Volume of Hyperboloid of One Sheet

Use the following substitutions to simplify the expression.

$x=8u$ , $y=9v$ , $z=10w$

$u^2+v^2-w^2=1$

Account for the Jacobian constant.

$J=\frac{\partial (x,y,z)}{\partial (u,v,w)}=\begin{vmatrix} 8 & 0 & 0 \\ 0 & 9 & 0\\ 0& 0 & 10 \end{vmatrix}=720$

Convert to cylindrical coordinates and account for the Jacobian i.e. $J=$ r.

$r^2-w^2=1$

$r=\pm \sqrt{w^2+1}$

Radius is always positive so $0\leq r\leq \sqrt{w^2+1}$ .

$\theta$ is defined for $0\leq \theta \leq 2\pi$ .

Because of the substitution $z=10w$

$0\leq 10w\leq 1$

$0\leq w\leq \frac{1}{10}$

Set up the integral.

$V=\int_{0}^{2\pi }\int_{0}^{\frac{1} {10}}\int_{0}^{\sqrt{w^2+1}}720r{d}r{d}w{d}\theta$

$=720\int_{0}^{2\pi }\int_{0}^{\frac{1}{10}}\left [ \frac{r^2}{2} \right ]_0^{\sqrt{w^2+1}}{d}w{d}\theta$

$=360\int_{0}^{2\pi }\int_{0}^{\frac{1}{10}}(w^2+1){d}w{d}\theta$

$=360\int_{0}^{2\pi }\left [ \frac{w^3}{3}+w \right ]_0^{ \frac{1}{10}}{d}\theta$

$=\frac{903}{25}\int_{0}^{2\pi }{d}\theta$

$=\frac{903}{25}\left [ \theta \right ]_0^{2\pi }$

$=\frac{1806\pi }{25}$

$\therefore A+B=1806+25=\mathbf{1831}$