Volume by Shells: Find Length of Radius?

[doppelganger007]

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

 

[quasar987]

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]

 

[tacman]

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!