Use different regions for area and revolution

The source graphing-calculator intersection is $(0.62389,0.71981)$. Denote its coordinates by $(a,b)$. Keep full calculator precision until the final answer; the source displays rounded limits. Although the printed domain includes both endpoints, $\tan x$ is undefined at $\pm\pi/2$: these are vertical asymptotes and the actual tangent domain here is $-\pi/2<x<\pi/2$.

For $a\le x\le1$, $\tan x$ is above $1/(1+x^2)$. The Teacher fills $A=\int_{0.62389}^1[\tan x-1/(1+x^2)]\,dx=0.179$ square units to three significant figures. Independently, an antiderivative is $-\ln(\cos x)-\tan^{-1}x$ on this interval.

This region lies to the left of the intersection, between the two curves and the y-axis. Rewrite the curves as $x=\tan^{-1}y$ and $x^2=1/y-1$. For $0\le y\le b$, the disc radius is $\tan^{-1}y$; for $b\le y\le1$, it is $\sqrt{1/y-1}$. The Teacher fills $V=\pi\int_0^{0.71981}(\tan^{-1}y)^2\,dy+\pi\int_{0.71981}^1(1/y-1)\,dy=0.480$ cubic units to three significant figures. These are adjacent stacks of discs with inner radius zero, so add their volumes.