1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Volume integration problem

  1. May 24, 2013 #1
    Volumes of revolution

    1. The problem statement, all variables and given/known data
    y=x^3 ; x=1; y=-1 axis:y=-1

    2. Relevant equations

    shell method ∫ 2pi*y*g(y)

    3. 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
    Last edited: May 24, 2013
  2. jcsd
  3. May 24, 2013 #2


    User Avatar
    Gold Member

    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.
  4. May 24, 2013 #3
    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
  5. May 24, 2013 #4


    User Avatar
    Gold Member

    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.
  6. May 24, 2013 #5
    i used the disk method here
    and this is how i set up my integral


    x^6 dx pi[x^(7)/7]=(pi/7)-(-pi/7) therefore giving my volume which is 2pi/7.
  7. May 24, 2013 #6


    User Avatar
    Gold Member

    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.
  8. May 24, 2013 #7
    How can I find the lower limit?

    Here's my attempt:

  9. May 24, 2013 #8
    Then whe I evaluate I from -1 to 1 I get 12pi/7 which is wrong
  10. May 24, 2013 #9
    What defines an inner and outer radius in this problem?
  11. May 24, 2013 #10


    User Avatar
    Gold Member

    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?
  12. May 24, 2013 #11
    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.
  13. May 24, 2013 #12
    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted