Calculus Made Easy Chapter 5, example 4 p. (30) [First question] 
In Chapter 5, example 4 of the Calculus Made Easy book I was having trouble understanding why the radius plugged in the formula is r=5.5 (as is correctly stated in the problem) yet when trying to check if the rate of variation is correct, the radius used is r(sub1)= 5 (and r(sub2)=6 but it is understandable since the problem aimed to find the rate of variation when r increases by 1 inch). When I attempted to manually verify the answer, it was correct. I want to understand why in checking, r(sub1) is always the r (used in the formula prior) minus 0.5.
 
 A: Perhaps the simplest, but not necessarily easiest, way to see the reason for using $r - 0.5$ for the lower value is to note that the derivative is a linear function of $r$. As such, the change in volume, given by the integral wrt $r$, over any distance, such as $1$ inch, would be the same as this distance times the rate of change at the middle of the range. This results in using the difference in volumes between $0.5$ in. more and $0.5$ in. less radius.
Another way to see is this is that it's due to the volume having a factor of the radius squared. Consider that, to check how the volume changes over a period of $1$ in. at $r_0 = 5.5$ we have
$$r_1 = r_0 - a, \text{ with } a \gt 0 \tag{1}\label{eq1}$$
$$r_2 = r_0 + b, \text{ with } b \gt 0  \tag{2}\label{eq2}$$
$$a + b = 1 \tag{3}\label{eq3}$$
In other words, $r_1$ and $r_2$ are $2$ radii which are $1$ inch apart on either side of $r_0$. Next, note that
$$V_1 = \pi\left(r_0 - a\right)^2 h = \pi\left(r_0^2 - 2ar_0 + a^2\right) h \tag{4}\label{eq4}$$
$$V_2 = \pi\left(r_1 + b\right)^2 h = \pi\left(r_0^2 + 2br_0 + b^2\right) h \tag{5}\label{eq5}$$
Thus, \eqref{eq5} - \eqref{eq4}, and using \eqref{eq3}, gives
$$V_2 - V_1 = \pi\left(2\left(a + b\right)r_0 + b^2 - a^2\right)h = \pi\left(2r_0 + b^2 - a^2\right)h \tag{6}\label{eq6}$$
Note this is equal to
$$\frac{dV}{dt} = 2\pi rh \tag{7}\label{eq7}$$
at $r = r_0$ only if $b^2 - a^2 = 0$. Thus, this requires that $a = b = \frac{1}{2}$.
Overall, this shows you can use $0.5$ at any value of $r_0$. However, note this only works due to having terms using just $r^2$, $r$ and/or a constant in the original function, so the derivative is a linear function of the radius $r$. If any different power of $r$ was involved, such as $3$ for the volume of a sphere, then usually the difference in volumes at $0.5$ offsets would not be equal to the rate of change in volume at $r_0$.
