5. Properties of Spheres
Sign In
An error ocurred, try again later!
Chapter 3
5.
Properties of Spheres
| Student Learning Objectives: |
|---|
|
Why
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.
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.
Share
Report error
Translate
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.
Share
Report error
Translate
Discussion
Spheres
Share
Report error
Translate
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
Share
Report error
Translate
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
Share
Report error
Translate
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.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":"About","formTextAfter":"ft^3","answer":{"text":["41.63"]}}
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.
Share
Report error
Translate
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.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":"About","formTextAfter":"miles","answer":{"text":["3958.8"]}}
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.
Share
Report error
Translate
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.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["72"]}}
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.
Share
Report error
Translate
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.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":"mm^3","answer":{"text":["8779.7"]}}
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.
Share
Report error
Translate
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.
Share
Report error
Translate
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.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":"in^2","answer":{"text":["25.7"]}}
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.
Share
Report error
Translate
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.
Share
Report error
Translate
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.
Share
Report error
Translate
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.
Share
Report error
Translate
Properties of Spheres
Exercise 2.1
To calculate the volume of the spherical cap, we can use the following formula. V=π h/6(3a^2+h^2) Find the volume of a spherical cap given the following information. Answer in exact form.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":""},"formTextBefore":"V=","formTextAfter":"ft^3","answer":{"text":["\\dfrac{52\\pi}{3}"]}}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":""},"formTextBefore":"V=","formTextAfter":"ft^3","answer":{"text":["1134\\pi"]}}
security
report
code
security
report
code
To calculate the volume of the spherical cap, we need to know the height h and the distance a. V=π h/6(3 a^2+ h^2) We already know the value of a, so we need to determine the height h.
Finding the Height of the Spherical Cap
From the exercise, we have only been given the value of a and the radius of the sphere. Let's add this information to the diagram. We will also label the distance between the center of the sphere and the center of the spherical cap as x.
Notice the right triangle with a hypotenuse of 5 and legs of 4 and x. This is a Pythagorean Triple. In this particular triangle, the shorter leg is 3 feet long. Let's add this information to the diagram.
Now we can determine the value of h as the difference between the sphere's radius and 3. h=5-3= 2
Finding the Volume of the Spherical Cap
Now we have everything we need to calculate the volume of the spherical cap.
The volume of the spherical cap is 52π3 cubic feet.
This time we know the height of the spherical cap. Since we also know the sphere's radius, we can find the distance between the center of the sphere and the center of the spherical cap. 17-9 = 8 ft Let's add this information to the diagram.
Again, we see a right triangle. This time, it has a hypotenuse of 17 and legs of 8 and a. This is also a Pythagorean Triple with legs of 8 and 15 feet. Let's add this information to the diagram.
Now we have everything we need to calculate the volume of the spherical cap.
The volume of the spherical cap is 1134π cubic feet.
security
report
code
E
Hint & Answer
S
Solution
The following sphere has a diameter of 4(x+3) centimeters and a surface area of 1024π square centimeters.
Determine x.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":""},"formTextBefore":"x=","formTextAfter":null,"answer":{"text":["5"]}}
security
report
code
security
report
code
From the exercise, we know that the sphere has a surface area of 1024 π square centimeters. Let's recall how we calculate the surface area of a sphere. S=4π r^2 We also know from the exercise that the sphere has a diameter of 4(x+3) centimeters. However, we need the radius which is half the diameter. We can find it by dividing the given diameter by 2. r=4(x+3)/2=2(x+3) To find x, we will substitute the radius and surface area into the formula for the surface area of a sphere and then solve for x.
Since 4(x+3) is the diameter of a sphere, it must be nonnegative. Therefore, the value of x must be 5.
security
report
code
E
Hint & Answer
S
Solution
A sphere is inscribed in a cube with a volume of 216 cubic inches.
What is the sphere's volume? Leave your answer in terms of π.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":""},"formTextBefore":"V=","formTextAfter":"in^3","answer":{"text":["36\\pi"]}}
security
report
code
security
report
code
To determine the volume of a sphere, we use the following formula. V=4/3π r^3 As we can see, we need to know the sphere's radius. We know that the sphere is inscribed in the cube. This means that the side of the cube equals the diameter of the sphere. If we do a cross-section across the midpoint of opposite sides of one of the cube's surfaces, it would look something like below.
Therefore, to find the radius of the sphere, we must first find the side of the cube.
Finding the Side of the Cube
Remember that we have been given the cube's volume. To calculate the volume of a cube, we cube its side length. V=s^3 By substituting 216 for V and solving for s, we can determine the cube's side.
The side of the cube is 6 inches.
Finding the Volume of the Sphere
Since we know the side of the cube, we also know the diameter of the sphere. With this information, we can calculate the radius of the sphere. r=6/2=3 in Now we have everything we need to find the volume of the sphere.
The volume is 36π cubic inches.
security
report
code
E
Hint & Answer
S
Solution
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":""},"formTextBefore":"r=","formTextAfter":"m","answer":{"text":["3"]}}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":""},"formTextBefore":"r=","formTextAfter":"m","answer":{"text":["18"]}}
security
report
code
security
report
code
Let r be the radius of a sphere where the surface area of the sphere equals its volume in cubic meters.
Let's substitute the formulas for finding the surface area and volume of a sphere in the equation S=V and solve for r.
The radius of a sphere with equal surface area in square meters and volume in cubic meters is 3 meters.
This time, we want to determine the radius of a sphere where the ratio of the surface area to the volume is 1:6. This means that if we divide the surface area by the volume, we should get a ratio of 16.
As in the previous section, we will substitute the formulas used to find the surface area and the volume into this proportionality and solve for r.
Therefore, the radius of the sphere is 18 meters.
security
report
code
E
Hint & Answer
S
Solution
Edit Lesson
≥
>
■2
■▢
e▢
7
8
9
×
÷■1
=
=
√▢
▢√
∫▢▢
4
5
6
+
−
≤
<
log
ln
log▢
1
2
3
()
∣
sin
cos
tan
0
.
π
x
y
Loading content