Find the area of the surface generated by revolving about the​ x-axis the portion of the astroid x Superscript two thirds Baseline plus y Su

Question

Find the area of the surface generated by revolving about the​ x-axis the portion of the astroid x Superscript two thirds Baseline plus y Superscript two thirds Baseline equals 9 Superscript two thirds equal to one x(3/2)+y(2/3)=1)

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Kinsley 2 weeks 2022-01-07T00:52:59+00:00 1 Answer 0 views 0

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    2022-01-07T00:54:18+00:00

    The correct question is:

    Find the area of the surface generated by revolving about the​ x-axis, the portion of the astroid

    x^\frac{3}{2} + y^\frac{2}{3} = 1

    Answer: The Surface Area is

    \frac{6}{5}\pi

    Step-by-step explanation:

    First, we rewrite the expression in terms of x,  because we are revolving about the x-axis, we want to integrate in terms of x. Doing that, we have

    y = \left(1 - x^\frac{2}{3}\right)^\frac{3}{2}

    Next, we differentiate y with respect to x

    \frac{dy}{dx} = \frac{3}{2}\left(1 - x^\frac{2}{3}\right)^\frac{1}{2}\left(-\frac{2}{3}\right)x^\frac{-1}{3}}\\ \\= -\frac{\sqrt{\left(1 - x\right)^\frac{2}{3}}}{x^\frac{1}{3}}

    Thus,

    \left(\frac{dy}{dx}\right)^2 = \frac{\left(1 - x\right)^\frac{2}{3}}{x^\frac{2}{3}}}

    and so

    1 + \left(\frac{dy}{dx}\right) ^2 \\ \\ = 1 +\frac{(1 - x)^\frac{2}{3}}{x^ \frac{2}{3}} \\ \\=\frac{1}{x^\frac{2}{3}}

    Therefore, the Surface Area is given as:

    \int_{0}^{1}2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2}dx \\ \\= \int_{0}^{1}2\pi\left(1 - x^\frac{2}{3}\right)^\frac{3}{2}\sqrt{\frac{1}{x^\frac{2}{3}}}dx \\ \\=\int_{0}^{1}2\pi\left(1-x^\frac{2}{3}\right)^\frac{3}{2}x^\frac{-3}{2}dx. \\ \\

    If we let

    u = 1-x^\frac{2}{3}

    then

    du = -\frac{2}{3}x^{-\frac{1}{3}}dx,

    so we see that

    = -\frac{3}{2}\int_{0}^{1}2\pi\left(1-x^\frac{2}{3}\right)^\frac{3}{2} - \frac{2}{3}x^{-\frac{1}{3}} dx \\ \\= -3\pi \int_{0}^{1}u^\frac{3}{2}du \\ \\= 3\pi\frac{2}{5}u^\frac{5}{2}\left \{ {{u=1} \atop {u=0}} \right.  \\ \\= \frac{6}{5}\pi

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