Find the Volume.

The volume $V = \pi\int_{0}^{1} (x^4 + bx^2 + c)^2 dx = \pi\int_{0}^{1} (x^8 + 2bx^6 + (b^2 + 2c)x^4 + 2bc x^2 + c^2)dx$

$= \pi(\dfrac{x^9}{9} + \dfrac{2b}{7}x^7 + \dfrac{b^2 + 2c}{5}x^5 + \dfrac{2bc}{3}x^3 + c^2x)_{0}^{1} =$ $\pi(\dfrac{1}{9} + \dfrac{2b}{7} + \dfrac{b^2 + 2c}{5} + \dfrac{2bc}{3} + c^2)$

$\implies \dfrac{\partial{V}}{\partial{b}} = \pi(\dfrac{2}{7} + \dfrac{2b}{5} + \dfrac{2c}{3}) = 0$ and $\dfrac{\partial{V}}{\partial{c}} = \pi(\dfrac{2}{5} + \dfrac{2b}{3} + 2c) = 0$

$\implies \dfrac{1}{5}b + \dfrac{1}{3}c = \dfrac{-1}{7}$ and $\dfrac{1}{3}b + c = \dfrac{-1}{5}$

Solving the system we obtain $b = \dfrac{-6}{7}$ and $c = \dfrac{3}{35}$ .

$\dfrac{\partial^2{V}}{\partial{b^2}} = \dfrac{2\pi}{5} > 0$ and $\dfrac{\partial^2{V}}{\partial{c^2}} = 2\pi > 0$ and $\dfrac{\partial}{\partial{b}}(\dfrac{\partial{V}}{\partial{c}}) = \dfrac{\partial}{\partial{c}}(\dfrac{\partial{V}}{\partial{b}}) = \dfrac{2\pi}{3}$

The hessian matrix $M= \begin{vmatrix}{\dfrac{2\pi}{5}} && {\dfrac{2\pi}{3}} \\ {\dfrac{2\pi}{3}} && {2\pi}\end{vmatrix}$ and $\det(M) > 0 \implies$ we have a minimum at $(\dfrac{-6}{7},\dfrac{3}{35}) \implies$

$V_{min} = \pi(\dfrac{1225 - 2700 + 1998 -540 + 81}{11025}) = \dfrac{64}{11025}\pi = \dfrac{m}{n}\pi$

$\implies m + n = \boxed{11089}$ .