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: Cylindrical shells question

  1. Jul 18, 2010 #1
    1. From Stewart Calculus and Concepts 4th edition, page 454 #15

    Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the specified axis.

    15. y=4x-x^2, y=; rotate about x=1

    2. Relevant equations



    3. I was able to find the volume of the function y=4x-x^2 by integrating from 0 to 4, but the line y=3 cuts the function off near the top, and I don't know how to take some of that volume "out". Are 2 integrals needed for this? I also tried the washer method of integrating the top function(y=4x-x^2 minus the bottom function(y=3) and I still didn't get it. Ive been stuck on this one for about 2 hours now. Any hints?
     
  2. jcsd
  3. Jul 18, 2010 #2
    Nope, we don't need 2 integrals. Since this is rotation about a vertical axis (x=1), we have

    [tex] V = \int_a^b 2\pi*r*h*dx [/tex]
    Where,
    r = distance from axis
    h = height of shell.

    That's all we need. As for hints:
    Since we're rotating about x=1, we're essentially "losing" 1 unit of x in the sense that r = x - 1.
    Also, h is the difference of the two y(x) functions (top minus bottom).

    Can you figure out the rest? I think you should end up with V = 4/3.
     
  4. Jul 18, 2010 #3
    The back of the book says the answer is 8*pi/3

    I know how to do some of these but this one I must just be missing something fundamental.

    I don't know how to do a definite integral, but its from 0 to 4 of[tex]2\pi\int(x-1)(4x-x^2-3)dx[/tex]
     
  5. Jul 18, 2010 #4
    ooops....yea, I forgot the factor of 2pi out in front. So 4/3 * 2pi = 8pi/3 is correct.

    Close!!! Everything inside the integral is good, but your limits are a bit off, i.e. NOT from 0 to 4. Remember, the limits of your integral will be the 2 x-values that bound the region. Between which two x-values does the region lie?
     
  6. Jul 18, 2010 #5
    all I know is the functions intersect at x=1 and x=3. Are those the limits? If so then I don't understand why.
     
  7. Jul 18, 2010 #6
    Okay I got the answer finally.

    The whole time I was trying to solve it I was thinking for some reason that they were looking for the bottom half of the curve, not the top part bisected by y=3. O well, lesson learned not to make blind assumptions! Thank you Raskolnikov
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook