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: 568
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!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top