Results 1 to 14 of 14

Thread: Arc Length and Area Surface by revolution

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    20

    Arc Length and Area Surface by revolution

    y=1/6(x) + 1/(2x) , 1 ≤ x ≤ 2

    I don't really understand how to solve this...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3702
    arc length ...

    $\displaystyle
    S = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx
    $

    surface area ...

    $\displaystyle
    A = \int_a^b r(x) \cdot \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx
    $
    where $\displaystyle r(x)$ is the radius of revolution as a function of $\displaystyle x$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2009
    Posts
    20
    ok for the arc length i have

    ∫√[1- [(x⁴-1)/(2x)]] dx from 1 to 2

    ∫√[1- [(x⁸-2x⁴+2)/(4x⁴)] dx from 1 to 2

    i can't get past that.

    For the srface area i have

    2∏∫ [(x)/6 + 1/(2x)] √[1- [(x⁸-2x⁴+2)/(4x⁴)] dx

    and I can't get past that either.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3702
    Quote Originally Posted by hlpplz View Post
    y=1/6(x) + 1/(2x) , 1 ≤ x ≤ 2
    $\displaystyle \frac{dy}{dx} = \frac{x^2}{2} - \frac{1}{2x^2}$

    $\displaystyle \left(\frac{dy}{dx}\right)^2 = \frac{x^4}{4} - \frac{1}{2} + \frac{1}{4x^4}$

    $\displaystyle 1 + \left(\frac{dy}{dx}\right)^2 = \frac{x^4}{4} + \frac{1}{2} + \frac{1}{4x^4} = \left(\frac{x^2}{2} + \frac{1}{2x^2}\right)^2$

    $\displaystyle \sqrt{\left(\frac{x^2}{2} + \frac{1}{2x^2}\right)^2} =
    \frac{x^2}{2} + \frac{1}{2x^2}$

    $\displaystyle S = \int_1^2 \frac{x^2}{2} + \frac{1}{2x^2} \, dx$

    $\displaystyle A = \pi \int_1^2 \left(\frac{x^3}{6} + \frac{1}{2x}\right)^2\left(\frac{x^2}{2} + \frac{1}{2x^2}\right) \, dx
    $

    can you finish ?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2009
    Posts
    20
    how did you get

    $\displaystyle

    A = \pi \int_1^2 \left(\frac{x^3}{6} + \frac{1}{2x}\right)^2\left(\frac{x^2}{2} + \frac{1}{2x^2}\right) \, dx
    $

    i thought the equation for surface area was

    2∏∫ f(x) √[1+f'(x)] dx

    wouldn't that make it be

    2∏$\displaystyle

    \int_1^2 \left(\frac{x^3}{6} + \frac{1}{2x}\right)\left(\frac{x^2}{2} + \frac{1}{2x^2}\right) \, dx
    $

    then,

    2∏$\displaystyle

    \int_1^2 \frac{1}{2}\left(\frac{x^3}{3} + \frac{1}{x}\right)\left(x^2 + \frac{1}{x^2}\right) \, dx
    $

    thus,

    ∏$\displaystyle

    \int_1^2 \left(\frac{x^3}{3} + \frac{1}{x}\right)\left(x^2 + \frac{1}{x^2}\right) \, dx
    $
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3702
    you're right ... I had area of a circle on the brain instead of circumference.

    corrected.

    $\displaystyle
    A = 2\pi \int_1^2 \left(\frac{x^3}{6} + \frac{1}{2x}\right)\left(\frac{x^2}{2} + \frac{1}{2x^2}\right) \, dx
    $
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Mar 2009
    Posts
    20
    so the i have

    ∏$\displaystyle

    \int_1^2 \left(\frac{x^3}{3} + \frac{1}{x}\right)\left(x^2 + \frac{1}{x^2}\right) \, dx
    $

    ∏$\displaystyle

    \int_1^2 \left(\frac{x^5}{3} + \frac{x^3}{3x^2} + \frac{x^2}{x} + \frac{x}{x^3}\right) \, dx
    $

    ∏$\displaystyle

    \int_1^2 \left(\frac{x^5}{3} + \frac{x}{3} + {x} + \frac{1}{x^3}\right) \, dx
    $

    ∏$\displaystyle

    \int_1^2 \left(\frac{x^5}{3} + \frac{4x}{3} + \frac{1}{x^3}\right) \, dx
    $

    ∏$\displaystyle

    \left(\int_1^2 \frac{x^5}{3} + \int_1^2 \frac{4x}{3} + \int_1^2 \frac{1}{x^3} \, dx \right)
    $

    am i right so far?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3702
    you can only multiply one of the two factors in the integrand by 2, not both.

    $\displaystyle \pi \int_1^2 \left(\frac{x^3}{3} + \frac{1}{x}\right)\left(\frac{x^2}{2} + \frac{1}{2x^2}\right) \, dx
    $
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Mar 2009
    Posts
    20
    ok from this

    $\displaystyle

    \int_1^2 \left(\frac{x^3}{6} + \frac{1}{2x}\right)\left(\frac{x^2}{2} + \frac{1}{2x^2}\right) \, dx
    $

    what i did was just multiply it all out and got

    2∏$\displaystyle

    \int_1^2 \left(\frac{x^5}{12} + \frac{x^3}{12x^2} + \frac{x^2}{4x} + \frac{1}{4x^3}\right) \, dx
    $

    2∏$\displaystyle

    \int_1^2 \left(\frac{x^5}{12} + \frac{x}{12} + \frac{x}{4} + \frac{1}{4x^3}\right) \, dx
    $

    2∏$\displaystyle

    \int_1^2 \left(\frac{x^5}{12} + \frac{x}{12} + \frac{3x}{12} + \frac{1}{4x^3}\right) \, dx
    $

    2∏$\displaystyle

    \int_1^2 \left(\frac{x^5}{12} + \frac{4x}{12} + \frac{1}{4x^3}\right) \, dx
    $

    2∏$\displaystyle

    \int_1^2 \frac{1}{4}\left(\frac{x^5}{3} + \frac{4x}{3} + \frac{1}{x^3}\right) \, dx
    $

    ∏/2$\displaystyle

    \int_1^2 \frac{1}{4}\left(\frac{x^5}{3} + \frac{4x}{3} + \frac{1}{x^3}\right) \, dx
    $

    is that better?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3702
    $\displaystyle
    2\pi \int_1^2 \left(\frac{x^3}{6} + \frac{1}{2x}\right)\left(\frac{x^2}{2} + \frac{1}{2x^2}\right) \, dx
    $

    $\displaystyle

    2\pi \int_1^2 \left(\frac{x^5}{12} + \frac{x^3}{12x^2} + \frac{x^2}{4x} + \frac{1}{4x^3}\right) \, dx
    $

    $\displaystyle

    2\pi \int_1^2 \left(\frac{x^5}{12} + \frac{x}{12} + \frac{x}{4} + \frac{1}{4x^3}\right) \, dx
    $

    $\displaystyle

    2\pi \int_1^2 \left(\frac{x^5}{12} + \frac{x}{3} + \frac{1}{4x^3}\right) \, dx
    $

    integrate at this point. don't mess with the algebra anymore ... too big a chance of making a bonehead error.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Mar 2009
    Posts
    20
    ok so for the final result i was able to get

    $\displaystyle
    \frac{41\pi}{16}
    $

    is that correct?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3702
    Quote Originally Posted by hlpplz View Post
    ok so for the final result i was able to get

    $\displaystyle
    \frac{41\pi}{16}
    $

    is that correct?
    my calculator says $\displaystyle \frac{47\pi}{16}$
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Newbie
    Joined
    Mar 2009
    Posts
    20
    and for the arc length i ended up getting

    $\displaystyle

    \frac{17}{12}
    $

    is that right?
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3702
    you're done.
    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: Dec 10th 2011, 03:57 AM
  2. Surface area of a revolution
    Posted in the Calculus Forum
    Replies: 7
    Last Post: Mar 27th 2011, 02:47 PM
  3. Replies: 3
    Last Post: Oct 29th 2010, 05:31 AM
  4. area of a surface of revolution
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 21st 2010, 04:03 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