Question

The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method. y2 − x2 = 1, y = 2; about the x-axis

Answer 1

The bounded region we're concerned with lies above the x-axis.

$y^2-x^2=1\implies y=\sqrt{1+x^2}$

This hyperbola intersects with the line $y=2$ for

$2=\sqrt{1+x^2}\implies4=1+x^2\implies x^2=3\implies x=\pm\sqrt3$

Using the washer method, the volume of the resulting solid is

$\displaystyle\pi\int_{-\sqrt3}^3(2^2-(\sqrt{1+x^2})^2)\,\mathrm dx=2\pi\int_0^{\sqrt3}(3-x^2)\,\mathrm dx=\boxed{4\pi\sqrt3}$