$1(1)2.$ The plane region $R$ is in the first quadrant and is under the rectangular hyperbola $x^2-r^2-1$ over the interval $1,x.$

a$)$ The solid $S$ is obtained by revolving $R$ about the $x-$axis. Find the volume of $S$ using washers.

b$)$ The solid $S$ is obtained by revolving $R$ about the $-$axis.

Find the volume of $S$ using washers.

$S^2-t^2=1=>S=\sqrt[\phantom{2}]{1+t^2},$ radius $=\sqrt[\phantom{2}]{1+t^2}$

$($a$)$

$dv=(\sqrt[\phantom{2}]{1+t^2}{)}^{2}dt$

$=(1+t^2)dt$

$v={\displaystyle \underset{1}{\overset{x}{}}}(1+t^2)dt=>$

$=[(x+\frac{x^3}{3})-\frac{4}{3}]$

$=>x(\frac{1+t^3}{3}){|}_1^x||$

$=>[(x+\frac{x^3}{3})-(1+\frac{1}{3})]$

$($b$)$ Revoluing arownd $y-$axis $,z=1\sqrt[\phantom{2}]{z},s=\sqrt[\phantom{2}]{1+x^2}$

$t=\sqrt[\phantom{2}]{s^2-1}$

$dv=(\sqrt[\phantom{2}]{s^2-1}{)}^{2}ds$

$=(s^2-1)ds$

$v={\displaystyle \underset{\sqrt[\phantom{2}]{2}}{\overset{\sqrt[\phantom{2}]{1+x^2}}{}}}(s^2-1)ds$

I got this answer wrong so this is not the answer. please provide me with the correct solution