Unlocking the Secrets of Spheres: Volume, Surface Area, and Cavalieri's Principle

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.

Circle
Volume
Cylinder
Volume of a cylinder
Cone
Volume of a cone

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.

Animation to fill a hemisphere using a cone

What is the relationship between the volume of the full sphere and the volume of the given cone?

Discussion

Spheres

A sphere is the set of all points in the space that are equidistant from a given point called the center of the sphere.

The distance from the center to any point on the sphere is called the radius of the sphere.

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 \cdot 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 .

1. Volume of a hemisphere twice the volume of a cone. 2. Volume of a sphere is four times the volume of a 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.

V_{Cone} = \frac{1}{3} \cdot V_{Cylinder}

Consequently, the volume of a sphere is four-thirds the volume of a cylinder with radius and height

r .

1. Volume of a cone is one-third the volume of a cylinder. 2. volume of a sphere is four-thirds the volume of a cylinder

Finally, the volume of a cylinder with height

r

is given by the formula

π r^{3} .

Substituting this expression into the last equation, an explicit formula to calculate the volume of a sphere is obtained.

$V_{Sphere} = \frac{4}{3} π r^{3}$

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 $r$ 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

$LHS = RHS$

y = \pm 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

$y = r^{2} - x^{2}$

A_{H} = π \cdot (r^{2} - x^{2})^{2}

PowSqrt

$(a)^{2} = a$

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} - \frac{1}{3} π r^{3}

Simplify right-hand side

NumberToFrac

$a = \frac{3 \cdot a}{3}$

V_{hemisphere} = \frac{3 π r ^{3}}{3} - \frac{1}{3} π r^{3}

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

V_{hemisphere} = \frac{3 π r ^{3}}{3} - \frac{π r ^{3}}{3}

SubFrac

Subtract fractions

V_{hemisphere} = \frac{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 \cdot V_{hemisphere} = 2 \cdot \frac{2}{3} π r^{3} = \frac{4}{3} π r^{3}

The formula for volume of a sphere has been developed and proven. It can now be put to use to solve a real-world problem!

Example

Volume of a Water Tank

Last weekend, Tearrik installed a spherical water tank for his house.

Spherical water tank on a table with a drainage tube on the bottom left and a filling tube on the top right. The water level is less than half full.

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.

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 = \frac{4}{3} π 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.

V = \frac{4}{3} π r^{3}

Substitute

$r = 2.15$

V = \frac{4}{3} π (2.15)^{3}

Simplify

CalcPow

Calculate power

V = \frac{4}{3} π \cdot 9.938375

MultFrac

Multiply fractions

V = \frac{4 \cdot π \cdot 9 . 9 3 8 3 7 5}{3}

Multiply

V = \frac{1 2 4 . 8 8 9 3 0 3 \dots}{3}

CalcQuot

Calculate quotient

V = 41.629767 \dots

RoundDec

Round to $2$ decimal place(s)

V \approx 41.63

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.

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 = \frac{4}{3} π r^{3}

Since the volume is given and the radius is needed, solve the formula for

r .

V = \frac{4}{3} π r^{3}

Solve for

r

MultEqn

$LHS \cdot 3 = RHS \cdot 3$

3 V = 4 π r^{3}

DivEqn

$LHS / 4 π = RHS / 4 π$

\frac{3 V}{4 π} = r^{3}

RearrangeEqn

Rearrange equation

r^{3} = \frac{3 V}{4 π}

RadicalEqn

$3 LHS = 3 RHS$

3 r^{3} = 3 \frac{3 V}{4 π}

RootPowToNumber

$3 a^{3} = a$

r = 3 \frac{3 V}{4 π}

Next, to find the Earth's radius, substitute the given volume into the derived formula and simplify.

r = 3 \frac{3 V}{4 π}

Substitute

$V = 259875159532$

r = 3 \frac{3 ( 2 5 9 8 7 5 1 5 9 5 3 2 )}{4 π}

UseCalc

Use a calculator

r = 3958.755865 \dots

RoundDec

Round to $1$ decimal place(s)

r \approx 3958.8

Consequently, the Earth has a radius of approximately

3958.8

miles.

Knowing how to find the volume of different geometric solids helps to model some scenarios in the real world. Although geometric solids could not perfectly model real-world objects, these geometric solids can provide an approximation that yields valuable information about the scenario.

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?

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
$V_{C} = π r^{2} h$	$V_{S} = \frac{4}{3} π r^{3}$

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.

V_{C} = π r^{2} h

SubstituteII

$r = 10$ , $h = 15$

V_{C} = π (10)^{2} (15)

Simplify right-hand side

CalcPow

Calculate power

V_{C} = π \cdot 100 \cdot 15

Multiply

V_{C} = 1500 π

Next, substitute

r = 2.5

into the formula for the volume of a sphere to find the volume of each balloon.

V_{S} = \frac{4}{3} π r^{3}

Substitute

$r = 2.5$

V_{S} = \frac{4}{3} π (2.5)^{3}

Simplify right-hand side

CalcPow

Calculate power

V_{S} = \frac{4}{3} π \cdot 15.625

Multiply

V_{S} = \frac{6 2 . 5}{3} π

Now, to determine how many water balloons will fill up the bucket, divide the volume of the bucket by the volume of a balloon.

\frac{V _{C}}{V _{S}} = \frac{1 5 0 0 π in ^{3}}{\frac{6 2 . 5}{3} π in ^{3}}

Simplifying the right-hand side will give the number of balloons needed.

\frac{V _{C}}{V _{S}} = \frac{1 5 0 0 π in ^{3}}{\frac{6 2 . 5}{3} π in ^{3}}

Simplify right-hand side

CrossCommonFac

Cross out common factors

\frac{V _{C}}{V _{S}} = \frac{1 5 0 0 π in ^{3}}{\frac{6 2 . 5}{3} π in ^{3}}

CancelCommonFac

Cancel out common factors

\frac{V _{C}}{V _{S}} = \frac{1 5 0 0}{\frac{6 2 . 5}{3}}

DivByFracD

$\frac{a}{b / c} = \frac{a \cdot c}{b}$

\frac{V _{C}}{V _{S}} = \frac{3 \cdot 1 5 0 0}{6 2 . 5}

Multiply

\frac{V _{C}}{V _{S}} = \frac{4 5 0 0}{6 2 . 5}

CalcQuot

Calculate quotient

\frac{V _{C}}{V _{S}} = 72

By multiplying the last equation by

V_{S}

the equation

V_{C} = 72 V_{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.

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.

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
$V_{1} = \frac{1}{3} π r^{2} h$	$V_{2} = π r^{2} h$	$V_{3} = \frac{2}{3} π r^{3}$

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.

V_{1} = \frac{1}{3} π r^{2} h

SubstituteII

$r = 4$ , $h = 12$

V_{1} = \frac{1}{3} π (4)^{2} (12)

Simplify right-hand side

CalcPow

Calculate power

V_{1} = \frac{1}{3} π (16) (12)

Multiply

V_{1} = \frac{1 9 2}{3} π

CalcQuot

Calculate quotient

V_{1} = 64 π

The pencil's tip has a volume of

64 π

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.

V_{2} = π r^{2} h

SubstituteII

$r = 4$ , $h = 168$

V_{2} = π (4)^{2} (168)

Simplify right-hand side

CalcPow

Calculate power

V_{2} = π (16) (168)

Multiply

V_{2} = 2688 π

It has been found that the pencil's body has a volume of

2688 π

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.

V_{3} = \frac{2}{3} π r^{3}

Substitute

$r = 4$

V_{3} = \frac{2}{3} π (4)^{3}

Simplify right-hand side

CalcPow

Calculate power

V_{3} = \frac{2}{3} π (64)

Multiply

V_{3} = \frac{1 2 8}{3} π

Consequently, the pencil's eraser has a volume of

\frac{1 2 8}{3} π

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

SubstituteValues

Substitute values

V_{P} = 64 π + 2688 π + \frac{1 2 8}{3} π

FactorOut

Factor out $π$

V_{P} = π (64 + 2688 + \frac{1 2 8}{3})

Rewrite

Rewrite $64$ as $\frac{1 9 2}{3}$

V_{P} = π (\frac{1 9 2}{3} + 2688 + \frac{1 2 8}{3})

Rewrite

Rewrite $2688$ as $\frac{8 0 6 4}{3}$

V_{P} = π (\frac{1 9 2}{3} + \frac{8 0 6 4}{3} + \frac{1 2 8}{3})

Simplify

AddFrac

Add fractions

V_{P} = π (\frac{1 9 2 + 8 0 6 4 + 1 2 8}{3})

AddTerms

Add terms

V_{P} = π (\frac{8 3 8 4}{3})

UseCalc

Use a calculator

V_{P} = 8779.704269 \dots

RoundDec

Round to $1$ decimal place(s)

V_{P} \approx 8779.7

In conclusion, the volume of the Ramsha's pencil is approximately

8779.7

cubed millimeters.

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 Justification

To 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} V_{pyramid} = \frac{1}{3} B h ⇓ = \frac{1}{3} B r

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.

\frac{A _{pyramid base}}{V _{pyramid}}

Evaluate

SubstituteExpressions

Substitute expressions

\frac{B}{\frac{1}{3} B r}

DivByFracD

$\frac{a}{b / c} = \frac{a \cdot c}{b}$

\frac{3 B}{B r}

CrossCommonFac

Cross out common factors

\frac{3 B}{B r}

CancelCommonFac

Cancel out common factors

\frac{3}{r}

This ratio is equal to the ratio of the areas of the bases of

n

congruent pyramids to their volumes.

\frac{n \cdot A _{pyramid base}}{n \cdot V _{pyramid}}

Evaluate

CancelCommonFac

Cancel out common factors

\frac{n \cdot A _{pyramid base}}{n \cdot V _{pyramid}}

SimpQuot

Simplify quotient

\frac{A _{pyramid base}}{V _{pyramid}}

SubstituteValues

Substitute values

\frac{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.

S A_{sphere} V_{sphere} = n \cdot A_{pyramid base} = n \cdot V_{pyramid}

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.

\frac{S A _{sphere}}{V _{sphere}} \frac{S A _{sphere}}{V _{sphere}} = \frac{n \cdot A _{pyramid base}}{n \cdot V _{pyramid}} ⇓ = \frac{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.

\frac{S A _{sphere}}{V _{sphere}} = \frac{3}{r}

MultEqn

$LHS \cdot V_{sphere} = RHS \cdot V_{sphere}$

S A_{sphere} = \frac{3}{r} \cdot V_{sphere}

Evaluate

Substitute

$V_{sphere} = \frac{4}{3} π r^{3}$

S A_{sphere} = \frac{3}{r} (\frac{4}{3} π r^{3})

MultFrac

Multiply fractions

S A_{sphere} = \frac{3 ( 4 π r ^{3} )}{r ( 3 )}

ReduceFrac

$\frac{a}{b} = \frac{a / r}{b / r}$

S A_{sphere} = \frac{3 ( 4 π r ^{2} )}{3}

ReduceFrac

$\frac{a}{b} = \frac{a / 3}{b / 3}$

S A_{sphere} = 4 π r^{2}

As stated previously, this is an informal justification for this formula and not a formal proof.

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.

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 = 4 π r^{2}

Next, substitute

r = 1.43

into the previous formula.

S = 4 π r^{2}

Substitute

r

for

1.43

and evaluate

Substitute

$r = 1.43$

S = 4 π (1.43)^{2}

CalcPow

Calculate power

S = 4 π (2.0449)

UseCalc

Use a calculator

S = 25.696971 \dots

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.

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.

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.

Next, bore a cylindrical hole through the center of each solid so that the radius of each cylinder is the radius of the corresponding circular base. The resulting solids are called napkin rings. Note that both cylinders have the same height.

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 Equal

To 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 $A B,$ which is equal to $O D .$ Since $△ O D E$ is a right triangle, by the Pythagorean Theorem, $O D$ can be found.

$O D = r^{2} - (\frac{h}{2})^{2} = A B$

Also, $A C$ is the radius of the outer circle and applying the Pythagorean Theorem to the right triangle $A O C,$ an expression for it can be deducted.

$A C = r^{2} - y^{2}$

Having the two radii, the area of the cross-section can be found.

A_{Cross - Section} = π A C^{2} - π A B^{2}

Substitute values and simplify

SubstituteII

$A C = r^{2} - y^{2}$ , $A B = r^{2} - (\frac{h}{2})^{2}$

A_{Cross - Section} = π (r^{2} - y^{2})^{2} - π ⎝ ⎜ ⎛ r^{2} - (\frac{h}{2})^{2} ⎠ ⎟ ⎞^{2}

PowSqrt

$(a)^{2} = a$

A_{Cross - Section} = π (r^{2} - y^{2}) - π (r^{2} - (\frac{h}{2})^{2})

FactorOut

Factor out $π$

A_{Cross - Section} = π (r^{2} - y^{2} - (r^{2} - (\frac{h}{2})^{2}))

Distr

Distribute $- 1$

A_{Cross - Section} = π (r^{2} - y^{2} - r^{2} + (\frac{h}{2})^{2})

SubTerms

Subtract terms

A_{Cross - Section} = π (- y^{2} + (\frac{h}{2})^{2})

CommutativePropAdd

Commutative Property of Addition

A_{Cross - Section} = π ((\frac{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.

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.

As can be seen, when a circle is rotated about a diameter, it generates a sphere.

	{{ 'ml-lesson-number-slides' \| message : article.intro.bblockCount}}
	{{ 'ml-lesson-number-exercises' \| message : article.intro.exerciseCount}}
	{{ 'ml-lesson-time-estimation' \| message }}

{{ article.displayTitle }}

{{ 'ml-heading-abilities-covered' | message }}

{{ 'ml-heading-lesson-settings' | message }}

Catch-Up and Review

Proof

Finding the Hemisphere's Cross-Sectional Area

Finding the Cylinder's Cross-Sectional Area

Conclusion

Hint

Solution

Hint

Solution

Hint

Solution

Hint

Solution

Volume of the Pencil's Tip

Volume of the Pencil's Body

Volume of the Pencil's Eraser

Pencil's Volume

Proof

Hint

Solution

Covering of the Baseball

Why