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Thread: [SOLVED] Volume of Revolution Question

  1. #1
    Member looi76's Avatar
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    [SOLVED] Volume of Revolution Question

    Question:
    Th region $\displaystyle R$ which is bounded by the curve $\displaystyle y = \frac{2}{x + 1}$, the $\displaystyle x-axis$, and the lines $\displaystyle x = 1$ and $\displaystyle x = 5$. Show that the volume of the solid formed when $\displaystyle R$ is rotated completely about the x-axis is $\displaystyle \frac{4}{3}\pi$

    Attempt:

    $\displaystyle y = \frac{2}{x+1}$

    $\displaystyle \pi\int^5_1\left(\frac{2}{(x + 1)^2}\right)^2dx$

    $\displaystyle \Rightarrow \pi\left[\frac{4}{(x+1)}\right]^5_1$

    $\displaystyle \Rightarrow \frac{2}{3}\pi - 2\pi$

    $\displaystyle = -\frac{4}{3}\pi$

    Why did I get the answer negative? Where did I go wrong?
    Last edited by looi76; Apr 25th 2008 at 03:40 AM. Reason: forgot to square [MATH]\pi\int^5_1\left(\frac{2}{(x + 1)^2}\right)^2dx[/MATH]
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  2. #2
    Moo
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    Hello,

    Actually, I think that the answer has to be :

    $\displaystyle \int_1^5 \int_{\color{red}0}^{\frac{2}{x+1}} dy dx$

    As you're bounded with the x-axis, that is to say y=0, you have to put it in the integral..

    $\displaystyle \pi$ or not, I don't know
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  3. #3
    Member Danshader's Avatar
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    integrating 2/(x+1)^2 gives you -2/(x+1)

    your working looks weird O.o"

    ok i found the mistake >.<

    you squared the function of y wrongly

    the integration should be

    integration from 1 to 5 of the function 2^2/(x+1)^2

    instead of

    integration from 1 to 5 of the function (2/(x+1)^2)^2
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