# Stuck on a solid of rev problem

• Feb 12th 2013, 12:21 PM
AZach
Stuck on a solid of rev problem
The problem is: Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. $y=1/x^2$ ,
$y = 0$, $x=1$, $x=5$ about $y=-3$

This is the image I have (although mine's on paper and it shows the washer's depth decreasing) Attachment 26982

I did $V = \pi \int_{1}^{5} (\frac{1}{x^2})^2-(3)^2 dx$

I've fiddled around with that a few times, but I feel like I'm now starting to go in circles. I'm hoping I could get an eyeball so I can see what I'm not visualizing correctly. I know that I need the area between y=0 and y, and I assumed the bounds were x=1 and x=5, so I figured I had enough to take the integral with that information.
• Feb 12th 2013, 12:26 PM
AZach
Re: Stuck on a solid of rev problem
Is the radius of the inner circle for this problem $r = y -3$ ?
• Feb 12th 2013, 12:42 PM
bkbowser
Re: Stuck on a solid of rev problem
Quote:

Originally Posted by AZach
Is the radius of the inner circle for this problem $r = y -3$ ?

The inner circle is a constant. $\pi(3)^2*4$ where 3 is the radius, and 4=5-1 is the height.

I think you can shift the curve 1/x^2 upwards 3 units. Find the volume of this new curve and then subtract the volume of the cylinder defining the inner circle.
• Feb 12th 2013, 07:10 PM
hollywood
Re: Stuck on a solid of rev problem
You can approach it more directly, but you have to make sure you put in the *radius*, not the x value:

outer radius = $y+3 = \frac{1}{x^2}+3$
inner radius = $3$

Area of "washer" = $\pi \left(\frac{1}{x^2}+3\right)^2 - \pi (3)^2$

$V = \pi \int_{1}^{5} \left(\frac{1}{x^2}+3\right)^2-(3)^2\,dx$

- Hollywood