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Math Help - Applications of Intergration

  1. #1
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    Applications of Intergration

    Hey guys I have a quick question.

    Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the specified axis. Sketch the region and a typical shell.

    y=4x-x^2 , y=3 about x=1

    For this question would i just sub in 3 into the y and just solve and find the interval. Then do the integration of 2pie x (4x-x^2) between the two intervals?
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  2. #2
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    Re: Applications of Intergration

    Your shells will extend from y = 0 , to y = 4x-x^2 . They will be too high.

    The integrand should be f(x) - g(x).

    You just have f(x).
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  3. #3
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    Re: Applications of Intergration

    that's all they gave me though, so how do i continue?
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  4. #4
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    Re: Applications of Intergration

    Hello, kashmoneyrecord3!

    I don't understand your game plan,
    but "just sub in 3" sounds like a bad idea . . .


    Use the method of cylindrical shells to find the volume generated by rotating
    the region bounded by the given curves about the specified axis.
    Sketch the region and a typical shell.

    . . y\:=\:4x-x^2,\; y\,=\,3,\;\text{ about }x=1

    Did you make a sketch?

    Code:
            |
            |  :   ..*..
            |  :.*:::::::*.
           3+  *- - - - - -*
            | *:           :*
            |  :           :
            |* :           : *
            |  :           :
            |  :           :
        ----*--+-----------+--*----
            |  1           3  4
            |

    Formula: . V \;=\;2\pi\! \int^b_a\!\text{(radius)(height)}\,dx


    We have: . \begin{Bmatrix}r \,=\, x-1 \\ h \,=\,(4x-x^2) - 3 \\ a \,=\, 1 \\ b \,=\, 3 \end{Bmatrix}

    Hence: . V \;=\;2\pi\! \int^3_1\!(x-1)(4x-x^2-3)\,dx

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  5. #5
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    Re: Applications of Intergration

    how did you get the intervals?
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  6. #6
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    Re: Applications of Intergration

    Quote Originally Posted by kashmoneyrecord3 View Post
    how did you get the intervals?
    Just the way you said in the Original Post, "just sub in 3 into the y".

    Solve 3 = 4x-x^2 , for x. That should give x = 1, x = 3
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