Volume of revolution of region bounded by $y = 1/x, y = 0, x = 1/2, x = b$ about $y$-axis where $b \gt 1/2$.
This is a easy question, but I have a problem with the formula. If I consider the graph to be split into 3 parts, then there are three regions to take the volume of, and the equation I get is,
$$\text{volume} = \pi \int (b^2 -(1/y)^2 - (1/2)^2) dy$$
But what would the range of the integral be? 
I have added a picture of the graph:


You need to separate it into two parts : 
$$V=\int_{\frac 1b}^{2}\left(\pi\left(\frac 1y\right)^2-\pi\left(\frac 12\right)^2\right)dy+\int_{0}^{\frac 1b}\left(\pi b^2-\pi\left(\frac 12\right)^2\right)dy$$
The volume can be expressed as
$$V=\int_{\frac 1b}^{2}\pi\left(\frac 1y\right)^2dy+\int_{0}^{\frac 1b}\pi b^2dy-\int_{0}^{2}\pi\left(\frac 12\right)^2dy$$
