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Thread: Volume and Average Value

  1. #1
    Junior Member
    Nov 2008

    Volume and Average Value

    Can someone please help me complete this problem and understand it?

    Find the volume of the solid created by rotating the region bounded by the curves below around the x-axis.

    $\displaystyle x=1, x=9, y=0, y= \frac{9}{\sqrt{x+5}}$

    So far I have my work is as follows

    $\displaystyle V=\pi \int_1^9 \frac{9}{\sqrt{x+5}^2}dx $

    $\displaystyle V=\pi \int_1^9 \frac{9}{x+5}dx $

    The answer is $\displaystyle 81\pi ln \frac{7}{3}$

    Please help
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  2. #2
    MHF Contributor
    Oct 2005
    Your setup is almost right. You need to square the whole expression, meaning the 9 as well. That's where the 81 comes from. Bring the constants out and you have $\displaystyle 81 \pi \int_{1}^{9} \frac{1}{x+5}dx$. The integral of 1/(x+5) is ln(x+5) and then all you do is plug in your bounds. The last part uses an identity that ln(a)-ln(b) = ln(a/b)
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