Unlock Real-World Applications of Three-Dimensional Modeling, Geometric Shapes, and Volume Calculations

a By subtracting the volume of the cone from the volume of the cylinder, the volume of the Aquarium

A

can be found.

\underline{Aquarium A} V_{A} = \underline{Cylinder} V_{1} - \underline{Cone ∣} V_{2}

Recall the formulas for the volumes of a cylinder and a cone.

	Volume of the Cylinder	Volume of the Cone
Formula	$V_{1} = π r^{2} h$	$V_{2} = \frac{1}{3} π r^{2} h$

Both the cone and the cylinder forming Aquarium $A$ have a $10$ foot diameter, and therefore both have a radius of $5$ feet. With that in mind, substitute $r = 5$ and $h = 5 .$

	Volume of the Cylinder	Volume of the Cone
Formula	$V_{1} = π r^{2} h$	$V_{2} = \frac{1}{3} π r^{2} h$
Substitute Values	$V_{1} = π (5)^{2} 5$	$V_{2} = \frac{1}{3} π (5)^{2} 5$
Calculate	$V_{1} = 125 π$	$V_{2} = \frac{1 2 5 π}{3}$

The difference between

V_{1}

and

V_{2}

will give the number of cubic feet of water that Aquarium

A

can hold.

V_{A} = V_{1} - V_{2}

SubstituteII

$V_{1} = 125 π$ , $V_{2} = \frac{1 2 5 π}{3}$

V_{A} = 125 π - \frac{1 2 5 π}{3}

▼

Evaluate right-hand side

NumberToFrac

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

V_{A} = \frac{3 7 5 π}{3} - \frac{1 2 5 π}{3}

SubFrac

Subtract fractions

V_{A} = \frac{2 5 0 π}{3}

UseCalc

Use a calculator

V_{A} = 261.799387 \dots

RoundInt

Round to nearest integer

V_{A} \approx 262

Aquarium

A

can hold about

262

cubic feet of water.

b Begin by examining the cross-sections of each aquarium. Then, write an expression for the area of the water’s surface at height

h .

Aquarium $A$

Since the cone inside the aquarium is a right cone, its vertex is directly above the center of its base. Furthermore, its height and radius have the same length. Therefore, an isosceles right triangle inside the cone can be formed, as indicated in the diagram.

With this in mind, consider the vertical and horizontal cross-sections of the aquarium.

Vertical and horizontal cross-sections of Aquarium A

As seen on the diagram, the vertical cross-sections of the water — the shaded triangles — are isosceles right triangles, and the horizontal cross-sections form two concentric circles. Using the horizontal cross-sections, the area of the water’s surface at height

h

will be found. To do so, subtract the area of the inner circle

A_{inner}

from the area of the outer circle

A_{outer} .

A_{outer} - A_{inner}

SubstituteValues

Substitute values

π (5)^{2} - π (5 - h)^{2}

▼

Evaluate

CalcPow

Calculate power

π (25) - π (5 - h)^{2}

CommutativePropMult

Commutative Property of Multiplication

25 π - π (5 - h)^{2}

ExpandNegPerfectSquare

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

25 π - π (25 - 10 h + h^{2})

Distr

Distribute $- π$

25 π - 25 π + 10 π h - π h^{2}

SubTerm

Subtract term

10 π h - π h^{2}

When Aquarium

A

is filled to a height of

h

feet, the area of the water’s surface is

10 π h - π h^{2}

square feet.

Aquarium $B$

Examine the vertical cross-sections of the hemisphere.

The diagram shows a right triangle

A B C .

Its hypotenuse and length of one of its legs is known in terms of

h .

Therefore, using the Pythagorean Theorem,

B C

can also be expressed in terms of

h .

A B^{2} + B C^{2} = A C^{2}

SubstituteII

$A B = 5 - h$ , $A C = 5$

(5 - h)^{2} + B C^{2} = 5^{2}

▼

Solve for

B C

SubEqn

$LHS - (5 - h)^{2} = RHS - (5 - h)^{2}$

B C^{2} = 5^{2} - (5 - h)^{2}

CalcPow

Calculate power

B C^{2} = 25 - (5 - h)^{2}

ExpandNegPerfectSquare

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

B C^{2} = 25 - (25 - 10 h + h^{2})

Distr

Distribute $- 1$

B C^{2} = 25 - 25 + 10 h - h^{2}

SubTerm

Subtract term

B C^{2} = 10 h - h^{2}

SqrtEqn

$LHS = RHS$

B C^{2} = 10 h - h^{2}

SqrtPowToNumber

$a^{2} = a$

B C = 10 h - h^{2}

Now, considering the horizontal cross-sections of Aquarium

B,

B C

will be the radius of the inner circle.

In this instance, the area of the water’s surface at height

h

is the area of a circle with a radius of

10 h - h^{2} .

π (10 h - h^{2})^{2}

PowSqrt

$(a)^{2} = a$

π (10 h - h^{2})

Distr

Distribute $π$

10 π h - π h^{2}

For Aquarium

B,

the area of the water’s surface at height

h

is the same as the other aquarium,

10 π h - π h^{2}

square feet.

c Recall what the Cavalieri Principle states.

Cavalieri Principle
Two solids with the same height and the same cross-sectional area at every altitude have the same volume.

Both aquariums have a height of $5$ feet, and the area of the water’s surface when filled to a height of $h$ feet is the same for each aquarium.

	Aquarium $A$	Aquarium $B$
Height $(ft)$	$5$
Cross-Sectional Area $(ft^{2})$	$10 π h - π h^{2}$

According to the Cavalieri Principle, the volumes of water must be equal when both aquariums are filled. Therefore, no further calculations are necessary — Aquarium

B

will hold about

262

cubic feet of water.

Type of Glass
Cocktail Glass	Collins Glass
$8 \cdot 6$	$5 \cdot 10$
$$ 48$	$$ 50$

Type of Glass

Cocktail Glass

Collins Glass

8 \cdot 6

5 \cdot 10

$ 48

$ 50

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Catch-Up and Review

Comparing Aquarium Volumes

Comparing Aquarium Volumes II

Hint

Solution

Aquarium $A$

Aquarium $B$

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Finding the Volume of a Grain

Finding the Mass of a Grain

Finding the Number of Grains

Modeling the Human Eye

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