# Homework Help: The volume of a solid rotating about a different axis

1. Oct 5, 2012

### atomibay

1. The problem statement, all variables and given/known data
$y = {\frac{1}{4+x^2}}$ on the interval [0,2], revolving about y = -1
Use either the disk/washer or shell method to find the volume.

2. Relevant equations
$v = pi\int (outer radius)^2-(inner radius)^2\,dx$
$v = 2pi\int (radius)(height)\,dy$
$x = \sqrt{\frac{1}{y}-4}$

3. The attempt at a solution
$v = 2pi\int (y+1) \sqrt{\frac{1}{y}-4}\,dy$ from $\frac{1}{4}$ to $\frac{1}{8}$

I'm just stuck on setting up the integral. I get confused easily from these washer/shell problems, and it's worse when the axis changes haha. So I don't know if this integral is set up correctly. And, I feel like there's something off about my limits. Do I have to add another integral to integrate from 0? Or would it just be easier to use the washer method?

2. Oct 5, 2012

### gabbagabbahey

If I were you, I'd start by plotting y(x) for the given interval... can you imagine what it wouldlook like as your rotate it about y=-1? What would be the radius of a disk at x? What would be the area of the disk?

3. Oct 5, 2012

### atomibay

Yeah, I've tried graphing it but even while I was looking at it, I'm still confused as to how to read the inner radius and the outer radius.
I figured that the inner radius is (y+1) and the outer is $\frac{1}{4+x^2}$, right?
I'm not sure about the area though.

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4. Oct 5, 2012

### gabbagabbahey

Why have you only shaded in part of the area between y(x) and y=-1? Does the problem tel you to rotate onlythe areabetween y(x) and the x-axis around y=-1? Ifnot, I would say your inner radius is zeroand your outer radius is |y(x)|

5. Oct 5, 2012

### atomibay

Well, the problem didn't specify anything like that. All I received was the equation, the interval and what axis it's rotating about, so I just presumed that's what the graph would look like for the problem.

I'll try working it out with inner radius as 0 though, thanks!