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Math Help - frustrated - area by slice

  1. #1
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    frustrated - area by slice

    I need to find the area of the ring defined by y=x^3, y=0, and x=1, with the center of the ring being x=2.

    Part of the problem I am having is that while I understand the concept of what I need to do, I am getting confused between how I figure out which way the slices go (parallel to x axis, or parallel to y axis), whether I need to set up the integral with respect to x or with respect to y. For example, I have y=x^3, do I plug that into the integral, or do I need to solve for x (getting x=y^\frac{1}{3} and then plug that into the integral? I'm not sure how to keep that all straight.

    I am getting:

    \pi\int_0^1(2-y^\frac{1}{3})^2-(2-1)^2 which becomes \pi\int_0^1(3-2y^\frac{1}{3}+y^\frac{2}{3}

    Take the integral and get:

    \pi\[_0^1(3y-\frac{3}{2}y^\frac{4}{3}+\frac{3}{5}y^\frac{5}{3}) (haven't quite got the syntax of the left bracket and intervals correct) , or \pi(\frac{30}{10}-\frac{15}{10}+\frac{6}{10})

    The answer I get is \frac{21\pi}{10} and the book has \frac{3\pi}{5}

    I'm pretty confident my setup is incorrect, but I've tried several different setups and not gotten to the correct one. I'm not sure what else to try at this point.

    I would appreciate some help with the setup and also some pointers on how to set up these problems in general.

    Thanks!
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  2. #2
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    Re: frustrated - area by slice

    If some idiot (namely me) were to square a binomial correctly, things would go much better.
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  3. #3
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    Re: frustrated - area by slice

    (2-y^\frac{1}{3})^2 = 4 - 4y^\frac{1}{3} + y^\frac{2}{3}

    You got that, right?
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