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Math Help - Volume by Shell Method

  1. #1
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    Volume by Shell Method

    The problem is to find the volume formed by rotating the region enclosed by:
    x=3.4y and y^3=x with y ≥ 0 about the y-axis

    I tried changing the equations to y=x/3.4 and y=x^(1/3) which intersect at x=6.269, so i used that for the upper limit of the intergral which looked like:

    2pi * Intergral (x(x^(1/3) - x/3.4))dx with defining limits [0,6.269]

    Sorry I don't have a good math writer. Anyway, after intergrating i ended up with an answer of 29.30141 which i checked with an online applet to be wrong, but i can't find what i did wrong. Is everything i did up to the intergral right?
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  2. #2
    MHF Contributor
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    Nope. Sorry.

    It appears your integral is correct. Something must have gone wrong in the integration process.

    Note: One of the most valuable tools I EVER invented for myself is a stubborn refusal to use ONLY the method prescribed in the problem statement. I recommend VERY HIGHLY doing EVERY problem along both axes, using different methods. Every one.

    Learn the derivation:

    Volume of a Right Circular Cyllinder: V = \pi r^{2}h

    \frac{dV}{dh}\;=\;\pi r^{2}

    \frac{dV}{dr}\;=\;2\pi r h
    Last edited by TKHunny; February 7th 2010 at 01:57 PM. Reason: Lecture #283
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