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Math Help - Calculus volume ..

  1. #1
    Newbie
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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
    Posts
    319
    To find the area enclosed by

    \begin{aligned}<br />
y&=1\\<br />
x&=4\\<br />
y&=\sqrt{x},<br />
\end{aligned}

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

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

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

    \begin{aligned}<br />
V&=\int_1^4 \pi(\sqrt{x} - 1)^2\,dx\\<br />
&=\pi\int_1^4(x-2\sqrt{x}+1)\,dx\\<br />
&=\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\\<br />
&=\frac{15\pi}{2}-\frac{28\pi}{3}+3\pi\\<br />
&=\frac{7\pi}{6}.<br />
\end{aligned}
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