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Calculus Definite Integrals: Volumes by Washer Method

  1. Mar 23, 2014 #1
    1. The problem statement, all variables and given/known data
    Using Washer Method: Revolve region R bounded by y=x^2 and y=x^.5 about y=-3



    2. Relevant equations
    V= integral of A(x) from a to b with respect to a variable "x"
    A(x)=pi*radius^2


    3. The attempt at a solution
    pi(integral of (x^.5-3)^2 -(x^2)^2-3) from 0 to 1 with respect to x

    The answer involves x^.5+3 instead of x^.5 -3. I don't understand why you would add instead of subtract 3; y decreased from 0 to -3 when revolved around y=-3
     
  2. jcsd
  3. Mar 23, 2014 #2

    BvU

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    Make a drawing. See that the outer radius of your washers is ##\sqrt x + 3## and the inner is ## x^2 + 3##
     
  4. Mar 23, 2014 #3

    LCKurtz

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    The length of a vertical line is ##y_{upper}-y_{lower}##. Also don't forget to square your second term.
     
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