I have this question due, but I don't know how to solve it. Can someone please help me?
What have you tried?
(a) Let's set it up. We are using volume by cylindrical shell. So, the volume is:
$\displaystyle V = \int_0^a 2\pi x y dx = \int_0^a 2\pi xf(x) dx$
Then the hint is $u =y= f(x), du = f'(x)dx = -xf(x)dx = -xudx$
So, $-xdx = \dfrac{du}{u}$ gives:
$\displaystyle V = \int_{f(0)}^{f(a)} -2\pi du = 2\pi \int_{f(a)}^1 du = 2\pi (1-f(a))$
(b) the diameter is 2a. So, if a approaches infinity, so too does 2a.
(c) As a approaches infinity, the volume approaches $2\pi$, as $f(a)$ approaches zero.