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Math Help - Chk Work/Answer - Volume of the solid obtained by rotating R around the line

  1. #1
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    Chk Work/Answer - Volume of the solid obtained by rotating R around the line

    The region R is bounded by the graphs of x-2y=3 and x=y^2. Set up (but do not evaluate) the integral that gives the volume of the solid obtained by rotating R around the line x = -1.

    Area [Math]A = \pi(1+2y)^2 - \pi(1+y^2)2[/tex]

    Volume V = \sum(\pi(1+2y)^2 - \pi(1+y^2)2*Dy = \int_{0}^{2}\pi(1+2y)^2 - \pi(1+y^2)^2
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  2. #2
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    Hello, FLTR!

    Your integral is slightly off . . .


    The region R is bounded by the graphs of x-2y=3 and x=y^2.
    Set up (but do not evaluate) the integral that gives the volume of the solid
    obtained by rotating R around the line x = -1.

    Did you make a sketch?
    Code:
              |
          :   |                 /
          :   |            ...o (9,3)
          :   |      ..*::::/
          :   |  .*:::::::/
          :   |*::::::::/
       ---+---*:-:-:-:/------------
        -1:   |*::::/
          :   |   o (1,-1)
          :   | /      *
          :   /               *
            / |

    We will integrate with respect to y and use "washers".

    The two functions are: . \begin{array}{cc}x\:=\:2y + 3 \\ x = y^2\end{array}

    The "outer radius" is: . 2y + 3) -(-1) \:=\:2y + 4" alt="r_o\:=\2y + 3) -(-1) \:=\:2y + 4" />
    The "inner radius" is: . r_i\:=\:y^2 - (-1) \:=\:y^2 + 1
    . . And y ranges from -1 to 3.

    . . . . V \;=\;\pi\int^3_{-1} \bigg[(2y + 4)^2 - (y^2 + 1)^2\bigg]\,dy

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  3. #3
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    It's fun to do these things both shells and washers. One is often times more difficult than the other, though.

    Here's shells:

    2{\pi}\left[\int_{0}^{9}(x+1)(\sqrt{x}-\frac{x-3}{2})dx-\int_{0}^{1}(x+1)(-\sqrt{x}-\frac{x-3}{2})dx\right]

    This method subtracts that tiny shaded area below -\sqrt{x} and above \frac{x-3}{2}

    Which leaves the hash-marked shade left to rotate.
    Last edited by galactus; November 24th 2008 at 06:39 AM.
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