Area of surface of revolution of a cuboid about its space diagonal

Label the vertices of the cuboid $A\;(0,0,0)$ , $B\;(10,0,0)$ , $C\;(0,15,0)$ , $D\;(10,15,0)$ , $E\;(0,0,30)$ , $F\;(10,0,30)$ , $G\;(0,15,30)$ , $H\;(10,15,30)$ . If $0 \le t \le 35$ is the distance along the space axis $AH$ , with $t=0$ corresponding to the point $A$ , let $F_{XY}(t)$ be equal to the perpendicular distance of the line segment $XY$ from the space axis at the point a distance $t$ along the space axis (or be equal to $0$ if no portion of the line segment $XY$ exists at that point. Then $F_{XY}(t) \; = \; \left\{\begin{array}{lll} \sqrt{\left|\mathbf{r}_{XY}(t)\right|^2 - \left(\mathbf{r}_{XY}(t)\cdot \mathbf{n}\right)^2} & \hspace{1cm} & \overrightarrow{AX}\cdot \mathbf{n} \le t \le \overrightarrow{AY}\cdot\mathbf{n} \\ 0 & & \mathrm{o.w.}\end{array}\right.$ where $\mathbf{r}_{XY}(t) \; = \; \frac{1}{\overrightarrow{XY}\cdot{n}}\left[\left(\overrightarrow{AY}\cdot \mathbf{n} - t\right)\overrightarrow{AX} + \left(t - \overrightarrow{AX} \cdot \mathbf{n}\right)\overrightarrow{AY}\right]$ and $\mathbf{n} \; =\; \frac{1}{7}\left(\begin{array}{c} 2 \\ 3 \\ 6 \end{array}\right)$ is a unit vector pointing along the space axis.

If we plot $F_{AB}$ , $F_{AC}$ , $F_{AE}$ , $F_{BD}$ , $F_{BF}$ , $F_{CD}$ , $F_{CG}$ , $F_{EF}$ , $F_{EG}$ , $F_{DH}$ , $F_{FH}$ and $F_{GH}$ against $t$ , we obtain a picture of the profile of the solid of the revolution of the cuboid, in that the maximum of these twelve functions is what is rotated about the $t$ axis to obtain the solid of revolution. Thus the solid of revolution is the volume of revolution of the function $G$ about the $t$ -axis, where $G(t) \; = \; \left\{ \begin{array}{lll} F_{AB}(t) & \hspace{1cm} & 0 \le t \le \tfrac{20}{7} \\ F_{BD}(t) & & \tfrac{20}{7} \le t \le \tfrac{65}{14} \\ F_{AC}(t) & & \tfrac{65}{14} \le t \le \tfrac{45}{7} \\ F_{CD}(t) & & \tfrac{45}{7} \le t \le \tfrac{65}{7} \\ F_{DH}(t) & & \tfrac{65}{7} \le t \le \tfrac{35}{2} \\ G(35-t) & & \tfrac{35}{2} \le t \le 35 \end{array}\right.$

To calculate the surface of revolution, we need to evaluate $\mathcal{S} \; = \; \int_0^{35} 2\pi G(t) \sqrt{1 + G'(t)^2}\,dt$ which is a relatively straightforward piecewise integral, equal to $\frac{1}{504}\left[\begin{array}{l}84240 \pi + 43200 \sqrt{5} \pi + 31000 \sqrt{10} \pi + 69575 \sqrt{13} \pi + 640 \sqrt{394} \pi \\ + 5832 \sqrt{10} \pi \sinh^{-1}(\tfrac13) + 6912 \sqrt{5} \pi \sinh^{-1}(\tfrac12) + 5832 \sqrt{10} \pi \sinh^{-1}(\tfrac{16}{27}) + 6912 \sqrt{5}\pi \sinh^{-1}(2)\end{array}\right]$ which rounds to $\boxed{3669}$ .