1
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ What have you tried yourself (other than trying to look up a solution)? Start by drawing a picture. $\endgroup$ – rogerl Oct 13 '14 at 0:02
  • $\begingroup$ 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. $\endgroup$ – Katie Oct 13 '14 at 0:08
  • $\begingroup$ Did you set up the definite integral? $\endgroup$ – HDE 226868 Oct 13 '14 at 0:10
  • $\begingroup$ What methods do you know for determining the volume of a solid of revolution? $\endgroup$ – rogerl Oct 13 '14 at 0:11
  • $\begingroup$ The simpler problems from the homework involve using the equation antiderivative of pi(f(x)^2). $\endgroup$ – Katie Oct 13 '14 at 0:13
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.