1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Another Volume

  1. Feb 14, 2007 #1

    tony873004

    User Avatar
    Science Advisor
    Gold Member

    Rotate the area bounded by
    [tex]y = 5,\,y = x + (4/x)[/tex]
    about x=-1

    [​IMG]

    Verify the limits of integration
    [tex]x + \left( {4/x} \right) = 5,\,\,x = 1\& 4[/tex]

    solve
    [tex]

    \begin{array}{l}
    \int\limits_1^4 {2\pi r\,h\,dx} \\
    h = 5 - \left( {x + \left( {4/x} \right)} \right),\,\,r = x + 1 \\
    2\pi \int\limits_1^4 {\left( {x + 1} \right)\left( {5 - \left( {x + \left( {4/x} \right)} \right)} \right)} \,dx \\
    \\
    2\pi \int\limits_1^4 {\left( {x + 1} \right)\left( {5 - x - 4x^{ - 1} } \right)} \,dx \\
    \\
    2\pi \int\limits_1^4 {\left( {5x - x^2 - 4 + 5 - x - 4x^{ - 1} } \right)} \,dx \\
    \\
    2\pi \int\limits_1^4 {\left( {4x - x^2 + 1 - 4x^{ - 1} } \right)} \,dx \\
    \\
    \end{array}

    [/tex]
    [tex]
    \begin{array}{l}
    2\pi \int\limits_1^4 {\left( {4x - x^2 + 1 - 4x^{ - 1} } \right)} \,dx \\
    \\
    2\pi \left( {\frac{{4x^2 }}{2} - \frac{{x^3 }}{3} + \frac{x}{1} - \frac{{4\ln x}}{1}} \right)_1^4 \\
    \\
    2\pi \left( {\frac{{2x^2 }}{1} - \frac{{x^3 }}{3} + \frac{x}{1} - \frac{{4\ln x}}{1}} \right)_1^4 \\
    \\
    2\pi \left( {\frac{{6x^2 }}{3} - \frac{{3x^3 }}{3} + \frac{{3x}}{3} - \frac{{12\ln x}}{3}} \right)_1^4 \\
    \\
    2\pi \left( {\left( {\frac{{6\left( 4 \right)^2 }}{3} - \frac{{3\left( 4 \right)^3 }}{3} + \frac{{3\left( 4 \right)}}{3} - \frac{{12\ln \left( 4 \right)}}{3}} \right) - \left( {\frac{{6\left( 1 \right)^2 }}{3} - \frac{{3\left( 1 \right)^3 }}{3} + \frac{{3\left( 1 \right)}}{3} - \frac{{12\ln \left( 1 \right)}}{3}} \right)} \right) \\
    \\
    2\pi \left( {\left( {\frac{{96}}{3} - \frac{{192}}{3} + \frac{{12}}{3} - \frac{{12\ln \left( 4 \right)}}{3}} \right) - \left( {\frac{6}{3} - \frac{3}{3} + \frac{3}{3} - \frac{0}{3}} \right)} \right) \\
    \\
    2\pi \left( {\frac{{96}}{3} - \frac{{192}}{3} + \frac{{12}}{3} - \frac{{12\ln \left( 4 \right)}}{3} - \frac{6}{3}} \right) = \\
    \\
    2\pi \left( {\frac{{90}}{3} - \frac{{12\ln 4}}{3}} \right) = 2\pi \left( { - 30 - 4\ln 4} \right) = - 4\pi \left( {15 - 2\ln 4} \right) \\
    \\
    {\rm{Book says: }}8\pi \left( {3 - \ln 4} \right) \\
    \end{array}

    [/tex]

    [tex]
    {\rm{Book says: }}8\pi \left( {3 - \ln 4} \right)

    [/tex]
    Obviously it can't be negative, but I don't know where I messed up.
     
    Last edited: Feb 15, 2007
  2. jcsd
  3. Feb 15, 2007 #2

    tony873004

    User Avatar
    Science Advisor
    Gold Member

    I edited the above post to reflect my latest attempt. It's still wrong, so any help would be appreciated.
     
    Last edited: Feb 15, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Another Volume
  1. Yet another volume (Replies: 6)

  2. Another volume problem (Replies: 6)

  3. Another volume problem (Replies: 1)

Loading...