Results 1 to 8 of 8

Math Help - Area of surfaces of revolution

  1. #1
    Newbie
    Joined
    Sep 2008
    Posts
    11

    Area of surfaces of revolution

    HELP! I'm in Calc BC, i need step by step help

    Problem: Set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the x-axis.

    41. y=((x^3)/6)+((1)/(2x)), [1,2]


    Set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the y-axis

    43. y=(x^1/3)+2

    x values [2,8]

    y values [3,4]
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Ok start with the formula for the SA of a solid of revolution around the x-axis. S.A. = \int_{a}^{b} 2\pi f(x)\sqrt{1+f'[x]^2}dx Can you setup the integrals and try them?
    Last edited by Jameson; September 11th 2008 at 09:05 AM. Reason: made a boo boo
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2008
    Posts
    11
    i know the eqn and all that...i just need help with the step by step...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    I'm on this to!

    I did it like this (i'd get a second opinion since i'm new to this topic):

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by Jameson View Post
    Ok start with the formula for the SA of a solid of revolution around the x-axis. S.A. = \int_{a}^{b} 2\pi{\color{red}f[x]}\sqrt{1+f'[x]^2}dx Can you setup the integrals and try them?
    You forgot that vital term!

    --Chris
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by vstexas09 View Post
    HELP! I'm in Calc BC, i need step by step help

    Problem: Set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the x-axis.

    41. y=((x^3)/6)+((1)/(2x)), [1,2]

    Let y=f(x)=\tfrac{1}{6}x^3+\frac{1}{2x}

    Thus, f'(x)=\tfrac{1}{2}x^2-\frac{1}{2x^2}

    Therefore, [f'(x)]^2=\left(\tfrac{1}{2}x^2-\tfrac{1}{2}x^{-2}\right)^2=\tfrac{1}{4}x^4-\tfrac{1}{2}+\tfrac{1}{4}x^{-4}

    Plugging this into the surface area equation, we get:

    \begin{aligned}S.A.&=2\pi\int_1^2 f(x)\sqrt{1+\left(\tfrac{1}{4}x^4-\tfrac{1}{2}+\tfrac{1}{4}x^{-4}\right)}\,dx\\<br />
&=2\pi\int_1^2 f(x)\sqrt{\tfrac{1}{4}x^4+\tfrac{1}{2}+\tfrac{1}{4  }x^{-4}}\,dx\end{aligned}

    Note that \tfrac{1}{4}x^4+\tfrac{1}{2}+\tfrac{1}{4}x^{-4} is a perfect square: \tfrac{1}{4}x^4+\tfrac{1}{2}+\tfrac{1}{4}x^{-4}=\left(\tfrac{1}{2}x^2+\tfrac{1}{2}x^{-2}\right)^2

    Thus, the integral becomes:

    \begin{aligned}S.A.&=2\pi\int_1^2 f(x)\sqrt{\left(\tfrac{1}{2}x^2+\tfrac{1}{2}x^{-2}\right)^2}\,dx\\<br />
&=2\pi\int_1^2 f(x) \left(\tfrac{1}{2}x^2+\tfrac{1}{2}x^{-2}\right)\,dx\end{aligned}

    Since, f(x)=\tfrac{1}{6}x^3+\frac{1}{2x}, we see that the integral transforms into

    \begin{aligned}S.A.&=2\pi\int_1^2\left(\tfrac{1}{1  2}x^5+\tfrac{1}{3}x^3+\tfrac{1}{4}x^{-3}\right)\,dx\\<br />
&=\frac{\pi}{6}\int_1^2 \left(x^5+4x^2+3x^{-3}\right)\,dx\end{aligned}

    Take it from here. You should be able to do this!

    I hope this helps!

    --Chris
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Quote Originally Posted by Chris L T521 View Post
    You forgot that vital term!

    --Chris
    You are right! That was a very bad typo. Thank you. Fixed now.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    darn it! That means i've just found the length of it!

    I worked it out again and got 115pi/36.

    I would post my working except I did it on a whiteboard.
    Last edited by Showcase_22; September 11th 2008 at 09:48 AM. Reason: needed to add something.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Surface Area of Revolution
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 10th 2011, 03:57 AM
  2. Calculus 3--parametric surfaces/surface area
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 18th 2009, 05:11 AM
  3. Area of a Surface of Revolution
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 15th 2009, 10:51 AM
  4. Arc Length and Surfaces of Revolution
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 6th 2009, 05:55 PM
  5. surface area of revolution
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 6th 2008, 10:31 PM

Search Tags


/mathhelpforum @mathhelpforum