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Math Help - Mth101 Questio No2

  1. #1
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    Mth101 Questio No2

    Find the volume of the solid that results when the region enclosed by the given curves is revolved about the x-axis.
    y=x^2+1, y=x+3
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  2. #2
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    Hello, TAHIR!

    I assume you are new to this type of problem.
    . . I'll give you a leisurely walk-through.


    Find the volume of the solid that results when the region enclosed by the given curves
    is revolved about the x-axis: . y\:=\:x^2+1,\;\; y\;=\;x+3
    First, make a sketch . . .

    The first is an up-opening parabola with y-intercept 1.
    The second is an up-sloping line with y-intercept 3.
    Code:
                      |             /
          *           |           *
                      |         /:.
                      |       /:::.
             *        |     /::::*.
                      |   /:::::  .
                      | /::::::   .
               *     3*::::::*    .
                    /:|:::::      .
                  *:::|:::*       .
                /   ::|::         .
              /   .   *1          .
            /     .   |           .
        - / - - - + - + - - - - - + -
        /        -1   |           2
                      |
    For their intersections, equate the functions.
    . . x^2 + 1 \:=\:x + 3\quad\Rightarrow\quad x^2 - x - 2 \:=\:0\quad\Rightarrow\quad(x + 1)(x - 2) \:=\:0

    Hence, they intersect at: . x\:=\:-1,\:2


    The "upper" function is: . y_2 \:=\:x + 3
    The "lower" function is: . y_1\:=\:x^2 + 1

    The "washer" formula is: . V \;=\;\pi\int^b_a\left(y_2^2 - y_1^2\right)\,dx

    Hence, we have: . V \;=\;\pi\int^2_{\text{-}1}\left[(x+3)^2 - (x^2 + 1)^2\right]\,dx

    Can you finish it now?

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