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Thread: Volume from Revolution

  1. #1
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    Volume from Revolution

    Hey everyone. I have to revolve the surface bounded by $\displaystyle y=x\sqrt{x+1}$ and y = 0 around the y-axis and find the volume underneath it. I just have to set up the integral from [a,b]. The problem is, I don't know how to take the inverse of that function. From there I can just do
    $\displaystyle \pi\int_a^b[ f^{-1}(x)]^2dx$
    What should i do?

    Thanks in advance!
    Last edited by lilaziz1; Jun 21st 2011 at 10:54 AM.
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  2. #2
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    Re: Volume from Revolution

    What are your $\displaystyle \displaystyle x $ bounds?
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  3. #3
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    Re: Volume from Revolution

    it wasn't given so i have to use constants, e.g., a for lower bound and b for upper bound. the surface i have to revolve is bounded by y = 0 and $\displaystyle y=x\sqrt{x+1}$
    Last edited by lilaziz1; Jun 21st 2011 at 11:15 AM.
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    Re: Volume from Revolution

    Quote Originally Posted by lilaziz1 View Post
    it wasn't given so i have to use constants, e.g., a for lower bound and b for upper bound. the surface i have to revolve is bounded by y = 0 and $\displaystyle y=x\sqrt{x+1}$
    $\displaystyle x\sqrt{x+1} = 0$ at $\displaystyle x = 0$ and $\displaystyle x = -1$

    the region between $\displaystyle y = x\sqrt{x+1}$ and $\displaystyle y = 0$ is in quad III

    rotating this region about the y-axis using cylindrical shells ...

    $\displaystyle V = 2\pi \int_{-1}^0 x^2\sqrt{x+1} \, dx$

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