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Math Help - Volume calculation, function rotation around the x-axis.

  1. #1
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    Volume calculation, function rotation around the x-axis.

    Hello everyone.

    I'm in need of some help with Volume calculations (with a function that is rotating around the x-axis).

    I got stuck on the following: y = 3/sqrt{2-x} interval: -1 =< x =< 1

    {1 {1
    \pi \int y^2 = \pi \int 9/(2-x) = /pi [Need help with this bit]
    {-1 {-1

    I think my big problem is to find correct Antiderivative.

    Did edit a little to make it more understandable.
    Sorry but i dont understand how latex work (so it looks ugly i know) and the help page on this site doesn't work

    I would be very thankful if anyone with more insight than me could give me some help. Primarily with Antiderivative.

    Have a nice day
    Last edited by helpwouldbenice; May 11th 2011 at 11:19 AM.
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  2. #2
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    You're right so far, except the bounds of your integral are -1 and 1, and you need to write y in terms of x...
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  3. #3
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    Now that you've got the integral to \displaystyle \int_{-1}^1{\frac{9\pi}{2 - x}\,dx} = 9\pi\int_{-1}^1{\frac{1}{2-x}\,dx}, solve using a u substitution.
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