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McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

6. Surface Areas and Volumes of Spheres

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Exercise 30 Page 853

For finding the volume, use the formula for the volume of a prism and for the volume of a sphere. For finding the area, use the formula for the area of a rectangle and for the area of a sphere.

Volume: About 1038.2 cubic centimeters
Surface Area: About 798.5 square centimeters

Practice makes perfect

Let's analyze the given composite solid.

It is a square prism from which a hemisphere has been cut out. We are asked to find its volume and its surface area.

Volume

Notice that the volume of the composite solid is equal to the difference of the volume of the square prism and the volume of the hemisphere. c Volume of Composite Solid = c Volume of Square Prism - c Volume of Hemisphere The volume of a rectangular prism is equal to the product of its dimensions. Since the square prism is 10 by 10 by 13 centimeters, its volume is {\color{#0000FF}{V_\text{prism}}}=10\cdot 10\cdot 13={\color{#0000FF}{1300}}\,\text{cm}^3. Now, let's use the formula for the volume of a sphere to find the volume of the hemisphere. The radius of the sphere is r= 102=5 centimeters.

{\color{#009600}{V_\text{hemisphere}}}=\dfrac{1}{2}V_\text{sphere}

▼

Substitute values and evaluate

V_\text{sphere}={\color{#0000FF}{{\color{#FF0000}{\dfrac{4}{3}\pi r^3}}}}

{\color{#009600}{V_\text{hemisphere}}}=\dfrac{1}{2}\cdot{\color{#FF0000}{\dfrac{4}{3}\pi r^3}}

Multiply fractions

{\color{#009600}{V_\text{hemisphere}}}=\dfrac{4}{6}\pi r^3

a/b=.a /2./.b /2.

{\color{#009600}{V_\text{hemisphere}}}=\dfrac{2}{3}\pi r^3

r= 5

{\color{#009600}{V_\text{hemisphere}}}=\dfrac{2}{3}\pi (\textcolor{darkorange}{5})^3

Calculate power and product

{\color{#009600}{V_\text{hemisphere}}}=\dfrac{2}{3}\cdot 125\pi

MoveRightFacToNum

a/c* b = a* b/c

{\color{#009600}{V_\text{hemisphere}}}=\dfrac{2\cdot 125\pi}{3}

Multiply

{\color{#009600}{V_\text{hemisphere}}}=\dfrac{250\pi}{3}

Use a calculator

{\color{#009600}{V_\text{hemisphere}}}=261.799379\ldots

Round to 2 decimal place(s)

{\color{#009600}{V_\text{hemisphere}}}\approx {\color{#009600}{261.80}}

Now let's find the volume of the composite solid, V_\text{solid}.

V_\text{solid}={\color{#0000FF}{V_\text{prism}}}-{\color{#009600}{V_\text{hemisphere}}}

▼

Substitute values and evaluate

{\color{#0000FF}{V_\text{prism}}}={\color{#0000FF}{1300}}, {\color{#009600}{V_\text{hemisphere}}}={\color{#009600}{261.8}}

V_\text{solid}={\color{#0000FF}{1300}}-{\color{#009600}{261.8}}

Subtract terms

V_\text{solid}=1038.2

Therefore, the volume of the solid is about 1038.2 cubic centimeters. We are asked to round the answer to the nearest tenth, but it is already rounded to one decimal.

Area

Notice that the surface area of the composite solid is equal to the sum of four parts.

The area of the bottom base of the square prism, A_1.
The lateral area of the square prism, A_2.
The area of the top base of the square prism, A_3.
The area of the hemisphere, A_4.

Area A_1

The base of the square prism is a square with side lengths of 10 centimeters. Therefore, its base is A_1=10^2=100 cubic centimeters.

Area A_2

The formula for the lateral surface area of a right prism is A_2=Ph, where P is the perimeter of the base and h is the height of the prism. This tells us that h=13 centimeters and P=4* 10=40 centimeters since the base is a square. Now, let's find A_2.

A_2=Ph

▼

Substitute values and evaluate

P= 40, h= 13

A_2=( 40)( 13)

Multiply

A_2=520

Therefore, the lateral area is A_2=520 square centimeters.

Area A_3

Let's analyze the top base of the given composite solid.

Notice that it is equal to the difference between the area of the square, A_(□), and the area of the circle, A_(∘). The area of the square with side lengths of 10 centimeters is A_(□)=10^2=100 square centimeters. Now, let's use the formula for the area of a circle.

A_(∘)=π r^2

▼

Substitute 5 for r and evaluate

r= 5

A_(∘)=π ( 5)^2

Calculate power and product

A_(∘)=25π

Use a calculator

A_(∘)=78.539816...

Round to 2 decimal place(s)

A_(∘)≈ 78.54

Therefore, the area of the circle is about 78.54 square centimeters. Now let's find the area of the top base, A_3.

A_3 = A_(□)-A_(∘)

▼

Substitute values and evaluate

A_(□)= 100, 3= 4Template:\large\circ{78.54}

A_3 = 100- 78.54

Subtract terms

A_3=21.46

This tells us that the top area is about A_3=21.46 square centimeters.

Area A_4

Let's use the formula for area of a sphere to find the area of the hemisphere, A_4.

A_4=1/2*Area of Sphere

▼

Substitute values and evaluate

Area of Sphere= 4π r^2

A_4=1/2* 4π r^2

MoveRightFacToNumOne

1/b* a = a/b

A_4=4π r^2/2

Calculate quotient

A_4=2π r^2

r= 5

A_4=2π ( 5)^2

Calculate power and product

A_4=50π

Use a calculator

A_4=157.079632...

Round to 2 decimal place(s)

A_4≈ 157.08

The area of the hemisphere is about 157.08 square centimeters.

Area of Composite Solid

The area of the composite solid, A_\text{solid}, is the sum of area A_1, A_2, A_3, and A_4.

A_\text{solid}=A_1+A_2+A_3+A_4

SubstituteValues

Substitute values

A_\text{solid}=100+520+21.46+157.08

Add terms

A_\text{solid}=798.54

The area of the solid is about 798.54 square centimeters. We are asked to round the area to the nearest tenth. Therefore, the answer is 798.5 square centimeters.

Subchapter links

Exercises p.852-855