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Thread: One more question :P

  1. #1
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    One more question :P

    How do you SET UP this question? I can probably do it if I set it up, but I just don't know how to set it up. Also, please EXPLAIN how to set it up so I can set up other problems like this.

    Let V be the region bounded by the $\displaystyle y-axis$, $\displaystyle y = 8$, $\displaystyle y = 1$, and $\displaystyle y = x^{.75}$. Find the exact volume by rotating V around the y-axis.

    My attempt:

    $\displaystyle \pi{\int^8}[(8-x^{.75}+1)^2]dx$
    . .$\displaystyle ^0$
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  2. #2
    Eater of Worlds
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    You can do it with shells or disks. But don't forget to add on the rectangular region from 0 to 1. See it?. It is $\displaystyle 7{\pi}$

    $\displaystyle {\pi}\int_{1}^{8}[(y^{\frac{4}{3}})^{2}-1]dy$

    or

    $\displaystyle 2{\pi}\int_{1}^{16}[x(8-x^{\frac{3}{4}})]dx$

    This is one way to go about it.
    Last edited by galactus; Nov 24th 2008 at 05:38 AM.
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  3. #3
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    Thanks for clearing that up. It makes a lot more sense now.

    One more question though...

    Let's say that you had $\displaystyle y = x^{.75}$, but this time the region is bounded by the the $\displaystyle x-axis$ and the lines $\displaystyle x = 1$ and $\displaystyle x = 8$. Of course you still rotate the region around the y-axis, but how would the problem change when you set it up?
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  4. #4
    Eater of Worlds
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    Then, since with shells the cross-sections are parallel to the axis you are revolving around you would have:

    $\displaystyle 2{\pi}\int_{1}^{8}x(x^{\frac{3}{4}})dx$
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