Math 1450H, Frequently Asked Questions:
 Y. L. asks: I have a question about question 37 on ch 2.4 (p 118) and was wondering if you could explain it? I think I pretty much get the logic of most of the steps, but I was wondering how 1/2 a was chosen as a restriction ( for the step reasoning  xa < 1/2 a ) ?
See additional explanations and
two possible
solutions to this problem.
 J. R. contributed a neat diagram called the "house of calculus" that relates the behavior
of a function with its derivatives and helps remember the derivative tests.
 B. N. helped us memorize trig formulas.
 B.N. asked whether the solution to 8.2 Exercise 1 (b) is correct as stated.
This part of the problem asks to find the surface area of the object generated by revolving
the graph of y=x^{4} about the yaxis. The standard form of the answer would
be the integral of 2*pi*y^{1/4} (1+ y^{3/2}/16)^{1/2} from y=0 to y=1.
Now substituting x=y^{1/4}, dx=(1/4)y^{3/4}dy, and eliminating the y
so that 4 x^{3} dx = dy then gives that the answer is indeed the integral
of 2*pi*x*(1+16 x^{6})^{1/2} from x=0 to x=1.
The deeper reason for this is that ds=(1+(dx/dy)^{2})^{1/2}dy OR
ds=(1+(dy/dx)^{2})^{1/2}dx and the integral of the surface area
is $\int $_{s(0)}^{s(1)}
2*pi*x(y(s))ds
OR the integral
$\int $_{s(0)}^{s(1)} 2*pi*x(s)ds.
 For anyone who wants a bit more information than the terse explanation of a proof by induction in Stewart's book,
see more illustrative examples.

