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Revolving regions horizontally and vertically

  1. Mar 2, 2014 #1
    1. The problem statement, all variables and given/known data
    The question asks me to revolve the the region bounded by the graphs y=√X y=0 and x=3 around a) the line x=3 and then b) around the line x=6


    2. Relevant equations

    I know I am supposed to use the disk method or the washer method for this, which is ∏ ∫ R(y)^2-r(y)^2 from y=c to y=d.


    3. The attempt at a solution

    for a, I set up R(y)=y^2. I set up the integral from y=0 to y= √3 ∏ ∫(y^2)^2
    but when I plugged in the values I got 9.79 units cubed. In the back of my calculus book, however, it gives a different answer. I can't figure out what I did wrong.

    For b, I set R(y)=6 and r(y)=(6-y^2). I set up the the integral from y=0 to y=√3
    ∏∫6^2-((6-y^2)^2)

    I got an answer of 55.5 units cubed. Again, this was not the correct answer given to me in the back of my book.


    I can't figure out what I did wrong. Any help would be greatly appreciated.
     
  2. jcsd
  3. Mar 2, 2014 #2
    For a, you don't quite have the correct value for the radius. R(y) = y2 is correct for the function, but remember that if you're revolving around x = 3, then the radius is going to be the distance between R(y) and 3.

    For b, I think it's the same type of thing. The outer radius should be the distance between x = 6 and R(y), and the inner radius should be the distance between x = 6 and x = 3 (which is the inner boundary).
     
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