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Volumes of revolution

  1. Dec 3, 2009 #1
    A log having the shape of a right circular cylinder of radius a is lying on its side. A wedge is removed from the log by making a vertical cut and another cut at an angle of 45 degrees, both cuts intersecting at the center of the log. Find the volume of the wedge.

    And as soon as I finish typing the problem, I find the scan already online. Oh well.

    http://www.mvla.net/teachers/JimM/Calculus%20BC-AP/Documents/Worksheet%205%204.pdf [Broken]

    #17 on the second page here.

    Obviously, I'm supposed to use the integral somehow, but my first approach was different. Because the wedge has an angle of 45, I know it is 1/8 of the sphere around it. The volume of a sphere is V= (4/3)pi*r^3, and in this case, you just replace r with a. Now that you have the volume of the sphere, I just multiply it by 1/8 to get (1/6)pi*r^3.

    However, this answer is totally different from the one in the book. The answer key has you take the integral from 0 to a of 2sqrt(a^2y^2) which equals (2/3)a^3

    I don't understand how that works and how my thinking was wrong....
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Dec 4, 2009 #2
    You mean 2ysqrt(a^2-y^2) which equals (2/3)a^3?

    That is not a part of the sphere if you look at it carefully

    How they got 2ysqrt(a^2-y^2) ?
    Using three integrals,
    z goes from y to a
    x goes from - sqrt(a^2-y^2) to sqrt(a^2-y^2)
    y goes from 0 to a

    solving this will give you 2ysqrt(a^2-y^2) = (2/3)a^3
     
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