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Thread: Calculus volume ..

  1. #1
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    May 2009
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    Calculus volume ..

    graph the area bounded by y= 1 x=4 and y = radical x

    fidn the volume of the solid formed by revolving the area about the line y=1

    find H.

    use disc/ washer method

    find all teh radiuses..


    my finals is in about 7 hours and im cramming alot.. and if anyone is willing to help im willing to thank them to unimaginable heights! i need to pass calc hopefully..
    this will make or break my grade >_>
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  2. #2
    Senior Member
    Joined
    Dec 2008
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    319
    To find the area enclosed by

    $\displaystyle \begin{aligned}
    y&=1\\
    x&=4\\
    y&=\sqrt{x},
    \end{aligned}$

    we note that $\displaystyle \sqrt{x}$ intersects $\displaystyle y=1$ at $\displaystyle x=1$ and attains $\displaystyle y=2$ at $\displaystyle x=4$. Thus, the region bounded is in the first quadrant with limits $\displaystyle x=1$ and $\displaystyle x=4$. Subtracting the area under $\displaystyle y=1$ from the area under $\displaystyle y=\sqrt{x}$, we obtain

    $\displaystyle \begin{aligned}
    \int_1^4 (\sqrt{x}-1)\,dx&=\left(\frac{2}{3}x^{\frac{3}{2}}-x\right)_1^4\\
    &=\left(\frac{16}{3}-4\right)-\left(\frac{2}{3}-1\right)\\
    &=\frac{5}{3}.
    \end{aligned}$

    To find the area of revolution, we integrate not along rectangles but along discs:

    $\displaystyle \begin{aligned}
    V&=\int_1^4 \pi(\sqrt{x} - 1)^2\,dx\\
    &=\pi\int_1^4(x-2\sqrt{x}+1)\,dx\\
    &=\pi\left(\frac{x^2}{2}\right)_1^4-2\pi\left(\frac{2}{3}x^{\frac{3}{2}}\right)_1^4+\p i(x)_1^4\\
    &=\frac{15\pi}{2}-\frac{28\pi}{3}+3\pi\\
    &=\frac{7\pi}{6}.
    \end{aligned}$
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