general formula for the area of a lune when θ is measured in degrees.
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The surface area of a sphere can be calculated by multiplying 4, π, and the square of the radius. S = 4π r^2 Let's investigate a few spherical lunes with different central angles.
From the diagram above, we can describe the portion of a sphere that the spherical lune represents by using the central angle.
Half of a Sphere:& 1/2=180^(∘)/360^(∘) [1em]
Third of a Sphere:& 1/3=120^(∘)/360^(∘) [1em]
Quarter of a Sphere:& 1/4=90^(∘)/360^(∘)
The general formula for a spherical lune with a central angle of θ then becomes as follows.
A = θ/360^(∘) * 4π r^2
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Solution
A Dyson sphere is a hypothetical spherical superstructure that completely encompasses a star for the purpose of capturing the star's total power output.
A scientist has calculated that, in order for a Dyson sphere built around our Sun to not burn up, it has to be at a distance of at least 20 million kilometers from the surface of the Sun. Suppose we decide to harvest material from the Moon to build one (not a good idea, by the way). How many full size Moons would we need to make a Dyson sphere 1 kilometer thick?
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The radius of the Moon and of the Sun are approximately 1 700 kilometers and 700 000 kilometers, respectively. Round the answer to three significant digits.
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There are a couple of things we must figure out:
- The volume of the Dyson sphere.
- The volume of the Moon.
Volume of the Dyson Sphere
To determine the volume of the Dyson sphere, we need to know its radius. The radius to the Dyson sphere's inner wall is the sum of the Sun's radius and the minimum distance from the Sun's surface. Let's illustrate this.
If we add these distances, we get the distance from the Dyson sphere's center to its inner wall. 20* 10^6+0.7* 10^6=20.7 * 10^6 km To calculate the volume of the Dyson sphere, we will subtract the volume of the sphere created by the inner wall that faces the Sun from the volume of the sphere created by the outer wall that faces the Earth.
Let's label the volumes of the outer and inner spheres as V_O and V_I. Consequently, we will label their corresponding radii as r_O and r_I. The difference between these volumes is the volume of the Dyson sphere V_(DS). We can write the following expression. V_(DS)= V_O- V_I ⇓ V_(DS)= 4/3π (r_O)^3- 4/3π (r_I)^3 ⇓ V_(DS)=4/3π ((r_O)^3-(r_I)^3) We know that the inner radius is r_I= 20.7* 10^6 kilometers. We also know that the outer radius is 1 kilometer greater than this. Therefore, the outer radius r_O= 20.7* 10^6+1. Now we can form an expression for the volume of the Dyson sphere.
Let's keep the volume in exact form for now.
Volume of the Moon
We know the radius of the Moon. With this information, we can figure out its volume.
We will keep the volume of the Moon in exact form as well.
Number of Moons Needed
Finally, we will divide the volume of the Dyson sphere by the volume of the Moon to calculate how many Moons we would need to make our Dyson sphere. Remember that we will be rounding our number to three significant digits.
We would need about 262 000 Moons to complete the Dyson sphere.
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Properties of Spheres
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