Use symmetry for a half-turn and translate the rotation axis

The Teacher emphasises that $e^{x^2}$ is even and $R$ occupies two quadrants symmetrically about the y-axis. At height $y\in[1,4]$, the horizontal strip runs from $-\sqrt{\ln y}$ to $\sqrt{\ln y}$. Rotating this whole strip through $180^\circ$ sweeps a complete disc: its positive and negative halves sweep the two complementary semicircles. Thus there is no extra factor $1/2$.

Since $x^2=\ln y$, $V=\pi\int_1^4x^2\,dy=\pi\int_1^4\ln y\,dy$. Using integration by parts, $V=\pi[(y\ln y)_1^4-\int_1^4(1/y)y\,dy]=\pi(4\ln4)-\pi[y]_1^4=\pi(4\ln4-3)$ cubic units.

The translated curve is $y=e^{x^2}-4$. Moving the whole region and its axis down by 4 preserves the generated volume: rotation of the original region about $y=4$ becomes rotation of the translated region about the x-axis. Its x-intercepts are $\pm a$, where $a=\sqrt{\ln4}\approx1.177410023$; its minimum is $-3$.

$V=\pi\int_{-a}^{a}(e^{x^2}-4)^2\,dx$. The source uses bounds approximately $\pm1.17741$ and gives $42.6$ cubic units to 3 significant figures. The squared function is even, so the Teacher also writes $V=2\pi\int_0^a(e^{x^2}-4)^2\,dx$. An independent numerical check with the exact bound gives approximately $42.56317594$, agreeing with the printed answer.