Given $D=[{(x;y) \in \mathbb R^2 : 1 \le y \le ax^2 +1, 0\le x \le 2/a}], 0 \lt a$, let $W$ be the region obtained by rotating $D$ around $Y$ axis.
A) Find the volume of $W$
B) Find, if possible, the $a$ values $\epsilon (0; + \infty)$ so that the volume of $W$ is a minimum, and a maximum
C) Find, if possible, the $a$ values $\epsilon (1/3; 3)$ so that the volume of $W$ is a minimum, and a maximum
Well, what I've done so far is finding the inner and outer radius to calculate the volume in terms of $Y$. The inner radius would be be $\sqrt{\frac{y-1}{a}}$ and the outer would be $\frac{4+a}{a}$, which is $a$ evaluated in the parabola. Then, the volume would be
$\int_1^\frac{4+a}{a} (\sqrt{\frac{y-1}{a}})^2 -(\frac{4+a}{a})^2 \,dx$
And that's the function that I have to differentiate to find its maxima and minima, which, after differentiating, is $1/a$. Is ok what am I doing? How can I go on?
The smallest value of $y$ is $1$ and its largest value is $f\left(\frac2a\right)=\frac{4+a}a$. For each $y$ in that interval, the possible values of $x$ go from $\sqrt{\frac{y-1}a}$ to $\frac2a$. Therefore, that volume is equal to$$\pi\int_1^{\frac{4+a}a}\left(\frac2a\right)^2-\left(\sqrt{\frac{y-1}a}\right)^2\,\mathrm dy=\pi\left(\frac{8}{a^3}+\frac{4}{a^2}+\frac{1}{2a}\right).$$This is a strictly decreasing function of $a$. Can you take it from here?