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Thread: volume of solid of revolution further integration

  1. #1
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    volume of solid of revolution further integration

    Question: The triangle formed by the three straight lines y =0, y=2x and y = 3-x is rotated about the side y = 0.
    Find the volume of the solid so generated, and the x-coordinate of its centre of gravity.
    my attempt:
    a) find the volume of the solid
    element of the volume = $\displaystyle \pi y^2 \mathrm{d} x $
    $\displaystyle v_{1} = \int_{0}^{1}\pi y^2 \mathrm{d} x $
    $\displaystyle v_{1} = \int_{0}^{1}\pi (2x)^2 \mathrm{d} x = \frac{4}{3} \pi $

    $\displaystyle v_{2} = \int_{1}^{3}\pi y^2 \mathrm{d} x $
    $\displaystyle v_{1} = \int_{1}^{3}\pi (x-3)^2 \mathrm{d} x = \frac{8}{3} \pi $
    $\displaystyle v_{1}+ v_{2} = 4 \pi $

    b) element of volume = [tex] x y^2 \pi \mathrm{d} x [\tex]

    $\displaystyle \bar{x}_{1}\int_{0}^{1} \pi y^2 \mathrm{d} x = \int_{0}^{1} \pi x y^2 \mathrm{d} x $
    $\displaystyle \bar{x}_{1}\int_{0}^{1} \pi (2x)^2 \mathrm{d} x = \int_{0}^{1} \pi x (2x)^2 \mathrm{d} x $
    $\displaystyle \bar{x}_{1}\int_{0}^{1} \pi 4x^2 \mathrm{d} x = \int_{0}^{1} \pi 4x^3 \mathrm{d} x $
    $\displaystyle \bar{x}_{1}\left ( \frac{4}{3} \right ) = 1 $
    $\displaystyle \bar{x}_{1} = \frac{3}{4} $

    $\displaystyle \bar{x}_{2}\int_{1}^{3} \pi y^2 \mathrm{d} x = \int_{1}^{3} \pi x y^2 \mathrm{d} x $
    $\displaystyle \bar{x}_{2}\int_{1}^{3} \pi (3-x)^2 \mathrm{d} x = \int_{2}^{3} \pi x (3-x)^2 \mathrm{d} x $
    $\displaystyle \bar{x}_{2}\int_{1}^{3} \pi (9-6x+x^2) \mathrm{d} x = \int_{1}^{3} \pi (9x-6x^2+x^3) \mathrm{d} x $
    $\displaystyle \bar{x}_{2}\left ( \frac{8}{3} \right ) = 4 $
    $\displaystyle \bar{x}_{2} = \frac{3}{2} $
    $\displaystyle \bar{x}_{1} + \bar{x}_{2} = \frac{3}{2} + \frac{3}{4} = \frac{9}{4} $
    the book gives an answer for part a) $\displaystyle 4 \pi $ b) $\displaystyle \frac{5}{4} $
    I applied the same method i used in part a if part b
    I want to know if there is another way to do part a
    Please can someone help me with part b
    Last edited by bigmansouf; Mar 20th 2019 at 06:38 PM.
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    Re: volume of solid of revolution further integration

    For part (b), the sum of the two $\bar{x}$ís do not carry the same weight as you have done.

    Instead,

    $\bar{x} = \dfrac{\frac{3}{4} \cdot \frac{4\pi}{3} + \frac{3}{2} \cdot \frac{8\pi}{3}}{4\pi} = \dfrac{5}{4}$
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    Member Cervesa's Avatar
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    Re: volume of solid of revolution further integration

    Part (a) may also be done using cylindrical shells w/r to $y$ ...

    $y=3-x \implies x=3-y$

    $y=2x \implies x = \dfrac{y}{2}$

    $2x=3-x \implies x =1 \implies y=2$


    $\displaystyle 2\pi \int_0^2 y \bigg[(3-y)-\dfrac{y}{2}\bigg] \, dy$

    $\displaystyle 6\pi \int_0^2 y - \dfrac{y^2}{2} \, dy$

    $6\pi \bigg[\dfrac{y^2}{2} - \dfrac{y^3}{6} \bigg]_0^2 = 6\pi \left(\dfrac{2}{3}\right) = 4\pi$
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    Re: volume of solid of revolution further integration

    Quote Originally Posted by Cervesa View Post
    For part (b), the sum of the two $\bar{x}$’s do not carry the same weight as you have done.

    Instead,

    $\bar{x} = \dfrac{\frac{3}{4} \cdot \frac{4\pi}{3} + \frac{3}{2} \cdot \frac{8\pi}{3}}{4\pi} = \dfrac{5}{4}$
    please can you explain your approach i want to under stand it so i can successfully tackle the question in the future

    thank you
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