Find Volume 2

I did the general case: $x^{4n} + bx^{2n} + c$ .

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

$= \pi(\dfrac{x^{8n + 1}}{8n + 1} + \dfrac{2b}{6n + 1}x^{6n + 1} + \dfrac{b^2 + 2c}{4n + 1}x^{4n + 1} + \dfrac{2bc}{2n + 1}x^{2n + 1} + c^2x)_{0}^{1} =$ $\pi(\dfrac{1}{8n + 1} + \dfrac{2b}{6n + 1} + \dfrac{b^2 + 2c}{4n + 1} + \dfrac{2bc}{2n + 1} + c^2)$

$\implies \dfrac{\partial{V}}{\partial{b}} = 2\pi(\dfrac{1}{6n + 1} + \dfrac{b}{4n + 1} + \dfrac{c}{2n + 1}) = 0$ and $\dfrac{\partial{V}}{\partial{c}} = 2\pi(\dfrac{1}{4n + 1} + \dfrac{b}{2n + 1} + c) = 0$

$\implies \dfrac{1}{4n + 1}b + \dfrac{1}{2n + 1}c = \dfrac{-1}{6n + 1}$ and $\dfrac{1}{2n + 1}b + c = \dfrac{-1}{4n + 1}$

Solving the system we obtain $b = \dfrac{-2(2n + 1)}{6n + 1}$ and $c = \dfrac{2n + 1}{(4n + 1)(6n + 1)}$ .

$\dfrac{\partial^2{V}}{\partial{b^2}} = \dfrac{2\pi}{4n + 1} > 0$ and $\dfrac{\partial^2{V}}{\partial{c^2}} = 2\pi > 0$ and $\dfrac{\partial}{\partial{b}}(\dfrac{\partial{V}}{\partial{c}}) = \dfrac{\partial}{\partial{c}}(\dfrac{\partial{V}}{\partial{b}}) = \dfrac{2\pi}{2n + 1}$

The hessian matrix $M= \begin{vmatrix}{\dfrac{2\pi}{4n + 1}} && {\dfrac{2\pi}{2n + 1}} \\ {\dfrac{2\pi}{2n + 1}} && {2\pi}\end{vmatrix}$ and $\det(M) = \dfrac{16n^2\pi^2}{(4n + 1)(2n + 1)^2} > 0 \implies$ we have a minimum at $(\dfrac{-2(2n + 1)}{6n + 1},\dfrac{2n + 1}{(4n + 1)(6n + 1})$

After simplifying we obtain: $V_{n} = \dfrac{64n^4\pi}{(8n + 1)(6n + 1)^2(4n + 1)^2}$ .

Using $n = 10 \implies V_{min} = \dfrac{640000\pi}{81 * 61^2 * 41^2} = \dfrac{640000\pi}{506655081} = \dfrac{m}{n}\pi \implies m + n = \boxed{507295081}$ .