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Math Help - spin solids. help integral

  1. #1
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    spin solids. help integral

    hello. i have problem with spin solid to find volume.

    this is google translate of problem with help of math writing online program:

    find the volume of the area bounded by: x = 1-y^4, x = 0 around x=2

    x = 1-y^4 is same as y=\sqrt[4]{1-x}, yes? so is ring method:

    \pi\int_{0}^{1}(x-1)^2 -(\sqrt[4]{1-x})^2 dx

    this is okay?

    thank you. sorry for bad english
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  2. #2
    MHF Contributor
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    First, x = 1 - y^{4} MAY be the same as y = \sqrt[4]{1-x}, depending on what you are doing. Be more careful.

    Second, my rule, ALWAYS do it both ways so that you gain experience and you can check your own answer.

    Washers: \int_{-1}^{1}\pi\cdot\left[2^{2}-(2-x)^{2}\right]\;dy Substitute for 'x' and evaluate. With consideration for symmetry, this could be 2\cdot\int_{0}^{1}\pi\cdot\left[2^{2}-(2-x)^{2}\right]\;dy

    Cyllinders: \int_{0}^{1}2\cdot\pi\cdot (2-x)\cdot\left[2\cdot y\right]\;dx. Substitute for 'y' and evaluate. The 2y is there ONLY because of symmetry, otherwise, it would have been uglier.
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  3. #3
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    sorry. i dont get.

    where is original function in integral? original function not needed?

    graph shows half circle from -1 to 1, yes? if axis of spin is x=2, then we know radius of both: big radius = 3, little radius = 1

    \pi\int_{-1}^{1}3^2 - 1^2 dx

    \pi\int_{-1}^{1}8dx

    8\pi{x}\mid_{-1}^{1} = 8\pi - (-8\pi) = 16\pi

    is this answer?

    i think i answer my question, yes? jajaja
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