It’s easy to lose sight of reality when you are wrapped up in the middle of a calculus problem. For instance, I once asked a student to calculate the volume of some solid which depended on two lengths, a and b. After completing a very nasty integral, she happily proclaimed that the answer was some constant times a squared times b squared. To her great chagrin, instead of congratulating her on her nice work, I simply asked her what dimensions volume were in, and what dimensions her answer was in.
This sort of dimensional analysis can be extremely useful in many areas of calculus. Most of the time, it’s relatively straightforward, however there are a number of things to watch out for.
Integrals add dimension. This is obvious in some cases, but not in others. For instance, when computing a volume using a shell method, you integrate pi * r^2. The r^2 is in square units, but when multiplied by dx (or dy), which has linear units, and integrating, you end up with cubic units (i.e. volume) as expected.
Derivatives lose a dimension. Algebraically, this is very clear, since every calculus student knows, for instance, that the derivative of x cubed is three x squared (the first being in units cubed, the latter in square units). It takes a bit more thinking to see why this is true from a physical perspective. Differentials are useful; for instance, if you increase the size of a sphere just slightly, dV/dr = (4/3 pi r^3)’ = 4 pi r^2, so in differentials, dV = 4 pi r^2 dr. The units in this equation make sense, so working backward, it was important to lose a dimension when we took the derivative of volume to get 4 pi r^2 (surface area).
– Dave Kung 1998-08-09