Question 1015808
Easiest may be to integrate in terms of dy using washers. Note that by integrating along y, we have to split into two integrals, one of which is determined by x=1 and x=1/y, and the other is determined by x=1 and x=2. The volume is


*[tex \large V = \pi \int_{\frac{1}{2}}^1 \left( \frac{1}{y} \right)^2 - (1)^2 \, \textrm{d}y + \pi \int_{0}^{\frac{1}{2}} (2)^2 - (1)^2 \, \textrm{d}y]


The second integral can be done without calculus, as it is the difference of two volumes of cylinders. I'll leave it to you to compute the integrals.