If we have the $y=\sec^{-1}x$ and we want to rewrite it as $x=\sec y$ for the purpose of finding the volume using the cylindrical shell technique due to the revolution of the region bounded by $y=\sec^{-1}x$ and $x=2$ about $x=-1$.
Now to calculate the volume of this curve we will have to know the equation of the curve in the $1^{st}$ quad (the one below) so how do I generally find whether it's positive or negative?
Although not sketched, the curve has both positive and negative branches symmetric to x-axis which becomes positive in squaring for volume in $\int \pi y^2 dx$. The closed area to the left of thicker line is required to be swept out.
For convenience to find volume interchange the axes and shift along new displaced $y$ axis by 1 unit. $x$ limits are $\pm \sec^{-1}2$
$$ V= \pi \int_{{-\sec^{-1}2}}^{\sec^{-1}2} (1+ \sec x)^2 dx $$
So $V$ is addition of indefinite integrals
$$\int dx,\; \int \sec^2x\; dx,\;\int 2 \sec x . $$