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Thread: Rotational volume problem

  1. #1
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    Smile Rotational volume problem

    Hi everyone!

    I am having a bit trouble with a question about rotational volume. So, this is the problem:

    "An area R in the xy plane is given by:

    0 ≤ x ≤ 1

    and

    0 ≤ y ≤ x3

    Find the volume when rotating R around the axis y = -1
    "

    Here is how I was thinking. First, I drew this figure:


    Rotational volume problem-graph-problem.jpg

    My idea was to draw the boundaries, and two radius in red, which I need to find in order to use the washer method. So, I thought the lower boundary must be r(y) = g(y) - c = x3 - (-1) = x3 + 1 (The second red line in the picture)

    But I am unsure what I should put as the upper boundary (The first radius in the picture).

    I hope someone can help and tell me in what way I am thinking wrong.

    Thanks!
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Rotational volume problem

    The region $\displaystyle R$ is bounded by $\displaystyle y=0,\,x=1,\,y=x^3$.

    An arbitrary slice is the washer:

    $\displaystyle dV=\pi(r_o^2-r_i^2)\,dx$

    The outer radius is:

    $\displaystyle r_o=x^3+1$

    The inner radius is:

    $\displaystyle r_i=1$

    The "stack" of washers runs from $\displaystyle x=0$ to $\displaystyle x=1$, so summing the washers in the stack, we find the volume is:

    $\displaystyle V=\pi\int_0^1 (x^3+1)^2-1^2\,dx$

    Expand and simplify the integrand, then use the anti-derivative form of the FTOC to compute the volume.
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  3. #3
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    Re: Rotational volume problem

    Thank you so much for the help, I understood it much better now, I also noticed my figure is completely wrong.... How embarrassing
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