Region bounded by $$x^2-y^2=16, y=0, x=8$$, about y-axis
Integrating with respect to y, my lower and upper bounds are $-4\sqrt{3}$ and $+4\sqrt{3}$, respectively.
$$V=\pi \int_{-4\sqrt{3}}^{4\sqrt{3}} 64-(16+y^2)\,dy$$
$$=2\pi \int_{0}^{4\sqrt{3}} 48-y^2\,dy$$
$$=2\pi (192\sqrt{3})$$...