Volume of Solid

**nick898** · February 28th 2009, 02:21 PM

The problem: The region between y=1/x and the lines x=1/H, y=H, and y=1/H is rotated around the x-axis. Show, using disks/washers that the volume of this solid is

V = pi(H^2-1)^2/H^3

First, started this problem by experimenting with different values of H. I let H=2 and H=3 just to see any patterns. I noticed that the graphs of y=H and y=1/x always intersect at the point x=1/H. I also noticed the graphs of y=1/x intersects y=1/H at H. So I think my limits of integration would be from 1/H to H.

Second, I used washers. It seemed like the solid would look a funnel and I let the outer radius = 1/x and the inner radius = 1/H

Thirdly, adding it all up together I wanted to find pi multiplied by the integral of ((1/x)^2 - (1/H)^2)dx from 1/H to H.

I wanted to fix this up a bit and find common demoninators so I found the integrand to be (H^2-x^2)/(x^2*H^2)

This is where I am stuck at. I think I am on the right track because I used my calculator to find the answer and I think I came up with the right answer. I just don't know how to integrate that function. Maybe I have to put 1/H in terms of x?

**skeeter** · February 28th 2009, 02:43 PM

$V = \pi \int_{\frac{1}{H}}^H \left(\frac{1}{x}\right)^2 - \left(\frac{1}{H}\right)^2 \, dx$

$\pi \left[-\frac{1}{x} - \frac{x}{H^2}\right]_{\frac{1}{H}}^H$

$\pi\left[\left(-\frac{1}{H} - \frac{1}{H}\right) - \left(-H - \frac{1}{H^3}\right)\right]$

$\pi\left[-\frac{2}{H} + \frac{H^4 + 1}{H^3}\right]$

$\pi\left[\frac{H^4 - 2H^2 + 1}{H^3}\right]<br />$

$\frac{\pi(H^2-1)^2}{H^3}$

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