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
$$dV = V'(5) \ \Delta r$$ $$= 4 \pi (5)^2 \cdot (0.25)$$ $$= 25 \pi$$ $$\approx 78.54 \ in^3$$