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 of Solid in revolution

  1. Jul 23, 2011 #1
    1. The problem statement, all variables and given/known data
    The question asked is to make a bowl out of polynomial equations rotated around the y axis. The bottom of the bowl has to have a maximum at the center and a minimum at some distance from the center.

    The equations I want to use are x^2+10, 1.3x^2 and -.7x^2 + 4.

    The problem I face is what to do about the bottom of the bowl, since it is comprised of 2 functions meeting to form a minimum. Should I treat those two as its own volume of revolution, find the volume and then subtract the interior (top of bowl) from that total?

    2. Relevant equations

    Disc method: pi*integ: r(y)^2 dy

    3. The attempt at a solution

    pi*Integ: ((y-10)^1/2)^2 - [pi*Integ: ((y/1.3)^1/2)^2 - ((y-4/.7)^1/2)^2]

    Is my attempt correct? The volume I am trying to find is bounded by x=0 (the y axis) and the 3 other equations
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

    • bowl.PNG
      File size:
      11.2 KB
  2. jcsd
  3. Jul 24, 2011 #2
    Does this exercise allow us our own equations?
    Why not simplify it with?
    "y= x^2/2+30,y=1/8 x^4 -x^2 +5"
    This allows us one integral with shell method.

    If that's not allowed, we'll need at least two separate integrations.
    Are you required disc/ washer method? I think this exercise is easier with shell method on your equations. Then it can be done with only two integrals.

    Disc/ washer method will require 3 separate integrals.
    all numbers are approximate, but must be found exactly or more precisely before integrating.
    The limits of integration
    y:2.5->4 washer method
    y:4->10 disk method
    y:10->45 washer method
    For washer method it's always pi[(outer radius)^2 - (inner radius)^2]
    Disk method is a simplification of the same formula, with inner radius =0.

    Either way, we definitely need limits of integration.
    Without limits of integration, I can't tell how close you are to an effective method.
    Last edited: Jul 24, 2011
  4. Jul 24, 2011 #3
    We can use our own, I going to take your advice and use those equations! I couldn't find a curve that looked like that, this is much easier to work with! Thank you so much, this makes my life much easier!
  5. Jul 24, 2011 #4
    I knew some 4th degree function had the desired behavior. I just played around with wolframAlpha.com until I had the right graph.

    An even function is symmetric on both sides of the y axis.
    So only x^2 & x^4, & constant terms.

    I first tried x^4 +x^2 with various coefficients but that did not give any of the needed humps. Then I realized x^4 term & x^2 needed to go the opposite direction. From there I just fine tuned the coefficients until it looked nice.

    WolframAlpha.com is the best thing since sliced bread. Nearly instant graphing, less strict syntax to learn.

    Have fun in your learning.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook