Exercise 26 Page 853

A puffer fish can be modeled by the following sphere.

When the fish is puffed its diameter is 5 inches. This tells us the radius of the sphere is r= 52= 2.5 inches. Let's find its surface area, S_\text{puffed}, using the formula for the surface area of a sphere.

S_\text{puffed}=4\pi {\color{#0000FF}{r}}^2

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Substitute 2.5 for r and evaluate

Substitute

r= 2.5

S_\text{puffed}=4\pi ({\color{#0000FF}{2.5}})^2

CalcPow

Calculate power

S_\text{puffed}=4\pi (6.25)

Multiply

S_\text{puffed}=25\pi

UseCalc

Use a calculator

S_\text{puffed}=78.53981633\ldots

RoundDec

Round to 2 decimal place(s)

\textcolor{darkorange}{S_\text{puffed}}\approx \textcolor{darkorange}{78.54}

When the fish is puffed, its surface area is 1.5 times its normal surface area, \textcolor{darkviolet}{S_\text{normal}}. Therefore, we have the equation \textcolor{darkorange}{S_\text{puffed}}=1.5\textcolor{darkviolet}{S_\text{normal}}. Let's substitute 78.54 for \textcolor{darkorange}{S_\text{puffed}} and find the normal surface area.

\textcolor{darkorange}{S_\text{puffed}}=1.5\textcolor{darkviolet}{S_\text{normal}}

Substitute

\textcolor{darkorange}{S_\text{puffed}}={\color{#0000FF}{\textcolor{darkorange}{78.54}}}

\textcolor{darkorange}{78.54}=1.5\textcolor{darkviolet}{S_\text{normal}}

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Solve for \textcolor{darkviolet}{S_\text{normal}}

RearrangeEqn

Rearrange equation

1.5\textcolor{darkviolet}{S_\text{normal}}=78.54

DivEqn

.LHS /1.5.=.RHS /1.5.

\textcolor{darkviolet}{S_\text{normal}}=\dfrac{78.54}{1.5}