# 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.

• What have you tried yourself (other than trying to look up a solution)? Start by drawing a picture. – rogerl Oct 13 '14 at 0:02
• 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. – Katie Oct 13 '14 at 0:08
• Did you set up the definite integral? – HDE 226868 Oct 13 '14 at 0:10
• What methods do you know for determining the volume of a solid of revolution? – rogerl Oct 13 '14 at 0:11
• The simpler problems from the homework involve using the equation antiderivative of pi(f(x)^2). – Katie Oct 13 '14 at 0:13

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.
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.