Volume by Shells: Find Length of Radius?

  • Thread starter doppelganger007
  • Start date
  • Tags
    Volume
In summary, the problem is about finding the volume of a solid generated by revolving a region bounded by given equations around a specific line. There are two ways to integrate using shells, either over x or over y. The challenge is finding the height function for the tube method or the disk area function for the disk method.
  • #1
doppelganger007
18
0
Hi,

I'm doing a problem about finding volume using shells, and I'm really confused about what the the width function should be. The question asks to find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the indicated line:

y=1/(x^2), y=0, x=1, x=4, and revolve that region around the y-axis.

Wouldn't the length of the radius just be x? That's what I think it is but I'm not sure. Help would be appreciated, thanks :smile:
 
Physics news on Phys.org
  • #2
There are two ways of integrating using shells: integrating over x or over y (adding up tubes of adding up disks).

If you want to integrate over x, then yea, the radius is just x and the volume is

[tex]V=\int_1^4 (2\pi x)(\mbox{height of tube})dx[/tex]

So the "challenge" with the tube method is really more of finding the height function h(x).

If you want to integrate over y then it's about finding a "disk area function" A(y) such that

[tex]V=\int_{1/4^2}^{1/1^2}A(y)dy[/tex]
 
  • #3


Hi there,

You're on the right track! The length of the radius for a volume by shells problem is indeed x, since we are revolving the region around the y-axis. This means that the distance from the y-axis to any point on the curve is equal to the value of x.

To set up the integral for this problem, we can use the formula for volume by shells:

V = 2π∫(radius)(height)(width)dx

In this case, the radius is x, the height is y (since we are revolving around the y-axis), and the width is the difference between the bounds of integration, which is 4-1 = 3.

So the integral would be:

V = 2π∫(x)(1/(x^2))(3)dx

Simplifying this, we get:

V = 6π∫(1/x)dx

And integrating, we get:

V = 6πln(x)

Evaluating this from x=1 to x=4, we get a final answer of:

V = 6πln(4) - 6πln(1) = 6πln(4) ≈ 23.56

I hope this helps! Let me know if you have any other questions. Good luck with your problem!
 

Related to Volume by Shells: Find Length of Radius?

1. How do you find the length of the radius for volume by shells?

To find the length of the radius for volume by shells, you will need to use the formula V = 2π∫r h(x) dx, where r is the radius and h(x) is the height of the shell at a given point. You can then solve for r by dividing both sides by 2π and integrating both sides.

2. What is the difference between volume by shells and volume by disks?

The main difference between volume by shells and volume by disks is the shape of the cross-sections used to calculate the volume. In volume by shells, the cross-sections are cylindrical shells, whereas in volume by disks, the cross-sections are circular disks. This means that the integral used to calculate the volume will differ between the two methods.

3. How do you determine the height of the shells in volume by shells?

The height of the shells in volume by shells is determined by the function h(x) that represents the height of the shell at a given point. This function can be determined by visualizing the solid and understanding how the shells are stacked or by using the known equations for the solid and solving for h(x).

4. Can volume by shells be used for all solid shapes?

No, volume by shells can only be used for solids with rotational symmetry around an axis. This includes shapes like cylinders, cones, and spheres. If a solid does not have rotational symmetry, then volume by shells cannot be used and other methods, such as volume by disks or cross-sections, must be used instead.

5. How can volume by shells be applied in real-world situations?

Volume by shells is a useful concept in real-world situations such as finding the volume of a water tower, determining the amount of material needed for a cylindrical storage tank, or calculating the volume of a conical volcano. It can also be used in physics to calculate the moment of inertia of a solid object.

Similar threads

Replies
2
Views
881
  • Calculus
Replies
4
Views
810
Replies
2
Views
1K
Replies
5
Views
2K
Replies
4
Views
693
Replies
5
Views
2K
  • Calculus
Replies
3
Views
1K
Replies
1
Views
771
Back
Top