I have a question that says:
Write a differential formula of a sphere when the radius of a balloon goes from r_{0} to _{r0}+ dr.
Can you walk me through this?
For a cube:
The volume of a cube of edge $\displaystyle a$ is $\displaystyle V=a^3$
Thus the change in volume $\displaystyle \mathrm dV$ as the edge goes from $\displaystyle a$ to $\displaystyle a + \mathrm da$ is given by:
$\displaystyle \mathrm dV = (a + \mathrm da)^3 - a^3$
Expanding and simplifying we get
$\displaystyle \mathrm dV = 3a^2\mathrm da + 3a(\mathrm da)^2 + (\mathrm da)^3 $
Ignoring non-linear terms of differentials (squares, cubes, etc.) we end up with
$\displaystyle \mathrm dV = 3a^2\mathrm da$
When the radius of the sphere "goes from r to r+ dr" the volume goes from $\displaystyle \frac{4}{3}\pi r^3$ to $\displaystyle \frac{4}{3}\pi (r+ dr)^3= \frac{4}{3}\pi(r^3+ 3r^2dr+3rdr^2+ dr^3)$. The difference is $\displaystyle \Delta y= \frac{4}{3}\pi(3r^2 dr+ 3r dr^2+ dr^3)$. If we take dr sufficiently small that we can ignore the much smaller $\displaystyle dr^2$ and $\displaystyle dr^3$ that would be the differential $\displaystyle dy= 3r^2 dr$.