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!

Surface area of a revolution

  1. Mar 5, 2007 #1
    (my first dealings with latex.. so bare with me if this looks a little messed up at first :rolleyes: )
    1. The problem statement, all variables and given/known data
    Find the surface area for the equation:
    [tex]x = 3y^{4/3} - \frac{3}{32}y^{2/3}[/tex]

    with bounds [tex]-216 \leq y \leq 216[/tex]

    rotated about the Y-axis.

    2. Relevant equations

    [tex]\int^a_b 2\pi f(y) \sqrt{1+(\frac{dx}{dy})^2}[/tex]

    3. The attempt at a solution

    well... going with that equation i get to this point:

    [tex]2\pi \int^{216}_{-216} (3y^{4/3} - \frac{3}{32}y^{2/3})(4y^{1/3} + \frac{1}{16}y^{-1/3}) [/tex]

    from there I tried to multiply out the equation and solve the integral with the bounds, but it isn't giving me the correct answer. I'm not sure what I'm doing wrong. I suspect I have to break the integral up smaller pieces but im not sure where to break it at.
     
    Last edited: Mar 5, 2007
  2. jcsd
  3. Mar 5, 2007 #2
    you have substituted dx/dy incorrectly into the equation

    dx/dy should be inside the root
     
  4. Mar 5, 2007 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Because of the symmetry, you can integrate from 0 to 216 and then multiply by 2. What do you get?
     
  5. Mar 5, 2007 #4
    ok... when I multiply out the integrand I get:

    [tex]2\pi [12y^{5/3} - \frac{3}{16}y^2 - \frac{9}{2048}y^{1/3}]^{216}_{0}[/tex]

    when I evaluate at 216 & 0, i get 7552892.305... I multiply that by 2 (symmetry) and then multiply that by 2pi, which gives me 94925010.28

    edit... nevermind, i made a stupid math mistake. Got it right now, thanks a ton HallsofIvy!
     
    Last edited: Mar 5, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Surface area of a revolution
Loading...