Originally Posted by
Soroban Hello, xje4bv!
If these "differential approximations" are confusing to you, don't give up.
It took me the looongest time to understand them.
We have a sphere with radius 15 cm.
It has a volume according to the formula: .$\displaystyle V \:=\:\frac{4}{3}\pi r^3$
By adding a layer of paint, the radius is increased by $\displaystyle 4\text{ mm } = 0.4\text{ cm}.$
The question is: by how much is the volume increased?
We could do the actual arithmetic,
. . but we can approximate the answer with differentials.
We have: .$\displaystyle V \:=\:\frac{4}{3}\pi r^3$
Take differentials: .$\displaystyle dV \:=\:4\pi r^2\,dr$
If we understand what each part means, we're all set!
. . $\displaystyle \underbrace{dV}_{\text{change in volume}} \;=\;4\pi\cdot\underbrace{r^2}_{\text{radius}^2}\c dot\underbrace{dr}_{\text{change in radius}} $
We know: .$\displaystyle r = 15,\;dr = 0.4$
Therefore: . $\displaystyle dV \;=\;4\pi(15^2)(0.4) \;=\;360\pi\text{ cm}^3$
The volume of the paint needed is approximately 1131 cm³