1. Not finding help here? Sign up for a free 30min 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!

Volume of the solid

  1. Mar 5, 2005 #1
    The region bounded by [tex]x = 1- y^4, x=0 [/tex] is rotate about the line [tex] x = 3 [/tex] The volume of the resulting solid is ......

    here's what i done:
    [tex]x = 1- y^4 [/tex] in terms of y => [tex] y = (1-x)^{1/4}[/tex]

    my integral:
    [tex]\int_0^{1} 2*pi*(3-x)(1-x)^{1/4}[/tex]

    anyone know what i have done incorrectly?
  2. jcsd
  3. Mar 5, 2005 #2


    User Avatar
    Homework Helper
    Gold Member

    This is a hollow body, you have to subtract the volume of the body with radius r = 3- x(y) in the picture from that of the cylinder of radius R=3. And integrate with respect to y as the rotational angle is parallel to the y axis.

    [tex]V=\pi \int_{-1}^1 {(3^2-r^2)dy} [/tex]

    Last edited: Jun 29, 2010
  4. Mar 5, 2005 #3


    User Avatar
    Science Advisor
    Homework Helper

    Normally, either the shell method or the washer method will work. In this case, you're going to have some problems with [tex] (1-x)^{\frac{1}{4}}[/tex].

    It's not a function since an even root has both a positive and negative root. For example:

    [tex]\sqrt {1-.84}= \pm .4[/tex]

    Without the bottom half of the graph, you can't find the volume. You could compensate just by finding the volume of the region bounded by x=(1-x)^(1/4), x=0, y=0, and then multiplying by two.

    It would be easier to just to use the washer method, as ehild suggested.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?