How Do You Solve a Volume of Revolution Problem Using the Shell Method?

[nick.martinez]

Volumes of revolution

Homework Statement

y=x^3 ; x=1; y=-1 axis:y=-1

Homework Equations

shell method ∫ 2pi*y*g(y)

The Attempt at a Solution

2pi *∫(y^(1/3)+1)*(1+y^(1/3))dy

limits of integration are from -1 to 1 i think

please help

 

[CAF123]

That's not much of a problem statement, but I think you want the volume generated by rotating the region enclosed by the curve y = x³ and x = 1 around y = -1. Did you draw a sketch of the region? This will help to identify the radius and height of a typical shell and the appropriate limits of integration.

 

[nick.martinez]

the books says using the sketch show how to approximate the volume of the solid by a riemann sum, hence find the volume. also, i drew a sketch and having trouble coming up with the radius and height. maybe the shel method is not as practical as others in this problem. if you could please show how you up the problem i can take care of it from there. thanks

 

[CAF123]

Per the forum rules, I cannot just set the integral up for you. Perhaps you could explain how you obtained the radius and height that you got in your attempt. I would agree that the disk method is maybe a little easier in this case to set up, but the final integrals are just as easy to evaluate.

 

[nick.martinez]

i used the disk method here
and this is how i set up my integral

pi(x^3)^2dx

x^6 dx pi[x^(7)/7]=(pi/7)-(-pi/7) therefore giving my volume which is 2pi/7.

 

[CAF123]

i used the disk method here
and this is how i set up my integral

pi(x^3)^2dx

x^6 dx pi[x^(7)/7]=(pi/7)-(-pi/7) therefore giving my volume which is 2pi/7.

You want to rotate the region around the line y =-1. The region does not extend outwith the first quadrant so the lower limit for x (-1) is incorrect. The fact that you rotate around y=-1 will affect the radius of the disk. The formula you should use is: $$\pi \int_{x_0}^{x_1} r_{outer}^2 - r_{inner}^2 dx$$ What is $r_{outer}$ and $r_{inner}$? Definitely refer to a sketch here.

 

[nick.martinez]

How can I find the lower limit?

Here's my attempt:

∫(X)^3-(1)^2dx

 

[nick.martinez]

Then whe I evaluate I from -1 to 1 I get 12pi/7 which is wrong

 

[nick.martinez]

What defines an inner and outer radius in this problem?

 

[CAF123]

The outer radius is the distance from the line of rotation to the boundary of the region (the curve y = x³). The lower radius is the distance from y =-1 to the bottom of the region. This will give a typical disk. Integrate up over all disks from x=0 to x =1, each with volume element $$\pi R^2 \Delta x,$$ where $R^2 = r_{outer}^2 - r_{inner}^2$

Does this make sense?

P.s why did you start a new thread with what appears to be the same question?

 

[nick.martinez]

So basically I set the integral up like this using your method integrating from -1 to 1.
(X^3)^2-(-1)^2 and when I takin the anti derivative I get pi((x^7)/7)-x after I evaluate I get -6pi /7. I'm getting the right answer using my method.

 

[nick.martinez]

Is it wrong to use the disk method here and use "integral" ( x^3+1)dx as my radius and multiplying it by pi and evaluating it from -1 to 1. I'm getting the correct answer when I use this method which