Check the picture below.
so the region is the one bounded by a hyperbola and two straight lines, one of which is the x-axis, or namely y = 0, since the hyperbola has a vertex at x = 4, then our bounds are from 4 to 5.
[tex]x^2-y^2=16\implies x^2-16=y^2\implies \pm\sqrt{x^2-16}=y \\\\\\ \stackrel{I~Quadrant}{+\sqrt{x^2-16}=y}\qquad \qquad \implies \stackrel{\textit{using the disk method}~\hfill }{\displaystyle\int_{4}^{5}~\pi \left( \sqrt{x^2-16} \right)^2\cdot dx\implies \pi \int_{4}^{5}(x^2-16)dx} \\\\\\ \pi \left( \left. \cfrac{x^3}{3} \right]_{4}^{5} ~~ - ~~\left. 16x\cfrac{}{} \right]_{4}^{5}\right)\implies \pi \left[\cfrac{61}{3}~~ - ~~16 \right]\implies \cfrac{13\pi }{3}[/tex]
