Question

Let R be the region in the first quadrant bounded by x2 − y2 = 16, the horizontal line y = 0, and the vertical line x = 5. What is the volume of the solid formed by revolving R about the x-axis?

Answer 1

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.

`x^2-y^2=16\implies x^2-16=y^2\implies \pm√(x^2-16)=y \\\\\\ \stackrel{I~Quadrant}{+√(x^2-16)=y}\qquad \qquad \implies \stackrel{\textit{using the disk method}~\hfill }{\displaystyle\int_(4)^(5)~\pi \left( √(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}`