Truncated Right Hexagonal Prism!

The average length of the lateral edges is $\dfrac{3a + 2}{3}$ .

Let $O':(0,0,0)$ be center of the hexagonal base and erect a normal at $O'$ so that the top surface is centered at $O: (0,0,\dfrac{3a + 2}{3})$ .

Using coordinates for the surface we have:

$F: (\dfrac{a}{2},-\dfrac{\sqrt{3}}{2}a,a + 1), A:(a,0,a + 1), B:(\dfrac{a}{2},\dfrac{\sqrt{3}}{2}a,a + 1)$

$C:(-\dfrac{a}{2},\dfrac{\sqrt{3}}{2}a,a + 1), D:(-a,0,a)$ and $E:(-\dfrac{a}{2},-\dfrac{\sqrt{3}}{2}a,a)$

$\implies FA = AB = BC = DE = a$ and $CD = EF = \sqrt{a^2 + 1}$

and the radii $OF =OA = OB = OC = \dfrac{\sqrt{9a^2 + 1}}{3}$ and $OD = OE = \dfrac{\sqrt{9a^2 + 4}}{3}$

$h = \dfrac{\sqrt{27a^2 + 4}}{6} \implies A_{\triangle{FOA}} = \dfrac{\sqrt{27a^2 + 4}}{12}a = A_{\triangle{AOB}} =$ $A_{\triangle{BOC}}$ .

$h^{*} = \dfrac{\sqrt{27a^2 + 16}}{6} \implies A_{\triangle{DOE}} = \dfrac{\sqrt{27a^2 + 16}}{12}a$ .

$|\vec{U} X \vec{V}| = |\vec{U}| |\vec{V}|\sin(\theta) = |\vec{U}| d \implies$ $d = \dfrac{|\vec{U} X \vec{V}|}{ |\vec{U}|}$

$\vec{U} =a \vec{i} + 0\vec{j} + \vec{k}$

$\vec{V} = \dfrac{a}{2}\vec{i} + \dfrac{\sqrt{3}}{2}a\vec{j} + \dfrac{2}{3}\vec{k}$

$\implies$

$\vec{U} X \vec{V} = -\dfrac{\sqrt{3}}{2}a\vec{i} - \dfrac{a}{6}\vec{j} + \dfrac{\sqrt{3}}{2}a^2\vec{k}$

$\implies |\vec{U} X \vec{V}| = \dfrac{\sqrt{27a^2 + 28}}{6}a$ and $|\vec{U}| = \sqrt{1 + a^2} \implies$ $d = \dfrac{1}{6}\sqrt{\dfrac{27a^2 + 28}{1 + a^2}}a \implies$ $A_{\triangle{COD}} = A_{\triangle{EOF}} =$

$\dfrac{1}{2}(\sqrt{1 + a^2})(\dfrac{1}{6})(\sqrt{\dfrac{27a^2 + 28}{1 + a^2}})a = \dfrac{\sqrt{27a^2 + 28}}{12}a$

$\implies A_{surface} = \dfrac{2a\sqrt{27a^2 + 28} + 3a\sqrt{27a^2 + 4} + a\sqrt{27a^2 + 16}}{12}$

$A_{hexagonal base} = 6(\dfrac{1}{2})(a)(\dfrac{\sqrt{3}}{2})a = \dfrac{3\sqrt{3}}{2}a^2$

For the area of the trapezoids we have:

$A_{FAF'A'} = A_{ABA'B'} = A_{BCB'C'} = \dfrac{1}{2}(2)(a + 1)(a) = (a + 1)a$

and

$A_{CDC'D'} = A_{EFE'F'} = \dfrac{1}{2}(2a + 1)a = \dfrac{2a + 1}{2}a$

and $A_{DED'E'} = \dfrac{1}{2}(2a)a = a^2$

$\implies$ The total area of the trapezoids $A_{T} = 3a^2 + 3a + 2a^2 + a + a^2 = 6a^2 + 4a$

$\implies$

$S(a) = \dfrac{2a\sqrt{27a^2 + 28} + 3a\sqrt{27a^2 + 4} + a\sqrt{27a^2 + 16}}{12} + 3\dfrac{\sqrt{3}}{2}a^2 + 6a^2 + 4a =$

$\dfrac{2a\sqrt{27a^2 + 28} + 3a\sqrt{27a^2 + 4} + a\sqrt{27a^2 + 16} + 18\sqrt{3}a^2 + 72a^2 + 48a}{12}$

$\implies$

$\displaystyle\int_{0}^{1} S(a) \:\ da =$

$\dfrac{1}{12}(\dfrac{2}{81}(55)^{\frac{3}{2}} + \dfrac{1}{27}(31)^{\frac{3}{2}} + \dfrac{1}{81}(43)^{\frac{3}{2}} + 6\sqrt{3} + 48 - \dfrac{112\sqrt{7}}{81} - \dfrac{8}{27} - \dfrac{64}{81} )$

$\approx \boxed{6.13272}$ .