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Volume integration using washermethod

  1. Sep 27, 2008 #1
    volume integration using "washer method"

    1. The problem statement, all variables and given/known data
    The region R bounded by y = x[tex]^{2}[/tex], y = 0, x = 1 and x = 4 is rotated about x = −1

    2. Relevant equations
    i know that these equations take on the form of [tex]\pi\int_a^b \\((outer radius)^{2} - (inner radius)^{2})\\,dx[/tex]

    3. The attempt at a solution
    so i set the problem up like this and still cant get the correct answer which is [tex]339\pi/2[/tex]

    -for the bounds i know that the integral i want to evaluate is between 0 and 16 based on the line y=0 and the function [tex]y=x^{2}[/tex] evaluated at x=4.

    -the outer radius is [tex]1+\sqrt{y}[/tex] and the inner radius is just 1

    [tex]\pi\int_0^{16}\\ (1+\sqrt{y})^2-(1)^2\\,dy[/tex]
    Last edited: Sep 27, 2008
  2. jcsd
  3. Sep 27, 2008 #2


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    Welcome to PF!

    Hi peercortsa! Welcome to PF! :smile:

    That's right … but then your thickness is dx, so that's what you must integrate over …

    ∫(blah blah) dx, where blah blah is entirely a function of x. :wink:
  4. Sep 27, 2008 #3
    so..... did i even set up the equation correctly cuz i still cant seem to get the right answer no matter what i try :confused:
  5. Sep 27, 2008 #4


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    Your π∫…dx is correct for the area

    but you still need to put the height inside the ∫ to make the volume.

    Have a go! :smile:
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