The problem is to find the minimum volume for the solid of revolution around y axes given by the curve
$f(x)=-8ax^2+a+\frac{3}{4}$
My first idea was to find what is the parameter a that minimizes the area in the first quadrant given by integral
$\int_0^{\sqrt{\frac{a+3/4}{8a}}}f(x) dx$
So then the volume of the solid of revolution given by this curve would be minimum. The problem is that my answer doesnt match with the reference =(
$$V=\int_{x_1}^{x_2}2\pi x\,f(x)dx$$ where $f(x)=\frac{4a+3}{4}-8ax^2$ where $a\geq 0$, $x_1=0$ and $x_2=\sqrt{\frac{4a+3}{32a}}$ which is the positive $x$-intercept. Then, $$V=\int_0^{\sqrt{\frac{4a+3}{32a}}}2\pi x(\frac{4a+3}{4}-8ax^2)dx=\frac{(4a+3)^2}{256a}\pi.$$ Minumum volume occurs when $\frac{dV}{da}=0$. The positive solution is $a=\frac34$.