# Thread: Stuck on a solid of rev problem

1. ## 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. $\displaystyle y=1/x^2$ ,
$\displaystyle y = 0$, $\displaystyle x=1$,$\displaystyle x=5$ about $\displaystyle y=-3$

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

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

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.

2. ## Re: Stuck on a solid of rev problem

Is the radius of the inner circle for this problem $\displaystyle r = y -3$ ?

3. ## Re: Stuck on a solid of rev problem

Originally Posted by AZach
Is the radius of the inner circle for this problem $\displaystyle r = y -3$ ?
The inner circle is a constant. $\displaystyle \pi(3)^2*4$ 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.

4. ## 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 = $\displaystyle y+3 = \frac{1}{x^2}+3$
inner radius = $\displaystyle 3$

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

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

- Hollywood