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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 and the height of the cone is also 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 and height to fill up a hemisphere of radius Therefore, the volume of the hemisphere is twice the volume of the 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 is four times the volume of a cone with radius and height
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.
Consequently, the volume of a sphere is four-thirds the volume of a cylinder with radius and height
Finally, the volume of a cylinder with height is given by the formula 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.
Discussion
Formula for the Volume of a Sphere
The volume of a sphere with radius 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 and is parallel to the bases of the solids.
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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 base and hypotenuse Here, is the distance between the center of the base of the hemisphere and the center of the cross sectional circle, is the radius of the cross sectional circle, and is the radius of the hemisphere.
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
Conclusion
It can be stated that both solids have the same cross-sectional area at every altitude.
Example
Volume of a Water Tank
Last weekend, Tearrik installed a spherical water tank for his house.
If the tank has a radius of 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.
It is given that the radius of the tank is 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 cubic feet of water.
Example
Earth's Radius
The volume of the Earth is approximately 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.
Since the volume is given and the radius is needed, solve the formula for
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 miles.
Example
School Competition
Davontay needs to fill a cylindrical bucket in a school competition by throwing water balloons from a distance of feet. Each balloon has a spherical shape with a radius of inches. The radius of the bucket is inches, with a height of inches.
What is the minimum number of balloons needed to fill up the bucket?
Using this information, to find the volume of the bucket, and can be substituted into the formula for the volume of a cylinder.
Next, substitute 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.
Simplifying the right-hand side will give the number of balloons needed.
By multiplying the last equation by the equation is obtained. This indicates that 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 |
|---|---|
▼
Simplify right-hand side
CrossCommonFac
Cross out common factors
CancelCommonFac
Cancel out common factors
Multiply
Multiply
CalcQuot
Calculate quotient
Example
Volume of a Pencil
Ramsha bought a standard pencil whose radius is millimeters and the length, not including the eraser, is millimeters. After a good sharpening, the tip turned into a 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.
The pencil's tip has a volume of cubed millimeters.
Next, substitute and into the formula for a cylinder's volume.
It has been found that the pencil's body has a volume of cubed millimeters.
In conclusion, the volume of the Ramsha's pencil is approximately cubed millimeters.
| Volume of a Cone | Volume of a Cylinder | Volume of a Hemisphere |
|---|---|---|
Volume of the Pencil's Tip
The tip of the pencil has a radius of millimeters and a height of millimeters. Substituting these values into the first formula will give the volume of the tip.Volume of the Pencil's Body
The body of the pencil has a radius of millimeters. To find the height of this cylinder, subtract the height of the pencil's tip from the original length of the pencil.Volume of the Pencil's Eraser
The eraser is a hemisphere with a radius of Therefore, to find its volume, substitute into the hemisphere volume formula. Consequently, the pencil's eraser has a volume of cubed millimeters.Pencil's Volume
Finally, the volume of the pencil is equal to the sum of the volume of its parts.▼
Substitute values
Discussion
Formula for the Surface Area of a Sphere
Consider the fact that baseballs commonly have a radius of 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 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 congruent pyramids. The area of the base of each pyramid is
▼
Evaluate
SubstituteExpressions
Substitute expressions
CrossCommonFac
Cross out common factors
CancelCommonFac
Cancel out common factors
▼
Evaluate
CancelCommonFac
Cancel out common factors
SimpQuot
Simplify quotient
SubstituteValues
Substitute values
Example
Surface Area of a Baseball
Find the amount of leather needed to make the lining of a baseball that has a radius of 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.
Next, substitute into the previous formula.
Consequently, to make the lining of a baseball, about square inches of leather are needed. 
▼
Substitute for and evaluate
Covering of the Baseball
Notice that the surface area of a sphere with radius 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 be the height of the cylinder, be the radius of the sphere, and 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 which is equal to Since is a right triangle, by the Pythagorean Theorem, can be found.
Also, is the radius of the outer circle and applying the Pythagorean Theorem to the right triangle an expression for it can be deducted.
▼
Substitute values and simplify
FactorOut
Factor out
Distr
Distribute
SubTerms
Subtract terms
CommutativePropAdd
Commutative Property of Addition
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.
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