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Thread: Volumes of solids by slicing help

  1. #1
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    Volumes of solids by slicing help

    An equation of the curve in the figure is y^2 = x^3

    Line 4, OAC is the shaded curve
    Code:
         B  |               C(4,8)
           8+ - - - - - -*
            |           *|
            |         *::|
            |       * :::|
            |     * :::::|
            |  *:::::::::|    
          --*------------+
        O  |              4  A
    Find the volume of the solid of revolution generated when the given region of the figure is revolved about the indicated line:
    revolved OAC about the line AC

    Help the only problem i had is to subtract the radiuses to get to that given area of OAC...
    ill handle the integration

    Can you check my solution?

    $\displaystyle \pi\int_{0}^{8} y^{\frac{2}{3}}$

    $\displaystyle \pi\frac{3}{5}y^{\frac{5}{3}}\big|_{0}^{8}$

    $\displaystyle \frac{96}{5}\pi$
    Last edited by ^_^Engineer_Adam^_^; Sep 21st 2007 at 06:09 AM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    An equation of the curve in the figure is y^2 = x^3

    Line 4, OAC is the shaded curve
    Code:
         B  |               C(4,8)
           8+ - - - - - -*
            |           *|
            |         *::|
            |       * :::|
            |     * :::::|
            |  *:::::::::|    
          --*------------+
        O  |              4  A
    Find the volume of the solid of revolution generated when the given region of the figure is revolved about the indicated line:
    revolved OAC about the line AC

    Help the only problem i had is to subtract the radiuses to get to that given area of OAC...
    ill handle the integration

    Can you check my solution?

    $\displaystyle \pi\int_{0}^{8} y^{\frac{2}{3}}$

    $\displaystyle \pi\frac{3}{5}y^{\frac{5}{3}}\big|_{0}^{8}$

    $\displaystyle \frac{96}{5}\pi$
    that's not correct i'm afraid.

    you are using the disk method, so to rotate about the interval $\displaystyle [a,b]$ (whether with respect to $\displaystyle x$ or $\displaystyle y$), the formula for the Disk method is:

    $\displaystyle V = \pi \int_{a}^{b} [R(x \mbox { or } y)]^2$

    where $\displaystyle R(x \mbox { or } y)$ is the radius.

    since you're doing $\displaystyle V = \pi \int_{a}^{b} [R( {\color {red} y})]^2~{\color {red} dy}$, where $\displaystyle [a,b]$ is the interval on the y-axis, with $\displaystyle b > a$

    your radius here is: $\displaystyle R(y) = \left(4 - y^{2/3} \right)$
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