Stuck on volume of a solid of revolution The problem I've been working on is where the curves are y=sqrt(x), y=1/x, and x=5, which are rotated around the y-axis. I am able to do this problem when rotating around the x-axis, but I have no clue how to calculate the inner radius when neither curve allows for me to determine the inner radius. I just need an idea of how to start this problem. I do not know how to post a graph or something here or if that's even allowed for a visual picture, so sorry about that! Thanks!
 A: $2π\int_1^5x(\sqrt x-1/x)\rm dx=2π[2/5x^{5/2}-x]_1^5=2π((10\sqrt5-5)-(-3/5))=2π(10\sqrt5-22/5)=(20\sqrt5-44/5)π$, using cylindrical shells as @TobyMak suggests.
A: I find it often helps to make a graph of the equations, even when I think I know what the inner and outer radius will be.
For the three equations $y= \sqrt x,$ $y = 1/x,$ $x = 5$ I get the region that is shaded in the figure below.

Now your problem, if you want to use the disk/washer method, is that the curves $y=\sqrt x$ and $y= 1/x$ are each closer to the $y$ axis than the curve $x = 5,$ but neither curve is the inner radius for the whole region. If you draw a horizontal line at $y=1/2$ the intersection with the region starts at $y=1/x$ and ends at $x=5$, but if you draw a horizontal line at $y=2$ the intersection starts at $y=\sqrt x$  instead.
One thing you can do in a case like this is to break the region in two pieces.

Now the lighter shaded region is bounded by $y=1/x,$ $x=5,$ and $y=1$, and the darker shaded region is bounded by $y=\sqrt x,$ $x=5,$ and $y=1$.
You can work out the volumes swept by each of the smaller regions and add them together to get the total volume swept by the whole region.
Alternatively, you can use the shell method as suggested by others.
This lets you compute the volume using just one integral.
Eventually, however, you will probably encounter a problem that cannot be done in just one integral by either the shell method or the washer method.
