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McGraw Hill Integrated II, 2012 View details

6. Surface Areas and Volumes of Spheres

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Exercise 28 Page 853

Use the formula for the surface area of a sphere.

About 9 feet

Practice makes perfect

The shown spherical tree house, or tree sphere, can be modeled by the following sphere with a diameter of 10.5 feet.

The radius of the sphere is half of its diameter. This tells us that r= 10.52= 5.25 feet. Let's find the volume of the spherical tree house, V_(house), using the formula for the volume of a sphere.

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

▼

Substitute 5.25 for r and evaluate

Substitute

r= 5.25

V_\text{house}=\dfrac{4}{3}\pi ({\color{#0000FF}{5.25}})^3

CalcPow

Calculate power

V_\text{house}=\dfrac{4}{3}\pi (144.703125)

Multiply

V_\text{house}=\dfrac{4}{3}\cdot 144.703125\pi

MoveRightFacToNum

a/c* b = a* b/c

V_\text{house}=\dfrac{4(144.703125\pi)}{3}

Multiply

V_\text{house}=\dfrac{578.8125\pi}{3}

UseCalc

Use a calculator

V_\text{house}=606.1310326\ldots

RoundDec

Round to 2 decimal place(s)

\textcolor{darkorange}{V_\text{house}}\approx \textcolor{darkorange}{606.13}

The volume of the current treehouse is 1.8 times the volume of the first tree sphere that was built, \textcolor{darkviolet}{V_\text{first}}. Therefore, we have the equation \textcolor{darkorange}{V_\text{house}}=1.8\textcolor{darkviolet}{V_\text{first}}. Let's substitute 606.13 for \textcolor{darkorange}{V_\text{house}}, and 43π r^3 for \textcolor{darkviolet}{V_\text{first}}, and solve the equation for r!

\textcolor{darkorange}{V_\text{house}}=1.5\textcolor{darkviolet}{V_\text{first}}

SubstituteII

\textcolor{darkorange}{V_\text{house}}={\color{#0000FF}{\textcolor{darkorange}{606.13}}}, \textcolor{darkviolet}{V_\text{first}}={\color{#009600}{\textcolor{darkviolet}{\dfrac{4}{3}\pi r^3}}}

606.13=1.8* 4/3π r^3

▼

Solve for r

MultEqn

LHS * 3=RHS* 3

1818.39=1.8* 4π r^3

Multiply

1818.39=7.2 π r^3

RearrangeEqn

Rearrange equation

7.2 π r^3=1818.39

DivEqn

.LHS /7.2π.=.RHS /7.2π.

r^3=1818.39/7.2π

UseCalc

Use a calculator

r^3=80.390488...

RoundDec

Round to 2 decimal place(s)

r^3≈ 80.39

RadicalEqn

sqrt(LHS)=sqrt(RHS)

r≈ sqrt(80.39)

UseCalc

Use a calculator

r≈ 4.3158599...

RoundDec

Round to 2 decimal place(s)

r≈ 4.32

We find that the radius of the first tree sphere was about 4.32 feet. Therefore, its diameter was about 2* 4.32=8.64 feet. We are asked also to round the answer to the nearest foot. 8.64≈ 9 Finally, the diameter of the first tree sphere was about 9 feet.

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Exercises p.852-855

Exercise 28 Page 853

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