Volume with Disc Method: Revolving Around the x- or y-Axis AB & BC

Revolving a region makes every cross section a circle, so the disc method is 8.7's integral with $A = \pi r^{2}$: $V = \pi\int_a^b (f(x))^{2}\,dx$ about the $x$-axis, and $\pi\int_c^d (g(y))^{2}\,dy$ about the $y$-axis after solving the boundary for $x$ in terms of $y$. For $y = \sqrt{x}$ on $[0, 4]$ this gives $8\pi$. Dropping $\pi$ gives $8$, dropping the square gives $\frac{16\pi}{3}$, and squaring after integrating gives $\frac{256\pi}{9}$, since $\int f^{2}$ and $\left(\int f\right)^{2}$ are different quantities.

The radius is the distance from the axis of revolution to the curve, and $r = f(x)$ is the special case in which that axis is $y = 0$. Because a distance is squared, a region below the axis gives the same solid as its reflection. A disc assumes the slice has no hole, which holds exactly when the region touches the axis: the region under $y = \sqrt{x}$ does, while the region between $y = \sqrt{x}$ and $y = 1$ does not, and revolving the second one about the $x$-axis leaves a cylindrical hole that a disc integral would silently fill in.