The volume of a solid rotating about a different axis

In summary, the homework statement revolving about y=-1 uses either the disk/washer or shell method to find the volume.
  • #1
atomibay
3
0

Homework Statement


[itex]y = {\frac{1}{4+x^2}}[/itex] on the interval [0,2], revolving about y = -1
Use either the disk/washer or shell method to find the volume.



Homework Equations


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



The Attempt at a Solution


[itex]v = 2pi\int (y+1)
\sqrt{\frac{1}{y}-4}\,dy[/itex] from [itex]\frac{1}{4}[/itex] to [itex]\frac{1}{8}[/itex]

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?
 
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  • #2
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
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 [itex]\frac{1}{4+x^2}[/itex], right?
I'm not sure about the area though.
 

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  • #4
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 [itex]\frac{1}{4+x^2}[/itex], 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)|
 
  • #5
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!
 

FAQ: The volume of a solid rotating about a different axis

1. How do you calculate the volume of a solid rotating about a different axis?

The volume of a solid rotating about a different axis can be calculated using the disk method or the shell method. The disk method involves slicing the solid into thin disks perpendicular to the axis of rotation and finding the volume of each disk. The shell method involves slicing the solid into cylindrical shells parallel to the axis of rotation and finding the volume of each shell.

2. What is the difference between the disk method and the shell method?

The main difference between the disk method and the shell method is the orientation of the slices used to calculate the volume. The disk method uses perpendicular slices, while the shell method uses parallel slices. The disk method is typically used for solids with circular cross-sections, while the shell method is used for non-circular cross-sections.

3. Can the volume of a solid rotating about a different axis be negative?

No, the volume of a solid rotating about a different axis cannot be negative. The volume is a measure of the amount of space occupied by a solid, and it is always a positive value. If the calculated volume is negative, it is likely due to an error in the calculations.

4. How does the choice of axis affect the volume of a solid?

The choice of axis can greatly affect the volume of a solid rotating about a different axis. If the axis of rotation is chosen to be on the edge of the solid, the volume will be zero. If the axis is chosen to be through the center of mass of the solid, the volume will be maximized. The volume will also vary depending on the shape and orientation of the solid.

5. Are there any real-life applications of calculating the volume of a solid rotating about a different axis?

Yes, there are many real-life applications of calculating the volume of a solid rotating about a different axis. Some examples include finding the volume of a cylindrical fuel tank or a rotating machine part. It is also used in various engineering and physics problems involving rotational motion and moments of inertia.

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