The volume of a solid rotating about a different axis

atomibay
Messages
3
Reaction score
0

Homework Statement


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.



Homework Equations


v = pi\int (outer radius)^2-(inner radius)^2\,dx
v = 2pi\int (radius)(height)\,dy
<br /> x = \sqrt{\frac{1}{y}-4}<br />



The Attempt at a Solution


v = 2pi\int (y+1)<br /> \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?
 
Physics news on Phys.org
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?
 
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.
 

Attachments

  • graph.png
    graph.png
    10.1 KB · Views: 565
atomibay said:
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.

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)|
 
gabbagabbahey said:
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)|
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!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top