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Thread: Why is that we use the boundaries from -1 to 1 instead of 0 to 1?

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    Why is that we use the boundaries from -1 to 1 instead of 0 to 1?

    The given curve is rotated about the y-axis. Find the area of the resulting surface.


    $\sqrt[3]{x^2}+\sqrt[3]{y^2}=1 , 0\leq y \leq1 $


    We set up the surface area integral:


     \int_{-1}^{1} 2\pi x ds


    I understand how to find the arc-length(ds) and set up the integral but why wouldn't out boundaries stay from 0 to 1? My reasoning is that  2\pi x
    is taking in account that our radius from 0 to 1 should be doubled. Shouldn't we only care about the arc-length from 0 to 1?


    For instance if we were to find the surface area of  y=1/3 \sqrt[2]{x^3}+$ ,  $0\leq y \leq12 rotating about the y-axis, our integral would be,


     \int_{0}^{12} 2\pi y ds
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    Re: Why is that we use the boundaries from -1 to 1 instead of 0 to 1?

    The figure is defined from 0 to 1 but the it is rotated around the y- axis so the is exactly the same figure from -1 to 0. The entire rotated figure will go from x= -1 to x= 1.
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    Re: Why is that we use the boundaries from -1 to 1 instead of 0 to 1?

    what if it is rotated around the x axis? i thought saying the radious times one side of the curve would give us the surface area? thank you for helping me
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    Re: Why is that we use the boundaries from -1 to 1 instead of 0 to 1?

    You know that $0\le y \le 1$. Without it being rotated around anything, let's calculate bounds for $x$:

    $\sqrt[3]{x^2} + \sqrt[3]{y^2} = 1 \Longrightarrow x = \pm \left(1-\sqrt[3]{y^2}\right)^{\tfrac{3}{2}}$

    So, using the bounds for $y$, we find $x \ge -\left(1-\sqrt[3]{0^2}\right)^{\tfrac{3}{2}} = -1$ and $x \le \left(1-\sqrt[3]{0^2}\right)^{\tfrac{3}{2} = 1$. So, $-1 \le x \le 1$.
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    Re: Why is that we use the boundaries from -1 to 1 instead of 0 to 1?

    I wonder if the original question meant to restrict x to 0 \leq x \leq 1. The function x^\frac{2}{3} + y^\frac{2}{3} = 1 looks like an inverted "v" for 0 \leq y \leq 1. If this inverted "v" is rotated about the y-axis, it will yield twice the usual surface area if one were to use -1 \leq x \leq 1. I believe that the OP is in the right to be skeptical of the bounds -1 \leq x \leq 1. See Example 2, p.4 [Stewart].

    M><M
    Last edited by majamin; Nov 6th 2017 at 06:22 PM. Reason: small typo
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