SOLUTION 24: The grapefruit has a diameter of $10$ inches, a radius of $5$ inches, and a peel $0.25$ inches thick. Let's assume that a grapefruit is a sphere, and recall that the volume of a sphere of radius $r$ is
$$ V = \displaystyle{ 4 \over 3 } \pi r^3 $$
and choose
$$ r-values: 5 \rightarrow 5+(0.25)= 5.25 $$
so that
$$ \Delta r = 5.25-5 = 0.25 $$
The derivative of $ V $ is
$$ V'= \displaystyle{ 4 \over 3 } \cdot 3 \pi r^2 = 4 \pi r^2 $$
$ \ \ \ \ $ a.) The exact volume of the peel is the exact change of $V-$values and is
$$ \Delta V = V(5.25) - V(5) $$
$$ = \displaystyle{ 4 \over 3 } \pi (5.25)^3 - \displaystyle{ 4 \over 3 } \pi (5)^3 $$
$$ = \displaystyle{ 4 \over 3 } \pi (144.703125) - \displaystyle{ 4 \over 3 } \pi (125) $$
$$ = \displaystyle{ 578.8125 \over 3 } \pi - \displaystyle{ 500 \over 3 } \pi $$
$$ = \displaystyle{ 78.8125 \over 3 } \pi $$
$$ \approx 82.53 \ in^3 $$
$ \ \ \ \ $ b.) An estimate for volume of the peel is the Differential of $V$ (since $ \Delta V \approx dV)$ and is
$$ dV = V'(5) \ \Delta r $$
$$ = 4 \pi (5)^2 \cdot (0.25) $$
$$ = 25 \pi $$
$$ \approx 78.54 \ in^3 $$
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