Find the volume rotating the region enclosed by the curves $xy=1$, $x = y^{1/2}$, and $y = 2$ about the $y$-axis 
Find the volume rotating the region enclosed by the curves $xy=1$, $x = y^{1/2}$, and $y = 2$ about the $y$-axis

I've looked up solutions but nothing looks like my problem. 
I did draw a picture. I've never seen a problem like this because our professor never did one, so I'm not even sure how to approach the problem.
 A: It is difficult for me to produce a diagram for you, but a diagram is critical. Draw the hyperbola $xy=1$ (the part in the first quadrant will be enough). Draw the curve $x=y^{1/2}$. Note that this is the first quadrant part of the familiar parabola $y=x^2$. The two curves meet at the point $(1,1)$. 
Now draw the line $y=2$. The region we are rotating is below the line $y=2$, to the right of $xy=1$, and to the left of the parabola $y=x^2$. It is a very roughly triangular region, except that two sides of the "triangle" are curvy. 
If you have trouble with the graph, perhaps Wolfram Alpha, or another graphing program, or a graphing calculator, would help, but the diagram is really not difficult. 
Now imagine rotating this region about the $y$-axis. We get a solid with a "hole" in it. 
A cross-section of the solid at $y$ is a "washer," a circle with a circular hole in it. The outer radius of the washer is given by the "$x$" of the parabola, that is, by $y^{1/2}$. The inner radius is the $x$ of the hyperbola, it is $\frac{1}{y}$. So the area of cross-section at $y$ is
$$\pi\left((y^{1/2})^2-\frac{1}{y^2}\right).$$
For the volume, "add up" (integrate) from $y=1$ to $y=2$. The integration will be very straightforward. 
