Volume with Disc Method: Revolving Around Other Axes AB & BC

A radius is the distance from the axis of revolution to the boundary, so revolving about $y = k$ gives $r = |f(x) - k|$ and $V = \pi\int_a^b (f(x) - k)^{2}\,dx$. The order of subtraction follows from which side the region lies on, and the square removes the sign either way. Writing $r = f(x)$ is the 8.9 reflex, correct only because the axis was $y = 0$ there; for $y = \sqrt{x}$ and $y = 2$ on $[0, 4]$ about $y = 2$, the true $\frac{8\pi}{3}$ becomes $8\pi$ under that reflex and $\frac{136\pi}{3}$ if the shift is added rather than subtracted.

Because the shift creates a two-term radius, the false expansion returns: $(2 - \sqrt{x})^{2}$ is $4 - 4\sqrt{x} + x$, not $4 - x$. Revolving about a vertical line $x = k$ makes the slices horizontal, so the integral is in $dy$ with $y$-limits and every boundary solved for $x$; a surviving $x$ means the conversion never happened. The check that catches all of it is to evaluate the radius where the region touches the axis, since the radius must vanish there: at $x = 4$ the correct $2 - \sqrt{x}$ gives $0$ while the reflex $\sqrt{x}$ gives $2$.