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Thread: 18 volume rotated around y axis

  1. #1
    Super Member bigwave's Avatar
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    18 volume rotated around y axis

    $\textsf{Find the volume of the solid generated by revolving the region
    about y-axis.Given the boundares of }\\$
    \begin{align*} \displaystyle
    y&=4x-x^2\\
    y&=x\\
    \end{align*}
    ok I presume since this is rotated around the y-axis that we have to rewrite the equations in terms of y
    $y=-(x^2-4x)$
    $4-y=(x-2)^2$
    $\sqrt{4-y}=x-2$
    $\sqrt{4-y}+2=x$
    and
    $x=y$

    but the graph is not complete
    and is this not the same volume if it were rotated around the x-axis
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  2. #2
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    Re: 18 volume rotated around y axis

    Let's use the cylindrical shell method:

    First, let's find the points of intersection:

    $4x-x^2 = x \Longrightarrow x(3-x) = 0 \Longrightarrow x=0, x=3$

    So, we have our points of intersection. Next, let's determine which is larger. We need only check one point in the middle (say at x=1). We find that $4(1)-1 > 1$, so that is our "larger" function. The height of our cylindrical shell will be $4x-x^2-x = 3x-x^2$. Our radius will be $x$. Then, our integral becomes:

    \displaystyle \int_0^3 2\pi x (3x-x^2)dx = 2\pi \int_0^3 (3x^2-x^3)dx = 2\pi \left[x^3-\dfrac{x^4}{4}\right]_0^3 = 2\pi\left(27-\dfrac{81}{4}\right) = \dfrac{27\pi}{2}
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    Re: 18 volume rotated around y axis

    To do it by the washer method, you need to look at the graph. You want to integrate with different washers over different intervals. We can break it down into the following integrals:

    \displaystyle \pi \int_0^3 \left[y^2-\left(2-\sqrt{4-y}\right)^2\right]dy + \pi \int_3^4 \left[ \left( 2+\sqrt{4-y} \right)^2 - \left(2-\sqrt{4-y}\right)^2\right]dy

    I am not gonna solve this by hand. I just plugged it into WolframAlpha, and it gave the same result:
    Wolfram|Alpha: Computational Knowledge Engine)

    This integral comes from the fact that if $y = 4x-x^2$, then $x = 2-\sqrt{4-y}$ when $0\le x \le 2$ and $x = 2+\sqrt{4-y}$ when $2 \le x \le 3$. So, you have three functions now. When $0\le y \le 3$, you have $y \ge 2-\sqrt{4-y}$ and for $3\le y \le 4$, you have $2+\sqrt{4-y} \ge 2-\sqrt{4-y}$. That gives the washers at each interval.
    Last edited by SlipEternal; Nov 8th 2017 at 01:57 PM.
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  4. #4
    Super Member bigwave's Avatar
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    Re: 18 volume rotated around y axis

    I thot this was supposed to revolve around the y axis?? which is 0 to 4
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  5. #5
    Super Member bigwave's Avatar
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    Re: 18 volume rotated around y axis

    18 volume rotated around y axis-18.png
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  6. #6
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    Re: 18 volume rotated around y axis

    Quote Originally Posted by bigwave View Post
    I thot this was supposed to revolve around the y axis?? which is 0 to 4
    Correct. In both of the methods I gave, I revolved it around the y-axis. For the cylindrical shell method you integrate with respect to the axis opposite the revolution (x from 0 to 3). For the washer method you integrate with respect to the same axis as the revolution (y from 0 to 4). In the latter method the "outside" function changes at y=3 which is why I broke it into the two separate integrals.
    Last edited by SlipEternal; Nov 8th 2017 at 03:13 PM.
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