\input amstex
\input cyracc.def
\font\tencyr=wncyr10
\def\cyr#1{{\tencyr\cyracc#1}}
\def\sgn{\,\mathrm{sgn}\,}
\def\dd{\,\mathrm{d}}
\def\sledi{\;\Longrightarrow\;}

$$z=\pm\sqrt{x^2+y^2+1}\;\Longrightarrow\;z\in\Bigl[-\sqrt{x^2+y^2+1},\sqrt{x^2+y^2+1}\,\Bigr]$$

$$\align
T&=\biggl\{(x,y,z)\biggm|z\in\Bigl[-\sqrt{x^2+y^2+1},\sqrt{x^2+y^2+1}\,\Bigr]\;\wedge\;x^2+y^2\in[0,4]\biggr\}\cr
D&=\biggl\{(x,y)\biggm|x^2+y^2\in[0,4]\biggr\}
\endalign$$

$$\align
V(T)&=\iiint_T\dd x\dd y\dd z=\iint_D\Bigl\{\int_{-\sqrt{x^2+y^2+1}}^{\sqrt{x^2+y^2+1}}\dd z\Bigr\}\dd x\dd y=2\iint_D\sqrt{x^2+y^2+1}\dd x\dd y=\dotsb\cr
&\fbox{$\displaystyle
x=\rho\cos\theta$,
$\displaystyle\quad
y=\rho\sin\theta$,
$\displaystyle\quad
x^2+y^2\leq4\sledi\rho^2\leq4\sledi\rho\leq2$,
$\displaystyle\quad
\theta\in[-\pi, \pi]$,
$\displaystyle\quad
J=\rho$}\cr
\dotsb&=2\int_0^2\Bigl\{\int_{-\pi}^{\pi}\rho\sqrt{\rho^2+1}\dd\theta\Bigr\}\dd\rho=4\pi\int_0^2\rho\sqrt{\rho^2+1}\dd\rho=\dotsb\cr
&\fbox{$\displaystyle
\rho^2+1=t$,
$\displaystyle\quad
\rho\dd\rho=\frac{1}{2}\dd t$,
$\displaystyle\quad
\varphi(0)=1$,
$\displaystyle\quad
\varphi(2)=5$}\cr
\dotsb&=2\pi\int_1^5\sqrt{t}\dd t=\frac{4}{3}\pi(5^{3/2}-1)\approx42{,}6433.
\endalign$$