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
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 if I compute the definite integral keeping in mind that $a$ is a constant, I get that the volume would be $\frac {8 \pi}{a^3}$.
For B), the indefinite integral of the volume would be $\frac {4y}{a^2} + \frac {2y-y^2}{2a} + c$. So, to find the minima and maxima, I have to differentiate it and after doing so I got $\frac {4}{a^2} + \frac {1-y}{a}$, which is a straight line so it only has one maximun. Is it ok?